Chapter 29 Alternating-Current Circuits Conceptual Problems A coil in an ac generator rotates at 60 Hz. How much time elapses 1 • between successive peak emf values of the coil? Determine the Concept Successive peaks are one-half period apart. Hence the 1 1 1 elapsed time between the peaks is T = = = 8.33 ms . 2 2 f 2 60 s -1
(
2
•
)
If the rms voltage in an ac circuit is doubled, the peak voltage is
(a) doubled, (b) halved, (c) increased by a factor of 2 , (d ) not changed. Picture the Problem We can use the relationship between V and V peak to decide the effect of doubling the rms voltage on the peak voltage.
Express the initial rms voltage in terms of the peak voltage:
V rms
Express the doubled rms voltage in terms of the new peak voltage V' peak :
2V rms
Divide the second of these equations by the first and simplify to obtain:
=
V peak
2
=
V' peak
2 V' peak
2V rms V rms
=
2 V peak
⇒ 2=
V' peak V peak
2 Solving for V' peak yields:
V' peak = 2V peak ⇒ (a) is correct.
3 • [SSM] If the frequency in the circuit shown in Figure 29-27 is doubled, the inductance of the inductor will (a) double, (b) not change, (c) halve, (d ) quadruple. Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent independent of the frequency of the the circuit. The
inductive reactance, on the other hand, is frequency dependent. (b) is correct. If the frequency in the circuit shown in Figure 29-27 is doubled, the 4 • inductive reactance of the inductor will (a) double, (b) not change, (c) halve, (d ) quadruple.
2739
2740
Chapter 29
Determine the Concept The inductive reactance of an inductor varies with the
frequency according to X L
=
L. Hence, doubling ω will double X L. (a) is
correct. If the frequency in the circuit in Figure 29-28 is doubled, the 5 • capacitive reactance of the circuit will (a) double, (b) not change, (c) halve, (d ) quadruple. Determine the Concept The capacitive reactance of an capacitor varies with the
frequency according to X C = 1 C . Hence, doubling ω will halve X C C . (c) is correct. (a) In a circuit consisting solely of a ac generator and an ideal 6 • inductor, are there any time intervals when the inductor receives energy from the generator? If so, when? Explain your answer. (b) Are there any time intervals when the inductor supplies energy back to the generator? If so when? Explain your answer. Determine the Concept Yes to both questions. (a) While the magnitude of the current in the inductor is increasing, the inductor absorbs power from the generator. (b) When the magnitude of the current in the inductor decreases, the inductor supplies power to the generator. 7 • [SSM] (a) In a circuit consisting of a generator and a capacitor, are there any time intervals when the capacitor receives energy from the generator? If so, when? Explain your answer. (b) Are there any time intervals when the capacitor supplies power to the generator? If so, when? Explain your answer. Determine the Concept Yes to both questions. (a) While the magnitude of the charge is accumulating on either plate of the capacitor, the capacitor absorbs power from the generator. ( b) When the magnitude of the charge is on either plate of the capacitor is decreasing, it supplies supp lies power to the generator.
(a) Show that the SI unit of inductance multiplied by the SI unit of 8 • capacitance is equivalent to seconds squared. (b) Show that the SI unit of inductance divided by the SI unit of resistance is equivalent to seconds. Determine the Concept (a) Substitute the SI units of inductance and capacitance and simplify to obtain:
V⋅s C ⋅ A V
=
s ⋅ C = s2 C s
Alternating-Current Circuits (b) Substitute the SI units of inductance divided by resistance and simplify to obtain:
V⋅s A Ω
=
V⋅s A V A
2741
= s
9 • [SSM] Suppose you increase the rotation rate of the coil in the generator shown in the simple ac circuit in Figure 29-29. Then the rms current (a) increases, (b) does not change, (c) may increase or decrease depending on the magnitude of the original frequency, (d ) may increase or decrease depending on the magnitude of the resistance, (e) decreases. Determine the Concept Because the rms current through the resistor is given by ε peak NBA ε ω , I rms is directly proportional to ω . (a) is correct. I rms = rms = = R 2 2
If the inductance value is tripled in a circuit consisting solely of a 10 • variable inductor and a variable capacitor, how would you have to change the capacitance so that the natural frequency of the circuit is unchanged? (a) triple the capacitance, (b) decrease the capacitance to one-third of its original value, ( c) You should not change the capacitance.(d ) You cannot determine how to change the capacitance from the data given. Determine the Concept The natural frequency of an LC circuit is given by f 0
= 1 2π
LC .
Express the natural frequencies of the circuit before and after the inductance is tripled: Divide the second of the these equations by the first and simplify to obtain:
f 0
=
1 2π LC
and f 0' =
1 2π L'C'
1 '
f 0
f 0
= 2π
L'C' 1
LC
=
L'C'
2π LC Because the natural frequency is unchanged: When the inductance is tripled:
1=
C' =
LC L'C' L
3 L
⇒
C =
LC
L'C'
= 1 ⇒ C' =
1 C ⇒ 3
(b )
L L'
C
is correct.
11 • [SSM] Consider a circuit consisting solely of an ideal inductor and an ideal capacitor. How does the maximum energy stored in the capacitor compare to the maximum value stored in the inductor? (a) They are the same and each equal to the total energy stored in the circuit. (b) They are the same and each
2742
Chapter 29
equal to half of the total energy stored in the circuit. (c) The maximum energy stored in the capacitor is larger than the maximum energy stored in the inductor. (d ) The maximum energy stored in the inductor is larger than the maximum energy stored in the capacitor. (e) You cannot compare the maximum energies based on the data given because the ratio of the maximum energies depends on the actual capacitance and inductance values. Determine the Concept The maximum energy stored in the electric field of the 1 Q2 capacitor is given by U e = and the maximum energy stored in the magnetic 2 C 1 field of the inductor is given by U m = LI 2 . Because energy is conserved in an 2 LC circuit and oscillates between the inductor and the capacitor, U e = U m = U total.
(a ) 12
is correct. •
True or false:
(a) A driven series RLC circuit that has a high Q factor has a narrow resonance curve. (b) A circuit consists solely of a resistor, an inductor and a capacitor, all connected in series. If the resistance of the resistor is doubled, the natural frequency of the circuit remains the same. (c) At resonance, the impedance of a driven series RLC combination equals the resistance R. (d ) At resonance, the current in a driven series RLC circuit is in phase with the voltage applied to the combination. (a) True. The Q factor and the width of the resonance curve at half power are related according to Q = 0 Δ ; i.e., they are inversely proportional to each other. (b) True. The natural frequency of the circuit depends only on the inductance L of the inductor and the capacitance C of the capacitor and is given by ω = 1 LC . (c) True. The impedance of an RLC circuit is given by Z = R 2 resonance X L = X C and so Z = R. (d ) True. The phase angle δ is related to X L and X C according to − ⎛ X − X C ⎞ δ = tan 1 ⎜ L ⎟ . At resonance X L = X C and so δ = 0. ⎝ R ⎠
+ ( X L − X C )2 . At
Alternating-Current Circuits 13
•
2743
True or false (all questions related to a driven series RLC circuit):
(a) Near resonance, the power factor of a driven series RLC circuit is close to zero. (b) The power factor of a driven series RLC circuit does not depend on the value of the resistance. (c) The resonance frequency of a driven series RLC circuit does not depend on the value of the resistance. (d ) At resonance, the peak current of a driven series RLC circuit does not depend on the capacitance or the inductance. (e) For frequencies below the resonant frequency, the capacitive reactance of a driven series RLC circuit is larger than the inductive reactance. ( f ) For frequencies below the resonant frequency of a driven series RLC circuit, the phase of the current leads (is ahead of) the phase of the applied voltage. (a) False. Near resonance, the power factor, given by cos δ =
R 2
( X L − X C ) + R
2
,
is close to 1. (b) False. The power factor is given by cos δ =
R
( X L − X C ) + R 2
2
.
(c) True. The resonance frequency for a driven series RLC circuit depends only on L and C and is given by ω res = 1 LC (d ) True. At resonance X L − X C = 0 and so Z = R and the peak current is given by I peak = V app, peak R .
(e) True. Because the capacitive reactance varies inversely with the driving frequency and the inductive reactance varies directly with the driving frequency, at frequencies well below the resonance frequency the capacitive reactance is larger than the inductive reactance. ( f ) True. For frequencies below the resonant frequency, the circuit is more capacitive than inductive and the phase constant φ is negative. This means that the current leads the applied voltage. You may have noticed that sometimes two radio stations can be heard 14 • when your receiver is tuned to a specific frequency. This situation often occurs when you are driving and are between two cities. Explain how this situation can occur.
2744
Chapter 29
Determine the Concept Because the power curves received by your radio from two stations have width, you could have two frequencies overlapping as a result of receiving signals from both stations. 15
•
True or false (all questions related to a driven series RLC circuit):
(a) At frequencies much higher than or much lower than the resonant frequency of a driven series RLC circuit, the power factor is close to zero. (b) The larger the resonance width of a driven series RLC circuit is, the larger the Q factor for the circuit is. (c) The larger the resistance of a driven series RLC circuit is, the larger the resonance width for the circuit is. (a) True. Because the power factor is given by cos δ =
R 2
, for
⎛ ω L − 1 ⎞ + R 2 ⎜ ⎟ ω C ⎠ ⎝ values of ω that are much higher or much lower than the resonant frequency, the term in parentheses becomes very large and cosδ approaches zero. (b) False. When the resonance curve is reasonably narrow, the Q factor can be approximated by Q = 0 Δ . Hence a large value for Q corresponds to a narrow resonance curve. (c) True. See Figure 29-21. An ideal transformer has N 1 turns on its primary and N 2 turns on its 16 • secondary. The average power delivered to a load resistance R connected across the secondary is P2 when the primary rms voltage is V 1. The rms current in the primary windings can then be expressed as (a) P2/V 1, (b) ( N 1/ N 2)(P2/V 1), (c) ( N 2/ N 1)(P2/V 1), (d ) ( N 2/ N 1)2(P2/V 1). Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. Assuming no loss of power in the transformer, we can equate the power in the primary circuit to the power in the secondary circuit and solve for the rms current in the primary windings.
= P2
Assuming no loss of power in the transformer:
P1
Substitute for P1 and P2 to obtain:
I 1, rmsV 1, rms = I 2, rmsV 2, rms
Alternating-Current Circuits Solving for I 1,rms and simplifying yields:
I 1, rms
(a ) 17
•
[SSM]
=
I 2, rmsV 2, rms V 1, rms
=
2745
P2 V 1, rms
is correct.
True or false:
(a) A transformer is used to change frequency. (b) A transformer is used to change voltage. (c) If a transformer steps up the current, it must step down the voltage. (d ) A step-up transformer, steps down the current. (e) The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If a European traveler wants her hair dryer to work properly in the United States, she should use a transformer that has more windings in its secondary coil than in its primary coil. ( f ) The standard household wall-outlet voltage in Europe is 220 V, about twice that used in the United States. If an American traveler wants his electric razor to work properly in Europe, he should use a transformer that steps up the current. (a) False. A transformer is a device used to raise or lower the voltage in a circuit. (b) True. A transformer is a device used to raise or lower the voltage in a circuit. (c) True. If energy is to be conserved, the product of the current and voltage must be constant. (d ) True. Because the product of current and voltage in the primary and secondary circuits is the same, increasing the current in the secondary results in a lowering (or stepping down) of the voltage. (e) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the voltage in order to make her hair dryer work properly. ( f ) True. Because electrical energy is provided at a higher voltage in Europe, the visitor would want to step-up the current (and decrease the voltage) in order to make his razor work properly.
Estimation and Approximation The impedances of motors, transformers, and electromagnets include 18 •• both resistance and inductive reactance. Suppose that phase of the current to a large industrial plant lags the phase of the applied voltage by 25° when the plant
2746
Chapter 29
is under full operation and using 2.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 60 Hz rms line voltage at the plant is 40 kV. The resistance of the transmission line from the substation to the plant is 5.2 Ω. The cost per kilowatt-hour to the company that owns the plant is $0.14, and the plant pays only for the actual energy used. (a) Estimate the resistance and inductive reactance of the plant’s total load. ( b) Estimate the rms current in the power lines and the rms voltage at the substation. ( c) How much power is lost in transmission? (d ) Suppose that the phase that the current lags the phase of the applied voltage is reduced to 18º by adding a bank of capac itors in series with the load. How much money would be saved by the electric utility during one month of operation, assuming the plant operates at full capacity for 16 h each day? (e) What must be the capacitance of this bank of capacitors to achieve this change in phase angle? Picture the Problem We can find the resistance and inductive reactance of the plant’s total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the rms voltage at the substation by applying Kirchhoff’s loop rule to the substation-transmission wires-load circuit. The power 2 lost in transmission can be found from Ptrans = I rms Rtrans . We can find the cost
savings by finding the difference in the power lost in transmission when the phase angle is reduced to 18°. Finally, we can find the capacitance that is required to reduce the phase angle to 18° by first finding the capacitive reactance using the definition of tanδ and then applying the definition of capacitive reactance to find C . Rtrans
ε substation f = 60 Hz
= 5.2 Ω
ε rms= 40 kV
∼
Z
δ = 25°
(a) Relate the resistance and inductive reactance of the plant’s total load to Z and δ :
R = Z cos δ and X L = Z sin δ
Express Z in terms of the rms current I rms in the power lines and the rms
Z =
voltage ε rms at the plant:
ε rms I rms
Alternating-Current Circuits
Express the power delivered to the
Pav
plant in terms of ε rms, I rms , and δ and
and
solve for I rms :
I rms
Substitute to obtain:
Substitute numerical values and evaluate Z : Substitute numerical values and evaluate R and X L:
= ε rms I rms cos δ =
Z =
Z =
R
Pav
(1)
ε rms cos δ
2 ε rms cos δ
Pav
(40 kV)2 cos 25° 2.3 MW
= 630 Ω
= (630 Ω ) cos 25° = 571Ω = 0.57 k Ω
and X L = (630 Ω )sin 25° = 266 Ω
= 0.27 k Ω (b) Use equation (1) to find the current in the power lines:
I rms
=
2.3 MW (40 kV )cos 25°
= 63.4 A
= 63 A Apply Kirchhoff’s loop rule to the circuit:
ε sub − I rms Rtrans − I rms Z tot = 0
Solve for ε sub:
ε sub = I rms ( Rtrans + Z tot )
Substitute numerical values and
ε sub = (63.4 A )(5.2 Ω + 630 Ω)
evaluate ε sub: (c) The power lost in transmission is:
= 40.3 kV 2 Ptrans = I rms Rtrans
= (63.4 A )2 (5.2 Ω )
= 20.9 kW = 21kW (d ) Express the cost savings ΔC in terms of the difference in energy consumption (P25° − P18°)Δt and the per-unit cost u of the energy:
ΔC = (P25° − P18° )Δtu
2747
2748
Chapter 29 P18°
= I 182 ° Rtrans
Find the current in the transmission lines when δ = 18°:
I 18°
=
Evaluate P18° :
P18°
= (60.5 A )2 (5.2 Ω ) = 19.0 kW
Express the power lost in transmission when δ = 18°:
2.3 MW (40 kV ) cos18°
= 60.5 A
Substitute numerical values and evaluate ΔC : ΔC
h ⎞ ⎛ d ⎞ ⎛ $0.14 ⎞ = (20.9 kW − 19.0 kW )⎛ ⎜16 ⎟ ⎜ 30 ⎟⎜ ⎟ = $128 ⋅ d month kW h ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1
(e) The required capacitance is given by:
C =
Relate the new phase angle δ to the inductive reactance X L, the reactance due to the added capacitance X C, and the resistance of the load R:
tan δ =
Substituting for X C yields:
C =
2π fX C X L
− X C R
⇒ X C = X L − R tan δ
1 2π f ( X L − R tan δ )
Substitute numerical values and evaluate C : C =
1 2π (60 s -1 )(266 Ω − (571Ω ) tan 18°)
= 33 μ F
Alternating Current in Resistors, Inductors, and Capacitors 19 • [SSM] A 100-W light bulb is screwed into a standard 120-V-rms socket. Find (a) the rms current, (b) the peak current, and (c) the peak power. Picture the Problem We can use Pav
= ε rms I rms to find I rms , I peak =
I peak , and P peak = I peak ε peak to find P peak .
(a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it:
Pav
= ε rms I rms ⇒ I rms =
Pav
ε rms
2 I rms to find
Alternating-Current Circuits Substitute numerical values and evaluate I rms :
I rms
=
100 W
(b) Express I peak in terms of I rms :
I peak =
2 I rms
Substitute for I rms and evaluate I peak :
I peak =
2 (0.8333 A ) = 1.1785 A
120 V
2749
= 0.8333 A = 0.833 A
= 1.18 A (c) Express the maximum power in terms of the maximum voltage and maximum current:
P peak = I peak ε peak
Substitute numerical values and evaluate P peak :
P peak = (1.1785 A ) 2 (120 V ) = 200 W
A circuit breaker is rated for a current of 15 A rms at a voltage of 20 • 120 V rms. (a) What is the largest value of the peak current that the breaker can carry? (b) What is the maximum average power that can be supplied by this circuit? Picture the Problem We can I peak =
2 I rms to find the largest peak current the
breaker can carry and Pav = I rmsV rms to find the average power supplied by this circuit. (a) Express I peak in terms of I rms :
I peak =
(b) Relate the average power to the rms current and voltage:
Pav
2 I rms
= 2 (15 A ) = 21 A
= I rmsV rms = (15 A)(120 V ) = 1.8 kW
21 • [SSM] What is the reactance of a 1.00-μ H inductor at (a) 60 Hz, (b) 600 Hz, and (c) 6.00 kHz? Picture the Problem We can use X L
=
L to find the reactance of the inductor at
any frequency. Express the inductive reactance as a function of f :
X L
=
L = 2π fL
(a) At f = 60 Hz:
X L
= 2 (60 s −1 )(1.00 mH) = 0.38 Ω
2750
Chapter 29
(b) At f = 600 Hz:
X L
= 2π (600 s −1 )(1.00 mH) = 3.77 Ω
(c) At f = 6.00 kHz:
X L
= 2π (6.00 kHz)(1.00 mH) = 37.7 Ω
An inductor has a reactance of 100 Ω at 80 Hz. (a) What is its 22 • inductance? (b) What is its reactance at 160 Hz? Picture the Problem We can use X L
=
L to find the inductance of the inductor
at any frequency. (a) Relate the reactance of the inductor to its inductance:
X L
Solve for and evaluate L:
L =
(b) At 160 Hz:
X L
=
L = 2π fL ⇒ L =
100 Ω 2π (80 s −1 )
X L
2π f
= 0.199 H = 0.20 H
= 2π (160 s −1 )(0.199 H) = 0.20 k Ω
At what frequency would the reactance of a 10-μ F capacitor equal the 23 • reactance of a 1.0-μ H inductor? Picture the Problem We can equate the reactances of the capacitor and the inductor and then solve for the frequency.
Express the reactance of the inductor: Express the reactance of the capacitor: Equate these reactances to obtain:
Substitute numerical values and evaluate f :
X L
=
X C =
L = 2π fL
1 1 = ω C 2π fC
fL = 2π
f =
1 1 1 ⇒ f = 2π 2π LC fC
1
1
2π
(10 μ F)(1.0 mH)
= 1.6 kHz
What is the reactance of a 1.00-nF capacitor at (a) 60.0 Hz, 24 • (b) 6.00 kHz, and (c) 6.00 MHz? Picture the Problem We can use X C = 1
capacitor at any frequency.
C to find the reactance of the
Alternating-Current Circuits
Express the capacitive reactance as a function of f :
X C =
(a) At f = 60.0 Hz:
X C =
(b) At f = 6.00 kHz:
X C =
(c) At f = 6.00 MHz:
X C =
2751
1 1 = ω C 2π fC 1 2π (60.0 s −1 )(1.00 nF)
= 2.65 MΩ
1 2π (6.00 kHz )(1.00 nF)
= 26.5 k Ω
1 2π (6.00 MHz )(1.00 nF)
= 26.5 Ω
25 • [SSM] A 20-Hz ac generator that produces a peak emf of 10 V is connected to a 20-μ F capacitor. Find (a) the peak current and (b) the rms current. Picture the Problem We can use I peak = ε peak / X C and X C = 1/ω C to express I peak as
a function of ε peak , f , and C . Once we’ve evaluate I peak , we can use I rms = I peak / 2 to find I rms .
Express I peak in terms of ε peak
I peak =
and X C :
ε peak X C
1 1 = ω C 2π fC
Express the capacitive reactance:
X C =
Substitute for X C and simplify to obtain:
I peak = 2π fC ε peak
(a) Substitute numerical values and evaluate I peak :
I peak = 2π 20 s
(b) Express I rms in terms of I peak :
(
−1
)(20 μ F)(10 V)
= 25.1 mA = 25 mA I rms
=
I peak
2
=
25.1 mA 2
= 18 mA
At what frequency is the reactance of a 10-μ F capacitor (a) 1.00 Ω, 26 • (b) 100 Ω, and (c) 10.0 mΩ?
2752
Chapter 29
Picture the Problem We can use X C = 1
C = 1 2π fC to relate the reactance of
the capacitor to the frequency. The reactance of the capacitor is given by: (a) Find f when X C = 1.00 Ω:
(b) Find f when X C = 100 Ω:
(c) Find f when X C = 10.0 mΩ:
X C =
1 1 1 = ⇒ f = ω C 2π fC 2π CX C 1
f =
2π (10 μ F)(1.00 Ω )
= 16 kHz
1 = 0.16 kHz 2π (10 μ F )(100 Ω )
f =
f =
1 2π (10 μ F)(10.0 mΩ )
= 1.6 MHz
A circuit consists of two ideal ac generators and a 25-Ω resistor, all 27 •• connected in series. The potential difference across the terminals of one of the generators is given by V 1 = (5.0 V) cos(ω t – α ), and the potential difference across the terminals of the other generator is given by V 2 = (5.0 V) cos(ω t + α ), where α = π /6. (a) Use Kirchhoff’s loop rule and a trigonometric identity to find the peak current in the circuit. (b) Use a phasor diagram to find the peak current in the circuit. (c) Find the current in the resistor if α = π /4 and the amplitude of V 2 is increased from 5.0 V to 7.0 V. Picture the Problem We can use the trigonometric identity cosθ + cosφ = 2 cos 12 (θ + φ )cos 12 (θ − φ )
to find the sum of the phasors V 1 and V 2 and then use this sum to express I as a function of time. In (b) we’ll use a phasor diagram to obtain the same result and in (c) we’ll use the phasor diagram appropriate to the given voltages to express the current as a function of time. (a) Applying Kirchhoff’s loop rule to the circuit yields:
V 1 + V 2
Solve for I to obtain:
I =
V 1
− IR = 0 + V 2
R
Alternating-Current Circuits
2753
Use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ ) to find V 1 + V 2: V 1 + V 2
= (5.0 V)[cos( t − α ) + cos( t + α )] = (5 V )[2 cos 12 (2 t )cos 12 (− 2α )] π
= (10 V )cos cos ω t = (8.66 V ) cos ω t 6
Substitute for V 1 + V 2 and R to obtain:
I =
(8.66 V ) cos
t
25 Ω
= (0.346 A ) cos ω t
= (0.35 A ) cos ω t where I peak = 0.35 A
(b) Express the magnitude of the current in R:
r
I =
V R
The phasor diagram for the voltages is shown to the right.
V 2
V
30° 30°
V 1
r
Use vector addition to find V :
r
r
V = 2 V 1 cos 30° = 2(5.0 V ) cos 30°
= 8.66 V r
Substitute for V and R to obtain:
I
=
8.66 V 25 Ω
= 0.346 A
and I = (0.35 A ) cos t where I peak = 0.35 A
2754
Chapter 29
(c) The phasor diagram is shown to the right. Note that the phase angle r r between V 1 and V 2 is now 90°.
V 2
α
V
o
90 − α δ
V 1
r
Use the Pythagorean theorem to
r
V =
r
find V :
V 1
2
r
2
+ V 2 = (5.0 V )2 + (7.0 V )2
= 8.60 V r
Express I as a function of t :
V
cos(ω t + δ ) R where δ = 45° − (90° − α ) = α − 45°
I =
⎛ 7.0 V ⎞ ⎟⎟ − 45° = tan −1 ⎜⎜ ⎝ 5.0 V ⎠ = 9.462° = 0.165 rad Substitute numerical values and evaluate I :
I =
8.60 V 25 Ω
cos(ω t + 0.165 rad )
= (0.34 A )cos(ω t + 0.17 rad ) Undriven Circuits Containing Capacitors, Resistors and Inductors (a) Show that 1 LC has units of inverse seconds by substituting SI units for inductance and capacitance into the expression. (b) Show that ω 0 L/R (the expression for the Q-factor) is dimensionless by substituting SI units for angular frequency, inductance, and resistance into the expression. 28
•
Picture the Problem We can substitute the units of the various physical
quantitities in 1 / LC and Q =
L R to establish their units.
0
Alternating-Current Circuits (a) Substitute the units for L and C in the expression 1 LC and simplify
1
1
=
H⋅F
s ⎞ (Ω ⋅ s )⎛ ⎜ ⎟
=
1 s
2
2755
= s −1
⎝ Ω ⎠
to obtain: (b) Substitute the units for ω 0, L, and R in the expression Q = 0 L R and simplify to obtain:
1 V⋅s ⋅ s A V A
=
1 V ⋅s ⋅ s A V A
= 1 ⇒ no units
29 • [SSM] (a) What is the period of oscillation of an LC circuit consisting of an ideal 2.0-mH inductor and a 20-μ F capacitor? (b) A circuit that oscillates consists solely of an 80-μ F capacitor and a variable ideal inductor. What inductance is needed in order to tune this circuit to oscillate at 60 Hz? Picture the Problem We can use T = 2π /ω and ω = 1 LC to relate T (and hence f ) to L and C .
(a) Express the period of oscillation of the LC circuit:
T =
2π ω
For an LC circuit:
ω =
1 LC
Substitute for ω to obtain:
T = 2π LC
Substitute numerical values and evaluate T :
T = 2π
(2.0 mH)(20 μ F) = 2
(b) Solve equation (1) for L to obtain:
L =
Substitute numerical values and evaluate L:
L =
(1)
T
2
4π C
=
1 4π f 2C 2
1 4π (60 s 2
1.3 ms
) (80 μ F)
−1 2
= 88 mH
An LC circuit has capacitance C 0 and inductance L. A second LC 30 • circuit has capacitance 12 C 0 and inductance 2 L, and a third LC circuit has capacitance 2C 0 and inductance
1 2
L. (a) Show that each circuit oscillates with the
same frequency. (b) In which circuit would the peak current be greatest if the peak voltage across the capacitor in each circuit was the same?
2756
Chapter 29
Picture the Problem We can use the expression f 0
= 1 2π
LC for the resonance
frequency of an LC circuit to show that each circuit oscillates with the same frequency. In (b) we can use I peak = Q0 , where Q0 is the charge of the capacitor at time zero, and the definition of capacitance Q0
= CV to express I peak in terms of
ω , C and V.
Express the resonance frequency for an LC circuit:
f 0
(a) Express the product of L and C 0 for each circuit:
=
1 2π LC
Circuit 1: L1C 1 = L1C 0 , Circuit 2: L2C 2 and Circuit 3: L3C 3
= (2 L1 )( 12 C 0 ) = L1C 1 , = ( 12 L1 )(2C 0 ) = L1C 1
Because L1C 1 = L2 C 2 = L3C 3 , the resonance frequencies of the three circuits are the same. I peak = Q0
(b) Express I peak in terms of the charge stored in the capacitor:
= CV
Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor:
Q0
Substituting for Q0 yields:
I peak = CV
or, for ω and V constant, I peak ∝ C . Hence, the circuit with capacitance 2C 0 has the greatest peak current. A 5.0-μ F capacitor is charged to 30 V and is then connected across an 31 •• ideal 10-mH inductor. (a) How much energy is stored in the system? ( b) What is the frequency of oscillation of the circuit? ( c) What is the peak current in the circuit? Picture the Problem We can use U = 12 CV 2 to find the energy stored in the
electric field of the capacitor, ω 0 Q0
= CV to find I peak .
= 2π f 0 = 1
LC to find f 0, and I peak = Q0 and
Alternating-Current Circuits
(a) Express the energy stored in the system as a function of C and V :
U = 12 CV 2
Substitute numerical values and evaluate U :
U =
1 2
(5.0 μ F)(30 V)2 =
(b) Express the resonance frequency of the circuit in terms of L and C :
ω 0
= 2π f 0 =
Substitute numerical values and evaluate f 0:
f 0
=
1 LC
2757
2.3 mJ
⇒ f 0 =
1 2π (10 mH)(5.0 μ F)
1 2π LC
= 712 Hz
= 0.71 kHz (c) Express I peak in terms of the
I peak = Q0
charge stored in the capacitor:
= CV
Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor:
Q0
Substituting for Q0 yields:
I peak = CV
Substitute numerical values and evaluate I peak :
I peak = 2π 712 s
(
−1
)(5.0 μ F)(30 V )
= 0.67 A
A coil with internal resistance can be modeled as a resistor and an 32 •• ideal inductor in series. Assume that the coil has an internal resistance of 1.00 Ω and an inductance of 400 mH. A 2.00-μ F capacitor is charged to 24.0 V and is then connected across coil. (a) What is the initial voltage across the coil? ( b) How much energy is dissipated in the circuit before the oscillations die out? (c) What is the frequency of oscillation the circuit? (Assume the internal resistance is sufficiently small that has no impact on the frequency of the circuit.) (d ) What is the quality factor of the circuit?
2758
Chapter 29
Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to find the initial voltage across the coil. ( b) The total energy lost via joule heating is the total energy initially stored in the capacitor. (c) The natural frequency of the circuit is
given by f 0
= 1 2π
LC . In Part (d ) we can use its definition to find the quality
factor of the circuit. (a) Application of Kirchhoff’s loop rule leads us to conclude that the initial voltage across the coil is 24.0 V . (b) Because the ideal inductor can not dissipate energy as heat, all of the energy initially stored in the capacitor will be dissipated as joule heat in the resistor: (c) The natural frequency of the circuit is:
1 CV 2 2
U =
1 2
= (2.00 μ F)(24.0 V )2
= 0.576 mJ
f 0
=
1 2π LC
=
1 2π (400 mH)(2.00 μ F)
= 178 Hz (d ) The quality factor of the circuit is given by: Substituting for ω 0 and simplifying yields:
Substitute numerical values and evaluate Q:
Q=
ω 0 L R
1
Q=
Q=
L LC R
1 1.00 Ω
=
1 L R
C
400 mH 2.00 μ F
= 447
33 ••• [SSM] An inductor and a capacitor are connected, as shown in Figure 29-30. Initially, the switch is open, the left plate of the capacitor has charge Q0. The switch is then closed. ( a) Plot both Q versus t and I versus t on the same graph, and explain how it can be seen from these two plots that the current leads the charge by 90º. (b) The expressions for the charge and for the current are given by Equations 29-38 and 29-39, respectively. Use trigonometry and algebra to show that the current leads the charge by 90º. Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff’s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect
Alternating-Current Circuits
2759
to time, an expression for the current as a function of time. We’ll use a spreadsheet program to plot the graphs. Apply Kirchhoff’s loop rule to a clockwise loop just after the switch is closed:
Q C
Because I = dQ dt :
L
+ L
dI
=0
dt
d 2 Q 2
dt
+
Q C
= 0 or
d 2Q dt 2
+
1
Q=0 LC
Q(t ) = Q0 cos( t − δ )
The solution to this equation is:
1 LC
where ω = Because Q(0) = Q0, δ = 0 and:
Q(t ) = Q0 cos t
The current in the circuit is the derivative of Q with respect to t :
I =
dQ dt
=
d
[Q0 cos ω t ] = −ω Q0 sin ω t
dt
(a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Q0 were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°. 1.2
1.2
Charge Current 0.6
0.6
) A
) C m 0.0 (
m 0.0 ( I
Q
-0.6
-0.6
-1.2
-1.2 0
2
4
6
8
10
t (s)
(b) The equation for the current is:
I = − Q0 sin t
The sine and cosine functions are related through the identity:
π ⎞ − sin θ = cos⎛ ⎜θ + ⎟ ⎝ 2 ⎠
(1)
2760
Chapter 29
Use this identity to rewrite equation (1):
⎛ ⎝
I = −ω Q0 sin ω t = ω Q0 cos⎜ ω t +
π ⎞
⎟
2 ⎠
Thus, the current leads the charge by 90°.
Driven RL Circuits A circuit consists of a resistor, an ideal 1.4-H inductor and an ideal 34 •• 60-Hz generator, all connected in series. The rms voltage across the resistor is 30 V and the rms voltage across the inductor is 40 V. (a) What is the resistance of the resistor? (b) What is the peak emf of the generator? Picture the Problem We can express the ratio of V R to V L and solve this expression for the resistance R of the circuit. In (b) we can use the fact that, in an LR circuit, V L leads V R by 90° to find the ac input voltage.
V L = IX L = I L
(a) Express the potential differences across R and L in terms of the common current through these components:
and V R = IR
Divide the second of these equations by the first to obtain:
V R
Substitute numerical values and evaluate R:
R
(b) Because V R leads V L by 90° in an LR circuit:
V peak =
2V rms
Substitute numerical values and evaluate V peak :
V peak =
2
V L
=
IR I ω L
=
⎛ V ⎞ ⇒ R = ⎜⎜ R ⎟⎟ω L ω L ⎝ V L ⎠ R
⎛ 30 V ⎞ ⎟⎟2π (60 s −1 )(1.4 H ) = 0.40 k Ω = ⎜⎜ ⎝ 40 V ⎠
= 2
V R2
+ V L2
(30 V )2 + (40 V )2 =
71 V
35 •• [SSM] A coil that has a resistance of 80.0 Ω has an impedance of 200 Ω when driven at a frequency of 1.00 kHz. What is the inductance of the coil? Picture the Problem We can solve the expression for the impedance in an LR circuit for the inductive reactance and then use the definition of X L to find L.
Alternating-Current Circuits
Express the impedance of the coil in terms of its resistance and inductive reactance:
Z = R 2 + X L2
Solve for X L to obtain:
X L
=
Express X L in terms of L:
X L
= 2π fL
Equate these two expressions to obtain:
2π fL = Z
Substitute numerical values and evaluate L:
(200 Ω )2 − (80.0 Ω )2 L = 2π (1.00 kHz )
2
Z
2761
− R 2
2
− R ⇒ L = 2
Z 2 − R 2
2π f
= 29.2 mH A two conductor transmission line simultaneously carries a 36 •• superposition of two voltage signals, so the potential difference between the two conductors is given by V = V 1 + V 2, where V 1 = (10.0 V) cos(ω 1t ) and V 2 = (10.0 V) cos(ω 2t ), where ω 1 = 100 rad/s and ω 2 = 10 000 rad/s. A 1.00 H inductor and a 1.00 k Ω shunt resistor are inserted into the transmission line as shown in Figure 29-31. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) What is the voltage ( V out) at the output of the transmission line? (b) What is the ratio of the low-frequency amplitude to the high-frequency amplitude at the output? Picture the Problem We can express the two output voltage signals as the product of the current from each source and R = 1.00 k Ω. We can find the currents due to each source using the given voltage signals and the definition of the impedance for each of them.
(a) Express the voltage signals observed at the output side of the transmission line in terms of the potential difference across the resistor:
V 1, out
= I 1 R
and V 2, out
= I 2 R
2762
Chapter 29
Evaluate I 1 and I 2: I 1
=
V 1 Z 1
=
(10.0 V )cos100t = (9.95 mA) cos100t 2 2 -1 (1.00 k Ω ) + [(100 s )(1.00 H )]
and I 2
=
V 2 Z 2
(10.0 V )cos10 4 t = = (0.995 mA )cos10 4 t 2 (1.00 k Ω )2 + [(10 4 s −1 )(1.00 H )]
Substitute for I 1 and I 2 to obtain:
V 1, out
= (1.00 k Ω)(9.95 mA) cos100t = (9.95 V ) cos100t
where ω 1 = 100 rad/s and ω 2 = 10 000 rad/s. and V 2, out = (1.00 k Ω )(0.995 mA) cos10 4 t
= (0.995 V )cos104 t (b) Express the ratio of V 1,out to V 2,out:
V 1, out V 2, out
=
9.95 V 0.995 V
= 10 : 1
A coil is connected to a 120-V rms, 60-Hz line. The average power 37 •• supplied to the coil is 60 W, and the rms current is 1.5 A. Find ( a) the power factor, (b) the resistance of the coil, and (c) the inductance of the coil. (d ) Does the current lag or lead the voltage? Explain your answer. (e) Support your answer to Part (d ) by determining the phase angle. Picture the Problem The average power supplied to coil is related to the power 2 R to find R. Because the factor by Pav = ε rms I rms cos δ . In (b) we can use Pav = I rms
inductance L is related to the resistance R and the phase angle δ according to X L = L = R tan δ , we can use this relationship to find the resistance of the coil. Finally, we can decide whether the current leads or lags the voltage by noting that the circuit is inductive. (a) Express the average power supplied to the coil in terms of the power factor of the circuit: Substitute numerical values and evaluate cosδ :
Pav
cos
= ε rms I rms cos δ ⇒ cos δ =
=
Pav
ε rms I rms
60 W = 0.333 = 0.33 (120 V )(1.5 A )
Alternating-Current Circuits (b) Express the power supplied by the source in terms of the resistance of the coil:
Pav
2 = I rms R ⇒ R =
60 W
Substitute numerical values and evaluate R:
R =
(c) Relate the inductive reactance to the resistance and phase angle:
X L
=
L =
R tan δ
Solving for L yields:
Substitute numerical values and evaluate L:
L =
(d ) Evaluate X L:
X L
(1.5 A )2
2763
Pav 2 I rms
= 26.7 Ω = 27 Ω
L = R tan δ
ω
=
R tan cos
−1
(0.333)
2π f
(26.7 Ω ) tan (70.5°) 2π (60 s −1 )
= 0.20 H
= (26.7 Ω) tan(70.5°) = 75.4 Ω
Because the circuit is inductive, the current lags the voltage.
= cos−1 (0.333) = 71°
(e) From Part (a): 38
••
A 36-mH inductor that has a resistance of 40
Ω is connected to an
ideal ac voltage source whose output is given by ε = (345 V) cos(150π t ), where t is in seconds. Determine (a) the peak current in the circuit, ( b) the peak and rms voltages across the inductor, (c) the average power dissipation, and (d ) the peak and average magnetic energy stored in the inductor. Picture
the
V L , peak = I peak X L
Problem
=
(a)
We
can
use
I peak = ε peak
R
2
+ (ω L )2 and
LI peak to find the peak current in the circuit and the peak
voltage across the inductor. (b) Once we’ve found V L, peak we can find V L , rms using V L , rms
= V L , peak
2 . (c) We can use Pav
2 = 12 I rms R to find the average power
2 dissipation, and (d ) U L , peak = 12 LI peak to find the peak and average magnetic energy
stored in the inductor. The average energy stored in the magnetic field of the inductor can be found using U L ,av = Pav dt .
∫
2764
Chapter 29
(a) Apply Kirchhoff’s loop rule to the circuit to obtain:
ε − IZ = 0 ⇒ I =
Substitute numerical values and evaluate I :
I =
(b) Because ε = (345 V) cos(150π t ) :
V L, peak = 345 V
Find V L ,rms from V L, peak :
ε =
Z
(345 V )cos(150π t ) 2 (40 Ω )2 + [(150π s −1 )(36 mH )] = (7.94 A )cos(150π t ) and I peak = 7.9 A .
V L , rms
=
V L , peak
2
=
345 V 2
Pav = I
Substitute numerical values and evaluate Pav :
Pav
(d ) The maximum energy stored in the magnetic field of the inductor is:
2 U L , peak = 12 LI peak =
U (t ) is given by:
Substitute for U (t ) to obtain:
= 244 V
2
(c) Relate the average power dissipation to I peak and R:
The definition of U L ,av is:
ε 2 R 2 + (ω L )
⎛ I peak ⎞ 2 ⎟⎟ R = 12 I peak R = ⎜⎜ R ⎝ 2 ⎠
2 rms
= 12 (7.94 A )2 (40 Ω) = 1.3 kW 1 2
(36 mH)(7.94 A)2
= 1.1J
U L ,av
=
1
T
∫ U (t )dt
T 0
1 2 U (t ) = L[ I (t )] 2
U L ,av
=
L
T
[ I (t )] dt 2T ∫ 2
0
Evaluating the integral yields:
Substitute numerical values and evaluate U L ,av :
⎡ 1 I 2 ⎤T = 1 LI 2 peak ⎥ peak 2T ⎢⎣ 2 4 ⎦ L
U L ,av
=
U L ,av
= (36 mH )(7.94 A )2 = 0.57 J
1
4
Alternating-Current Circuits
2765
39 •• [SSM] A coil that has a resistance R and an inductance L has a power factor equal to 0.866 when driven at a frequency of 60 Hz. What is the coil’s power factor it is driven at 240 Hz? Picture the Problem We can use the definition of the power factor to find the relationship between X L and R when the coil is driven at a frequency of 60 Hz and then use the definition of X L to relate the inductive reactance at 240 Hz to the inductive reactance at 60 Hz. We can then use the definition of the power factor to determine its value at 240 Hz. R
Using the definition of the power factor, relate R and X L:
cos δ =
Square both sides of the equation to obtain:
cos δ =
Solve for X L2 (60 Hz) :
Z
R
= R
2
(1)
+ X
2 L
R 2
2
R
2
+ X L2 ⎛ 1 − 1 ⎞ ⎜ 2 ⎟ ⎝ cos δ ⎠
X L (60 Hz ) = R
2
Substitute for cosδ and simplify to obtain:
X L (60 Hz ) = R
2
Use the definition of X L to obtain:
X L2 ( f ) = 4π f 2 L2 and X L2 ( f' ) = 4π f' 2 L2
Dividing the second of these equations by the first and simplifying yields:
X L ( f' )
2
2
2
X L2 ( f )
=
⎛ 1 ⎞ 1 2 ⎜⎜ ⎟⎟ = 3 R − 1 2 0 . 866 ( ) ⎝ ⎠
4π f' 2 L2 4π f 2 L2
=
2
f'
f 2
or 2
2 L
X
Substitute numerical values to obtain:
⎛ f' ⎞ ( f' ) = ⎜⎜ ⎟⎟ X L2 ( f ) ⎝ f ⎠ 2
⎛ 240 s −1 ⎞ 2 ⎟ X (240 Hz ) = ⎜⎜ −1 ⎟ X L (60 Hz ) ⎝ 60 s ⎠ 1 ⎞ 16 = 16⎛ ⎜ R 2 ⎟ = R 2 ⎝ 3 ⎠ 3 2 L
Chapter 29
2766
Substitute in equation (1) to obtain:
R
(cos δ )240 Hz = R 2
16 2 R 3
+
=
3 19
= 0.397 40
••
A resistor and an inductor are connected in parallel across an ideal ac
voltage source whose output is given by ε = ε peak cosω t as shown in Figure 29-32. Show that (a) the current in the resistor is given by I R = ε peak / R cos ω t , (b) the current in the inductor is given by I L = ε peak / X L cos(ω t – 90º), and (c) the current in the voltage source is given by I = I R + I L = I peak cos(ω t – δ ), where I peak = ε max/ Z . Picture the Problem Because the resistor and the inductor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through the resistor and the current through the inductor. Because these two currents are not in phase, we’ll need to use phasors to calculate their sum. The amplitudes of the applied voltage and the currents are equal to the magnitude
r
of the phasors. That is
ε
= ε peak ,
r
r
I = I peak , I R
(a) The ac source applies a voltage given by ε ε peak cos t . Thus, the
=
r
= I R , peak , and I L = I L , peak .
ε peak cos
t = I R R
voltage drop across both the load resistor and the inductor is: The current in the resistor is in phase with the applied voltage:
Because I R , peak =
ε peak R
:
(b) The current in the inductor lags the applied voltage by 90°: Because I L , peak =
ε peak X L
:
(c) The net current I is the sum of the currents through the parallel branches:
I R
= I R, peak cos
I R
=
I L
= I L, peak cos( t − 90°)
I L
=
ε peak R
ε peak X L
I = I R + I L
t
cos ω t
cos(ω t − 90° )
Alternating-Current Circuits Draw the phasor diagram for the circuit. The projections of the phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, and the current in the inductor lags the applied voltage by 90°. The net current phasor is the sum of the r r r branch current phasors I = I L + I R .
(
Z is the impedance of the combination:
From the phasor diagram we have:
ε I R I
δ
ω t
ω t − δ
90°− ω t
)
The peak current through the parallel combination is equal to ε peak Z , where
2767
I L
I = I peak cos(ω t − δ ) ,
where I peak =
ε peak Z
2 I peak = I R2, peak + I L2, peak 2
2
⎛ ε ⎞ ⎛ ε ⎞ = ⎜⎜ peak ⎟⎟ + ⎜⎜ peak ⎟⎟ ⎝ R ⎠ ⎝ X L ⎠ 2 1 ⎞ ε peak 2 ⎛ 1 = ε peak ⎜⎜ 2 + 2 ⎟⎟ = 2 ⎝ R X L ⎠ Z where
1 2 Z
=
1 2 R
+
1 2 X L
. Solving for I peak yields:
From the phasor diagram:
I peak =
ε peak Z
where Z −2 = R −2 + X L−2
I = I peak cos(ω t − δ )
where
ε peak tan δ
=
I L , peak I R , peak
=
X L
ε peak
=
R X L
R
41 •• [SSM] Figure 29-33 shows a load resistor that has a resistance of R L = 20.0 Ω connected to a high-pass filter consisting of an inductor that has inductance L = 3.20-mH and a resistor that has resistance R = 4.00-Ω. The output
of the ideal ac generator is given by ε = (100 V) cos(2π ft ). Find the rms currents in all three branches of the circuit if the driving frequency is ( a) 500 Hz and (b) 2000 Hz. Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is ( c) 500 Hz and (d ) 2000 Hz.
2768
Chapter 29
Picture the Problem ε = V 1 + V 2 , where V 1 is the voltage drop across R and V 2 is
r
r
r
= V + V is the r
the voltage drop across the parallel combination of L and R L.
ε
relation for the phasors. For the parallel combination I = I R
+ I L . Also, V 1 is in
r
r
L
1
2
phase with I and V 2 is in phase with I R L . First draw the phasor diagram for the currents in the parallel combination, then add the phasors for the voltages to the diagram. The phasor diagram for the currents in the circuit is:
I R
L
I
δ
I L
Adding the voltage phasors to the diagram gives:
ε V 2
I R
L
δ
I
δ
V 1
ω t
I L
The maximum current in the inductor, I 2, peak , is given by:
I 2, peak =
V 2, peak
(1)
Z 2
where Z 2−2 = R L−2 + X L−2 tan δ is given by:
tan δ
= =
Solve for δ to obtain:
δ
I L , peak I R , peak R L X L
=
=
(2)
V 2, peak X L V 2, peak RL
R L
ω L
⎛ R ⎞ = tan −1 ⎜⎜ L ⎟⎟ ⎝ 2π fL ⎠
=
R L
2π fL
(3)
Alternating-Current Circuits
2769
Apply the law of cosines to the triangle formed by the voltage phasors to obtain: 2 ε peak = V 12, peak + V 22, peak + 2V 1, peak V 2, peak cos δ
or 2
2
I peak Z
2 2 2 2 = I peak R + I peak Z 2 + 2 I peak RI peak Z 2 cos δ
= R 2 + Z 22 + 2 RZ 2 cos δ
Dividing out the current squared yields:
Z
Solving for Z yields:
Z = R 2 + Z 22
The maximum current I peak in the circuit is given by: I rms is related to I peak according
to: (a) Substitute numerical values in equation (3) and evaluate δ :
Solving equation (2) for Z 2 yields:
Substitute numerical values and evaluate Z 2:
2
I peak =
I rms
δ
=
+ 2 RZ 2 cos δ
ε peak
(5)
Z
1 2
(4)
(6)
I peak
⎛ ⎞ 20.0 Ω ⎟⎟ = tan −1 ⎜⎜ ⎝ 2π (500 Hz)(3.20 mH) ⎠ ⎛ 20.0 Ω ⎞ ⎟⎟ = 63.31° = tan −1 ⎜⎜ 10 . 053 Ω ⎝ ⎠
Z 2
=
Z 2
=
1 R L−2 + X L−2
1
(20.0 Ω)−2 + (10.053 Ω)−2
= 8.982 Ω Substitute numerical values and evaluate Z : Z =
(4.00 Ω)2 + (8.982 Ω)2 + 2(4.00 Ω)(8.982 Ω) cos 63.31° = 11.36 Ω
Substitute numerical values in equation (5) and evaluate I peak :
I peak =
100 V 11.36 Ω
= 8.806 A
2770
Chapter 29
Substitute for I peak in equation
I rms
(6) and evaluate I rms :
1
=
2
(8.806 A ) =
6.23 A
V 2, peak = I peak Z 2
The maximum and rms values of V 2 are given by:
= (8.806 A )(8.982 Ω ) = 79.095 V and V 2,rms
1
=
2 1
= The rms values of I R L ,rms and
I RL ,rms
I L,rms are:
2
=
V 2, peak
(79.095 V ) = 55.929 V
V 2,rms R L
55.929 V 20.0 Ω
= 2.80 A
55.929 V 10.053 Ω
= 5.53 A
=
and I L ,rms
=
V 2,rms X L
=
= 40.2 Ω , δ = 26.4° , Z 2 = 17.9 Ω , Z = 21.6 Ω , I peak = 4.64 A , and
(b) Proceed as in (a) with f = 2000 Hz to obtain:
X L
= 3.28 A , V 2,max = 83.0V , V 2, rms = 58.7 V ,
I rms
I R L ,rms
= 2.94 A , and I L,rms = 1.46 A
(c) The power delivered by the ac source equals the sum of the power dissipated in the two resistors. The fraction of the total power delivered by the source that is dissipated in load resistor is given by: P R L P R L
+ P R
−1
⎛ P ⎞ ⎛ I 2 R ⎞ = ⎜⎜1 + R ⎟⎟ = ⎜⎜1 + 2 rms ⎟⎟ ⎝ P R ⎠ ⎝ I R ,rms RL ⎠ L
−1
L
Substitute numerical values for f = 500 Hz to obtain: −1
P R L P R L
+ P R
f =500 Hz
⎛ (6.23 A )2 (4.00 Ω ) ⎞ ⎟ = 0.502 = 50.2% = ⎜⎜1 + 2 ⎟ ⎝ (2.80 A ) (20.0 Ω ) ⎠
Alternating-Current Circuits
2771
(d ) Substitute numerical values for f = 2000 Hz to obtain: −1
P R L P R L
42
••
+ P R
f = 2000 Hz
⎛ (3.28 A )2 (4.00 Ω ) ⎞ ⎟⎟ = 0.800 = 80.0% = ⎜⎜1 + 2 Ω ( ) ( ) 2 . 94 A 20 . 0 ⎝ ⎠
An ideal ac voltage source whose emf ε 1 is given by (20 V) cos(2 π ft )
and an ideal battery whose emf ε 2 is 16 V are connected to a combination of two resistors and an inductor (Figure 29-34), where R1 = 10 Ω, R2 = 8.0 Ω, and L = 6.0 mH. Find the average power delivered to each resistor if (a) the driving frequency is 100 Hz, (b) the driving frequency is 200 Hz, and (c) the driving frequency is 800 Hz. Picture the Problem We can treat the ac and dc components separately. For the
dc component, L acts like a short circuit. Let ε 1, peak denote the peak value of the voltage supplied by the ac voltage source. We can use P = ε 22 R to find the power dissipated in the resistors by the current from the ideal battery. We’ll apply Kirchhoff’s loop rule to the loop including L, R1, and R2 to derive an expression for the average power delivered to each resistor by the ac voltage source. (a) The total power delivered to R1 and R2 is:
The dc power delivered to the resistors whose resistances are R1 and R2 is: Express the average ac power delivered to R1: Apply Kirchhoff’s loop rule to a clockwise loop that includes R1, L, and R2: Solving for I 2 yields:
P1
= P1, dc + P1, ac
and P2 = P2, dc
P1,dc
P1, ac
=
=
ε 22 R1
+ P2, ac and P2,dc
=
(2)
=
ε 22 R2
ε 12, rms ε 12, peak =
R1
R1 I 1 − Z 2 I 2
I 2
(1)
R1 Z 2
I 1
2 R1
=0
=
R1
ε 1, peak ε 1, peak
Z 2 R1
=
Z 2
2772
Chapter 29
Express the average ac power delivered to R2:
2
P2, ac
= =
Substituting in equations (1) and (2) yields:
P1
1 2
⎛ ε 1, peak ⎞ ⎟⎟ R2 R2 = ⎜⎜ Z ⎝ 2 ⎠
2 2, rms
I
1 2
ε 12, peak R2 2 Z 22
2 ε 12, peak ε 2 = +
R1
2 R1
and P2
Substitute numerical values and evaluate P1:
2 ε 12, peak R2 ε 2 = +
R2
2 Z 22
(16 V) 2
(20 V)2 P1 = + = 10 Ω 2(10 Ω )
46 W
Substitute numerical values and evaluate P2: P2
=
(16 V )2 8.0 Ω
(20 V )2 (8.0 Ω ) + = 2 2 2 [(8.0 Ω ) + (2π {100 s −1 }{6.0 mH}) ]
(b) Proceed as in (a) to evaluate P1 and P2 with f = 200 Hz:
(c) Proceed as in (a) to evaluate P1 and P2 with f = 800 Hz:
52 W
P1
= 25.6 W + 20.0 W = 46 W
P2
= 32.0 W + 13.2 W = 45 W
P1
= 25.6 W + 20.0 W = 46 W
P2
= 32.0 W + 1.64 W = 34 W
An ac circuit contains a resistor and an ideal inductor connected in 43 •• series. The voltage rms drop across this series combination is 100-V and the rms voltage drop across the inductor alone is 80 V. What is the rms voltage drop across the resistor? Picture the Problem We can use the phasor diagram for an RL circuit to find the voltage across the resistor.
Alternating-Current Circuits
2773
r
V L
The phasor diagram for the voltages in the circuit is shown to the right:
ε
rms
r
V R
Use the Pythagorean theorem to express V R:
V R
2 = ε rms − V L2
Substitute numerical values and evaluate V R:
V R
= (100 V ) 2 − (80 V ) 2 = 60 V
Filters and Rectifiers The circuit shown in Figure 29-35 is called an RC high-pass filter 44 •• because it transmits input voltage signals that have high frequencies with greater amplitude than it transmits input voltage signals that have low frequencies. If the input voltage is given by V in = V in peak cos ω t, show that the output voltage is V out = V H cos(ω t – δ ) where V H
= V in peak
−2
1 + (ω RC ) . (Assume that the output is
connected to a load that draws only an insignificant amount of current.) Show that this result justifies calling this circuit a high-pass filter. Picture the Problem The phasor diagram for the RC high-pass filter is r
V R
r
shown to the right. V app and V R are the phasors for V in and V out, respectively. Note that tan δ = − X C R. That δ is
ω t − δ
r
δ
negative follows from the fact that V app
ω t
r
lags V R by r
δ .
V app
The projection of
V app onto the horizontal axis is
r
V C
V app = V in, and the projection of V R
onto the horizontal axis is V R = V out. Express V app :
= V app, peak cos t where V app, peak = V peak = I peak Z and Z 2 = R 2 + X C 2 (1)
Because δ < 0:
ω t + δ
V app
= ω t − δ
2774
Chapter 29
V R is given by:
Solving equation (1) for Z and substituting for X C yields:
Because V out
= V R :
= V R, peak cos( t − δ ) where V R , peak = V H = I peak R V R
Z = R
V out
V out
1 ⎞ + ⎛ ⎜ ⎟ ⎝ ω C ⎠
2
(2)
= V R, peak cos( t − δ ) = I in peak R cos(ω t − δ ) =
Using equation (2) to substitute for Z yields:
2
=
V in peak R cos(ω t − δ ) Z V in peak
⎛ 1 ⎞ ⎟ ⎝ ω C ⎠
2
R cos(ω t − δ )
R 2 + ⎜
Simplify further to obtain:
V out
=
V in peak −2
1 + (ω RC )
cos(ω t − δ )
or V out
=
V H cos(ω t − δ )
where
As ω → ∞:
V H
=
V H
→
V in peak −2
1 + (ω RC ) V in peak
1 + (0)
2
= V in peak showing that
the result is consistent with the high pass name for this circuit. (a) Find an expression for the phase constant δ in Problem 44 in terms 45 •• of ω , R and C . (b) What is the value of δ in the limit that ω → 0? (c) What is the value of δ in the limit that ω → ∞? (d ) Explain your answers to Parts ( b) and (c).
Alternating-Current Circuits Picture the Problem The phasor diagram for the RC high-pass filter is
below.
V R
r
r
shown
V app and
V R are
the V app
phasors for V in and V out, respectively. r The projection of V app onto the
δ ω t
horizontal axis is V app = V in, and the r projection of V R onto the horizontal axis is V R = V out. r
V C
r
(a) Because V app lags V R by δ .
Use the definition of X C to obtain:
Solving for δ yields:
2775
tan δ = −
V C V R
=−
IX C IR
=−
X C R
1 1 tan δ = − ω C = − R ω RC
⎡ ⎣
δ = tan −1 ⎢−
(b) As ω → 0:
δ →
(c) As ω → ∞:
δ → 0
⎤ ω RC ⎥⎦ 1
− 90°
r
r
(d ) For very low driving frequencies, X C >> R and so V C effectively lags V in by r
90°. For very high driving frequencies, X C << R and so V R is effectively in phase r
with V in . Assume that in Problem 44, R = 20 k Ω and C = 15 nF. (a) At what 46 •• frequency is V H = 12 V in peak ? This particular frequency is known as the 3 dB frequency, or f 3dB for the circuit. (b) Using a spreadsheet program, make a graph of log10(V H) versus log10( f ), where f is the frequency. Make sure that the scale extends from at least 10% of the 3-dB frequency to ten times the 3-dB frequency. (c) Make a graph of δ versus log10( f ) for the same range of frequencies as in Part (b). What is the value of the phase constant when the frequency is equal to the 3-dB frequency? Picture the Problem We can use the results obtained in Problems 44 and 45 to find f 3 dB and to plot graphs of log(V out) versus log( f ) and δ versus log( f ).
2776
Chapter 29
(a) Use the result of Problem 44 to express the ratio V out/V in peak :
V in peak V out V in peak
When V out
= V in peak / 2 :
=
1 + (ω RC ) V in peak
1 1 + (ω RC )
Square both sides of the equation and solve for ω RC to obtain: Substitute numerical values and evaluate f 3 dB:
=
V out
=
B
(b) From Problem 44 we have:
In Problem 45 it was shown that:
Rewrite these expressions in terms of f 3 dB to obtain:
−2
=
1 1 + (ω RC )
2 1 ⇒ f 3 dB RC
=
1 2π RC
1 = 0.53 kHz 2π (20 k Ω )(15 nF) V in peak
1 + (ω RC )
−2
⎡ 1 ⎤ ⎣ ω RC ⎥⎦
δ = tan −1 ⎢−
V out
=
V in peak
⎛ 1 ⎞ ⎟⎟ 1 + ⎜⎜ ⎝ 2π fRC ⎠
2
V peak
=
⎛ f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠
and
⎡ 1 ⎤ −1 ⎡ f 3 dB ⎤ = tan ⎥ ⎢− f ⎥ fRC π 2 ⎣ ⎦ ⎣ ⎦
δ = tan −1 ⎢−
A spreadsheet program to generate the data for a graph of V out versus f and δ versus f is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 A8
−2
1
=
RC = 1 ⇒ ω =
f 3 dB
−2
Formula/Content 2.00E+03 1.50E−08 1 531 53
Algebraic Form R C V in peak f 3 dB 0.1 f 3 dB
2
Alternating-Current Circuits C8
$B$3/SQRT(1+(1($B$4/A8))^2)
V in peak
⎛ f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠
1 2 3 4 5 6 7 8 9 10 11
2
D8 E8
LOG(C8) ATAN(−$B$4/A8)
log(V out) ⎡ f ⎤ tan −1 ⎢− 3 dB ⎥ ⎣ f ⎦
F8
E8*180/PI()
δ in degrees
A B R= 2.00E+04 C = 1.50E−08 V in peak = 1 f 3 dB= 531
C ohms F V Hz
D
E
F
f 53 63 73 83
log( f ) 1.72 1.80 1.86 1.92
V out 0.099 0.118 0.136 0.155
log(V out) −1.003 −0.928 −0.865 −0.811
delta(rad) delta(deg) −1.471 −84.3 −1.453 −83.2 −1.434 −82.2 −1.416 −81.1
55 56 57
523 533 543
2.72 2.73 2.73
0.702 0.709 0.715
−0.154 −0.150 −0.146
−0.793 −0.783 −0.774
−45.4 −44.9 −44.3
531 532 533 534
5283 5293 5303 5313
3.72 3.72 3.72 3.73
0.995 0.995 0.995 0.995
−0.002 −0.002 −0.002 −0.002
−0.100 −0.100 −0.100 −0.100
−5.7 −5.7 −5.7 −5.7
The following graph of log(V out) versus log( f ) was plotted for V in peak = 1 V. 0.0
-0.2 ) t
u o
V
( g o l
-0.4
-0.6
-0.8
-1.0 1.5
2.0
2.5
3.0
log( f )
3.5
4.0
2777
2778
Chapter 29
A graph of δ (in degrees) as a function of log( f ) follows. 0 -10 -20 -30 ) g e d ( a t l e d
-40 -50 -60 -70 -80 -90 1.5
2.0
2.5
3.0
3.5
4.0
log( f )
Referring to the spreadsheet program, we see that when f = f 3 dB, δ ≈
− 44.9°.
This result is in good agreement with its calculated value of −45.0°. 47 •• [SSM] A slowly varying voltage signal V (t ) is applied to the input of the high-pass filter of Problem 44. Slowly varying means that during one time constant (equal to RC ) there is no significant change in the voltage signal. Show that under these conditions the output voltage is proportional to the time derivative of V (t ). This situation is known as a differentiation circuit . Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. Because the voltage drop across the resistor is small compared to the voltage drop across the capacitor, we can express the voltage drop across the capacitor in terms of the input voltage.
Apply Kirchhoff’s loop rule to the input side of the filter to obtain:
Substitute for V (t ) and I to obtain:
Because Q = CV C:
V (t ) − V C − IR = 0
where V C is the potential difference across the capacitor. V in peak cos ω t − V c − R dQ dt
Substitute for dQ/dt to obtain:
=
d
dQ dt
=0
[CV C ] = C dV C
dt
dt
V peak cos ω t − V C − RC
dV C dt
=0
the differential equation describing the potential difference across the capacitor.
Alternating-Current Circuits Because there is no significant change in the voltage signal during one time constant: Substituting for RC
dV C dt
yields:
dV C dt
= 0 ⇒ RC
dV C dt
V in peak cos t − V C
2779
=0
=0
and V C = V in peak cos t
Consequently, the potential difference across the resistor is given by:
V R
= RC
dV C
=
dt
RC
d
[V
in peak
dt
cos ω t ]
We can describe the output from the high-pass filter from Problem 44 48 •• using a decibel scale: β = (20 dB) log10 V H V in peak ) , where β is the output in decibels. Show that for V H V H
=
1 2
=
1 2
V in peak , β = 3.0 dB. The frequency at which
V in peak is known as f 3dB (the 3-dB frequency). Show that for f << f 3dB, the
output β drops by 6 dB if the frequency f is halved. Picture the Problem We can use the expression for V H from Problem 44 and the definition of β given in the problem to show that every time the frequency is halved, the output drops by 6 dB.
From Problem 44:
V H
V in peak
=
1 + (ω RC )
or V H V in peak
Express this ratio in terms of f and f 3 dB and simplify to obtain:
For f << f 3dB: B
V H V peak
V H V peak
From the definition of β we have:
=
−2
1 1 + (ω RC ) 1
=
⎛ f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠
2
−2
f
≈ 2 3 dB
f
⎛ f 2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ f 3 dB ⎠
⎛ V H ⎞ ⎟ ⎜ V peak ⎟ ⎝ ⎠
β = 20 log10 ⎜
f
= 2 3 dB
f
=
⎛ f 2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ f 3 dB ⎠
f f 3 dB
Chapter 29
2780
Substitute for V H/V peak to obtain:
⎛ f ⎞ ⎟⎟ f ⎝ 3 dB ⎠
β = 20 log10 ⎜⎜
Doubling the frequency yields:
⎛ 12 f ⎞ ⎟⎟ β ' = 20 log10 ⎜⎜ f ⎝ 3 dB ⎠
The change in decibel level is:
Δ β
= β ' − β ⎛ 12 f ⎞ ⎛ ⎞ ⎟⎟ − 20 log10 ⎜⎜ f ⎟⎟ = 20 log10 ⎜⎜ ⎝ f 3 dB ⎠ ⎝ f 3 dB ⎠ = 20 log10 ( 12 ) ≈ − 6 dB
Show that the average power dissipated in the resistor of the V in2 peak high-pass filter of Problem 44 is given by Pave = . −2 ⎡ ⎤ ω 2 R 1 + RC 49
••
[SSM]
(
⎣
)
⎦
Picture the Problem We can express the instantaneous power dissipated in the resistor and then use the fact that the average value of the square of the cosine function over one cycle is ½ to establish the given result.
The instantaneous power P(t ) dissipated in the resistor is:
P(t ) =
The output voltage V out is:
V out
= V H cos( t − δ )
V H
=
From Problem 44:
Substitute in the expression for P(t ) to obtain:
2 V out
R
V in peak
1 + (ω RC )
−2
2
P(t ) =
V H R
cos 2 (ω t − δ ) 2
= Because the average value of the square of the cosine function over one cycle is ½:
Pave
=
V in peak
[
R 1 + (ω RC )
−2
]
cos 2 (ω t − δ )
V in2 peak
[
2 R 1 + (ω RC )
−2
]
Alternating-Current Circuits
2781
One application of the high-pass filter of Problem 44 is a noise filter 50 •• for electronic circuits (a filter that blocks out low-frequency noise). Using a resistance value of 20 k Ω, find a value for the capacitance for the high-pass filter that attenuates a 60-Hz input voltage signal by a factor of 10. That is, so V H = 101 V in peak . Picture the Problem We can solve the expression for V H from Problem 44 for the required capacitance of the capacitor.
From Problem 44:
V H
We require that:
=
V H V in peak
V in peak
1 + (ω RC )
=
−2
1 1 + (ω RC )
−2
=
1 10
or
1 + (ω RC ) Solving for C yields:
1
C =
Substitute numerical values and evaluate C :
C =
−2
99ω R
= 10
=
1 2π 99 Rf 1
2
99 (20 k Ω )(60 Hz)
= 13 nF
51 •• [SSM] The circuit shown in Figure 29-36 is an example of a low pass filter. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) If the input voltage is given by V in = V in peak cos ω t, show that the output voltage is V out = V L cos(ω t – δ ) where V L
= V in peak
(
)
2
1+ ω RC . (b) Discuss the trend of the output voltage in the
limiting cases ω → 0 and ω →
∞.
Picture the Problem In the phasor diagram for the RC low-pass filter,
V R
r
r
V app and V C are the phasors for V in and
V out, respectively. The projection of
V app
r
V app onto the horizontal axis is V app = r
V in, the projection of V C onto the
horizontal r
axis
is
V C
=
V out,
V peak = V app , and φ is the angle between r
V C and the horizontal axis.
ω t
φ δ V C
2782
Chapter 29
(a) Express Vapp :
V out
= V C is given by:
If we define δ as shown in the phasor diagram, then:
= V in peak cos t where V in peak = I peak Z and Z 2 = R 2 + X C2 V app
V out
= V C , peak cos φ = I peak X C cos φ
V out
= I peak X C cos( t − δ ) =
V in peak Z
Solving equation (1) for Z and substituting for X C yields:
Z = R
Using equation (2) to substitute for Z and substituting for X C yields:
V out
2
V out
2
(2)
V in peak
=
=
X C cos(ω t − δ )
1 ⎞ + ⎛ ⎜ ⎟ ⎝ ω C ⎠
R
Simplify further to obtain:
(1)
2
1 ⎞ + ⎛ ⎜ ⎟ ⎝ ω C ⎠
V in peak
1 + (ω RC )
2
2
1 cos(ω t − δ ) ω C
cos(ω t − δ )
or V out
=
V L cos(ω t − δ )
where V L
=
V in peak
1 + (ω RC )
2
(b) Note that, as ω → 0, V L → V peak . This makes sense physically in that, for low frequencies, X C is large and, therefore, a larger peak input voltage will appear across it than appears across it for high frequencies. Note further that, as ω → ∞, V L → 0. This makes sense physically in that, for high frequencies, X C is small and, therefore, a smaller peak voltage will appear across it than appears across it for low frequencies. Remarks: In Figures 29-19 and 29-20, is defined as the phase of the voltage drop across the combination relative to the voltage drop across the resistor.
Alternating-Current Circuits
2783
(a) Find an expression for the phase angle δ for the low-pass filter of 52 •• Problem 51 in terms of ω , R and C . (b) Find the value of δ in the limit that ω → 0 and in the limit that ω → ∞. Explain your answer. Picture the Problem The phasor diagram for the RC low-pass filter is
V R
r
r
shown to the right. V app and V C are the phasors for V in and V out, respectively. r The projection of V app onto the
V app
ω t
horizontal axis is V app = V in and the r projection of V C onto the horizontal
δ
r
axis is V C = V out . V peak = V app . (a) From the phasor diagram we have: Use the definition of X C to obtain:
Solving for δ yields:
V C
tan δ =
tan δ =
V R V C
=
R
1 C
I peak R I peak X C
=
R X C
= ω RC
δ = tan −1 (ω RC )
(b) As ω → 0, δ → 0° . This behavior makes sense physically in that, at low frequencies, X C is very large compared to R and, as a consequence, V C is in phase with V in . As ω → ∞, δ → 90° . This behavior makes sense physically in that, at high frequencies, X C is very small compared to R and, as a consequence, V C is out of phase with V in . Remarks: See the spreadsheet solution in the following problem for additional evidence that our answer for Part (b) is correct.
Using a spreadsheet program, make a graph of V L versus input 53 •• frequency f and a graph of phase angle δ versus input frequency for the low-pass filter of Problems 51 and 52. Use a resistance value of 10 k Ω and a capacitance value of 5.0 nF.
2784
Chapter 29
Picture the Problem We can use the expressions for V L and δ derived in Problems 51 and 52 to plot the graphs of V L versus f and δ versus f for the low pass filter of Problem 51. We’ll simplify the spreadsheet program by expressing both V L and δ as functions of f f 3 dB.
From Problems 51 and 52 we have:
V L
=
V in peak
1 + (ω RC )
2
and δ = tan −1 (ω RC ) Rewrite each of these expressions in terms of f to obtain:
V L
=
V in peak
1 + (2π fRC )
2
and δ = tan −1 (2π fRC ) A spreadsheet program to generate the data for graphs of V L versus f and δ versus f for the low-pass filter is shown below. Note that V in in peak has been arbitrarily set equal to 1 V. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B8
Formula/Content 2.00E+03 5.00E−09 1 $B$3/SQRT(1+((2*PI()*A8* 1000*$B$1*$B$2)^2))
Algebraic Form R C V in in peak V in peak
C8
ATAN(2*PI()*A8*1000*$B$1*$B$2)
tan −1 (2π fRC )
D8
C8*180/PI()
δ in degrees
A
1 2 3 4 5 6 7 8 9 10
B C R= 1.00E+04 ohms C = 5.00E−09 F Vin peak = 1 V
(kHz) f (kHz) 0 1 2 3
V out out 1.000 0.954 0.847 0.728
(rad) δ (rad) 0.000 0.304 0.561 0.756
1 + (2π fRC )
D
δ (deg) 0.0 17.4 32.1 43.3
2
Alternating-Current Circuits 54 55 56 57
47 48 49 50
0.068 0.066 0.065 0.064
1.503 1.505 1.506 1.507
2785
86.1 86.2 86.3 86.4
A graph of V L as a function of f follows: 1.0
0.8
) V (
0.6
L
V
0.4
0.2
0.0 0
10
20
30 f
40
50
40
50
( kHz)
A graph of δ as a function of f follows: 90
) g e d ( a t l e d
60
30
0 0
10
20
30 f
(kHz)
54 ••• A rapidly varying voltage signal V (t ) is applied to the input of the low pass filter of Problem 51. Rapidly varying means that during one time constant (equal to RC ) there are significant changes in the voltage signal. Show that under these conditions the output voltage is proportional to the integral of V (t ) with respect to time. This situation is known as an integration circuit . Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We’ll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. The solution to these simultaneous equations will yield the amplitude of the output voltage.
2786
Chapter 29
Apply Kirchhoff’s loop rule to the input side of the filter to obtain:
V (t ) − IR − V C
=0
where V C C is the potential difference across the capacitor.
Substitute for V (t ) and I to obtain:
V in peak cos ω t − R
Because Q = CV C C:
dQ dt
Substitute for dQ/dt to obtain:
V C C is given by:
The fact that V (t ) varies rapidly means that ω >> 1 and so:
=
d
dQ dt
− V C = 0
[CV C ] = C dV C
dt
dt
V in peak cos ω t − RC
dV C
− V C = 0 dt the differential equation describing the potential difference across the capacitor. V C = IX c
=
I
ω C
V C ≈ 0
and dV V peak cos ω t − RC C dt
Separating the variables in this differential equation and solving for V C C yields:
V C =
=0
1 V peak cos ω t dt RC
∫
55 ••• [SSM] The circuit shown in Figure 29-37 is a trap filter . (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) Show that the trap filter acts to reject signals in a band of frequencies
centered at ω = 1/ LC . (b) How does the width of the frequency band rejected depend on the resistance R? Picture the Problem The phasor diagram for the trap filter is shown below. r
r
V app and V L r
r
+ V C are the phasors for V in and
V out out, respectively. The projection of r
V app onto the horizontal axis is V app = V in in, and the projection of V L
r
+ V C onto the
horizontal axis is V L + V C impedance of the trap be zero out. Requiring that the impedance C = V out will yield the frequency at which the circuit rejects signals. Defining the bandwidth as
Δω = ω − ω trap and requiring that Z trap = R will yield an expression
for the bandwidth and reveal its dependence on R.
Alternating-Current Circuits V app
V L + V C
V L
V R
δ
t
ω
ω t − δ
V C
(a) Express Vapp :
= V app, peak cos t where V app, peak = V peak = I peak Z V app
and Z 2 = R 2 + ( X L − X C )
2
V out is given by:
= V out, peak cos( t − δ ) where V out, peak = I peak Z trap and Z trap = X L − X C
Solving equation (1) for Z yields:
Z = R 2 + ( X L − X C )
Because V out
= V L + V C :
(1)
V out
2
V out
(2)
= V out, peak cos( t − δ ) = I peak Z trap cos(ω t − δ ) =
V peak Z trap cos(ω t − δ ) Z
Using equation (2) to substitute for Z yields:
V out
Noting that V out = 0 provided
Z trap
= X L − X C = 0
ω L −
1 = 0 ⇒ ω = ω C
=
V peak R
2
+ Z
2 trap
Z trap cos(ω t − δ )
Z trap = 0, set Z trap = 0 to obtain:
Substituting for X L and X C yields:
(b) Let the bandwidth Δω be:
Δω = ω − ω trap
1 LC
(3)
2787
2788
Chapter 29
Let the frequency bandwidth be defined by the frequency at which Z trap
ω L −
= R . Then:
Because ω trap
=
1
1 = R ⇒ 2 LC − 1 = RC ω C
2
LC
⎛ ω ⎞ ⎜ ⎟ ⎜ ω trap ⎟ − 1 = ω RC ⎝ ⎠
:
For ω ≈ ω trap:
2 ⎛ ω 2 − ω trap ⎞ ⎜ ⎟ ⎜ ω trap ⎟ ≈ ω trap RC ⎝ ⎠
2 Solve for ω 2 − ω trap :
ω
Because ω ≈ ω trap,
+
trap
≈2
trap
Substitute in equation (3) to obtain:
:
2
2 − ω trap = (ω − ω trap )(ω + ω trap )
2
2 − ω trap ≈ 2ω trap (ω − ω trap )
ω
Δω = ω − ω trap =
2 RC ω trap
2
=
R
2 L
56 ••• A half-wave rectifier for transforming an ac voltage into a dc voltage is shown in Figure 29-38. The diode in the figure can be thought of as a one-way valve for current. It allows current to pass in the forward direction (the direction of the arrowhead) only when V in is at a higher electric potential than V out by 0.60 V (that is, whenever V in − V out ≥ +0.60 V ). The resistance of the diode is
effectively infinite when V in − V out is less than +0.60 V. Plot two cycles of both input and output voltages as a function of time, on the same graph, assuming the input voltage is given by V in = V in peak cos ω t. Picture the Problem For voltages greater than 0.60 V, the output voltage will mirror the input voltage minus a 0.60 V drop. But when the voltage swings below 0.60 V, the output voltage will be 0. A spreadsheet program was used to plot the following graph. The angular frequency and the peak voltage were both arbitrarily set equal to one.
Alternating-Current Circuits 1.0
2789
1.0
V_in V_out
) V ( n i
0.5
0.5
0.0
0.0
-0.5
-0.5
) V ( t u o
V
V
-1.0
-1.0 0
2
4
6 t
8
10
12
(s)
57 ••• The output of the rectifier of Problem 56, can be further filtered by putting its output through a low-pass filter as shown in Figure 29-39a. The resulting output is a dc voltage with a small ac component (ripple) shown in Figure 29-39b. If the input frequency is 60 Hz and the load resistance is 1.00 k Ω, find the value for the capacitance so that the output voltage varies by less than 50 percent of the mean value over one cycle. Picture the Problem We can use the decay of the potential difference across the capacitor to relate the time constant for the RC circuit to the frequency of the input signal. Expanding the exponential factor in the expression for V C will allow us to find the approximate value for C that will limit the variation in the output voltage by less than 50 percent (or any other percentage).
The voltage across the capacitor is given by:
V C = V in e −t RC
Expand the exponential factor to obtain:
e−
For a decay of less than 50 percent:
1−
Because the voltage goes positive every cycle, t = 1/60 s and:
t RC
≈1−
1 t RC
1 2 t ≤ 0.5 ⇒ C ≤ t RC R
C ≤
2 ⎛ 1 s ⎞ = 33 μ F ⎜ ⎟ 1.00 k Ω ⎝ 60 ⎠
Driven LC Circuits 58
••
The generator voltage in Figure 29-40, is given by
ε = (100 V) cos (2π ft ). (a) For each branch, what is the peak current and what is the phase of the current relative to the phase of the generator voltage? (b) At the resonance frequency there is no current in the generator. What is the angular frequency at resonance? (c) At the resonance frequency, find the current in the inductor and what is the current in the capacitor. Express your results as
2790
Chapter 29
functions of time. (d ) Draw a phasor diagram showing the phasors for the applied voltage, the generator current, the capacitor current, and the inductor current for the case where frequency is higher than the resonance frequency. Picture the Problem We know that the current leads the voltage across and capacitor and lags the voltage across an inductor. We can use I L , peak = ε peak X L
and I C , peak = ε peak X C to find the amplitudes of these currents. The current in the generator will vanish under resonance conditions, i.e., when I L
=
I C . To find the
currents in the inductor and capacitor at resonance, we can use the common potential difference across them and their reactances together with our knowledge of the phase relationships mentioned above. (a) Express the amplitudes of the currents through the inductor and the capacitor:
I L , peak =
ε peak ε peak =
X L
2π fL
and I C , peak =
ε peak X C
=
ε peak 1 2π fC
= 2π fC ε peak
Substitute numerical values to obtain: I L , peak =
100 V (4.00 H )2π f
=
25.0 V/H , lagging ε by 90° 2π f
and I C , peak = (25.0 μ F)(100 V )ω =
(2.50 mV ⋅ F) 2π f , leading ε by 90°
(b) Express the condition that I = 0:
I L
Solve for ω to obtain:
ω =
Substitute numerical values and evaluate ω :
=
ω =
I C or
ε = ε = ω C ε
ω L
1 C
1 LC
1
(4.00 H )(25.0 μ F)
= 100 rad/s
Alternating-Current Circuits (c) Express the current in the inductor at ω = ω 0:
I L
2791
⎛ 25.0 V/H ⎞ ⎛ π ⎟ cos⎜ ω t − ⎞⎟ = ⎜⎜ −1 ⎟ 2 ⎠ ⎝ 100 s ⎠ ⎝ π ⎞ = (250 mA)cos⎛ ⎜ ω t − ⎟ 2 ⎠ ⎝
where ω = 100 rad/s.
⎛ ⎝ π ⎞ = − (250 mA) cos⎛ ⎜ ω t + ⎟ 2 ⎠ ⎝
Express the current in the capacitor at ω = ω 0:
(
)
I C = (2.50 mV ⋅ F) 100 s −1 cos⎜ ω t +
π ⎞
⎟
2 ⎠
where ω = 100 rad/s. (d ) The phasor diagram for the case where the inductive reactance is larger than the capacitive reactance is shown to the right.
I C
ε max I
I L
A circuit consist of an ideal ac generator, a capacitor and an ideal 59 •• inductor, all connected in series. The charge on the capacitor is given by Q = (15 μ C) cos(ω t + π 4 ), where ω = 1250 rad/s. (a) Find the current in the circuit as a function of time. ( b) Find the capacitance if the inductance is 28 mH. (c) Write expressions for the electrical energy U e, the magnetic energy U m, and the total energy U as functions of time. Picture the Problem We can differentiate Q with respect to time to find I as a
function of time. In (b) we can find C by using ω = 1 LC . The energy stored in the magnetic field of the inductor is given by U m the electric field of the capacitor by U e
=
1 2
Q
2
C
.
= 12 LI 2 and the energy stored in
2792
Chapter 29
(a) Differentiate the charge with respect to time to obtain the current:
⎛ ω t + π ⎞⎤ = −(15 μ C)(1250 s −1 )sin ⎛ ω t + π ⎞ ( ) 15 C cos μ ⎜ ⎟ ⎜ ⎟ ⎢ dt dt ⎣ 4 ⎠⎥⎦ 4 ⎠ ⎝ ⎝ π ⎞ π ⎞ ⎛ = −(18.75 mA )sin⎛ ⎜ ω t + ⎟ = − (19 mA)sin ⎜ ω t + ⎟ 4 ⎠ 4 ⎠ ⎝ ⎝
I (t ) =
dQ
=
d ⎡
where ω = 1250 rad/s (b) Relate C to L and ω :
ω =
Solve for C to obtain:
C =
Substitute numerical values and evaluate C :
C =
1 LC
1 ω 2 L
1
(1250 s − ) (28 mH) 1 2
= 22.86 μ F
= 23 μ F (c) Express and evaluate the magnetic energy U m (t ) : U m (t ) = 12 LI
2
π ⎞ π ⎞ ⎛ = 12 (28 mH )(18.75 mA)2 sin 2 ⎛ ⎜ ω t + ⎟ = (4.9 μ J )sin 2 ⎜ ω t + ⎟ 4 ⎠ 4 ⎠ ⎝ ⎝
where ω = 1250 rad/s Use U e
=
1 2
Q2
to find the electrical
C energy stored in the capacitor as a function of time:
U e (t ) =
1 2
(15 μ F)2 22.86 μ F
⎛ ⎝
cos 2 ⎜ ω t +
π ⎞
⎟
4 ⎠
π ⎞ = (4.92 μ J )cos 2 ⎛ ⎜ ω t + ⎟ 4 ⎠ ⎝ π ⎞ = (4.9 μ J )cos 2 ⎛ ⎜ ω t + ⎟ 4 ⎠ ⎝
where ω = 1250 rad/s The total energy stored in the electric and magnetic fields is the sum of U m (t ) and U e (t ) :
⎛ ⎝
U = (4.92 μ J )sin 2 ⎜ ω t +
where ω = 1250 rad/s.
π ⎞
π ⎞ ⎛ ⎟ + (4.92 μ J )cos 2 ⎜ ω t + ⎟ = 4.9 μ J 4 ⎠ 4 ⎠ ⎝
Alternating-Current Circuits
2793
60 ••• One method for determining the compressibility of a dielectric material uses a driven LC circuit that has a parallel-plate capacitor. The dielectric is inserted between the plates and the change in resonance frequency is determined as the capacitor plates are subjected to a compressive stress. In one such arrangement, the resonance frequency is 120 MHz when a dielectric of thickness 0.100 cm and dielectric constant κ = 6.80 is placed between the plates. Under a compressive stress of 800 atm, the resonance frequency decreases to 116 MHz. Find the Young’s modulus of the dielectric material. (Assume that the dielectric constant does not change with pressure.) Picture the Problem We can use the definition of the capacitance of a dielectricfilled capacitor and the expression for the resonance frequency of an LC circuit to derive an expression for the fractional change in the thickness of the dielectric in terms of the resonance frequency and the frequency of the circuit when the dielectric is under compression. We can then use this expression for Δt /t to calculate the value of Young’s modulus for the dielectric material.
Use its definition to express Young’s modulus of the dielectric material: Letting t be the initial thickness of the dielectric, express the initial capacitance of the capacitor: Express the capacitance of the capacitor when it is under compression: Express the resonance frequency of the capacitor before the dielectric is compressed: When the dielectric is compressed:
Y =
stress strain
C 0
=
C c
=
ω 0
=
ω c
=
=
ΔP Δt t
(1)
κ ∈0 A t
κ ∈0 A t − Δt
1 C 0 L
1 C c L
1
=
=
κ ∈0 AL t
1 κ ∈0 AL t − Δt
Express the ratio of ω c to ω 0 and simplify to obtain:
κ ∈0 AL ω c ω 0
=
t κ ∈0 AL t − Δt
= 1−
Δt t
2794
Chapter 29
Expand the radical binomially to obtain:
Solve for Δt /t :
Δt t
Substitute in equation (1) to obtain:
12
Δt ⎞ Δt = ⎛ ⎜1 − ⎟ ≈ 1 − 2t ω 0 ⎝ t ⎠ provided Δt << t . ω c
⎛ ω ⎞ = 2⎜⎜1 − c ⎟⎟ ⎝ ω 0 ⎠ ΔP
Y =
⎛
2⎜⎜1 −
⎝
Substitute numerical values and evaluate Y :
Y =
ω c ⎞
⎟
ω 0 ⎠⎟
(800 atm )(101.325 kPa/atm) ⎛ 116 MHz ⎞ ⎟⎟ 120 MHz ⎝ ⎠
2⎜⎜1 −
= 1.22 ×109 N/m 2 61 ••• Figure 29-41 shows an inductor in series with a parallel plate capacitor. The capacitor has a width w of 20 cm and a gap of 2.0 mm. A dielectric that has a dielectric constant of 4.8 can be slid in and out of the gap. The inductor has an inductance of 2.0 mH. When half the dielectric is between the capacitor plates (when x = 12 w ), the resonant frequency of this combination is 90 MHz.
(a) What is the capacitance of the capacitor without the dielectric? (b) Find the resonance frequency as a function of x for 0 ≤ x ≤ w . Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel. Let C 1 be the capacitance of the dielectric-filled capacitor and C 2 be the capacitance of the air-filled capacitor. We’ll derive expressions for the capacitances of the parallel capacitors and add these expressions to obtain C ( x). We can then use the given resonance frequency when x = w/2 and the given value for L to evaluate C 0. In Part (b) we can use our result for C ( x) and the relationship between f , L, and C ( x) at resonance to express f ( x).
(a) Express the equivalent capacitance of the two capacitors in parallel:
C ( x ) = C 1 + C 2
Express A2 in terms of the total area of a capacitor plate A, w, and the distance x:
A2 A
=
x w
=
κ ∈0 A1
⇒ A2 = A
d
x w
+
∈0 A2 d
(1)
Alternating-Current Circuits Express A1 in terms of A and A2:
Substitute in equation (1) and simplify to obtain:
A1
x ⎞ = A − A2 = A⎛ ⎜1 − ⎟ ⎝ w ⎠
C ( x ) =
κ ∈0 A ⎛ x ⎞ ∈0 A x ⎜1 − ⎟ + d ⎝ w ⎠ d w
∈0 A ⎡ ⎛ x ⎞ x ⎤ κ ⎜1 − ⎟ + d ⎢⎣ ⎝ w ⎠ w ⎥⎦ κ − 1 ⎤ = κ C 0 ⎡⎢1 − x ⎣ κ w ⎥⎦ ∈ A where C 0 = 0 =
d
⎛ w ⎞ = κ C ⎡1 − κ − 1 w ⎤ ⎟ 0⎢ ⎝ 2 ⎠ ⎣ κ w 2 ⎥⎦ κ − 1⎤ = κ C 0 ⎡⎢1 − ⎣ 2κ ⎥⎦ κ + 1 = C 0
Find C (w/2):
C ⎜
2
Express the resonance frequency of the circuit in terms of L and C ( x):
f ( x ) =
1
⎛ w ⎞ = ⎟ ⎝ 2 ⎠ 2π
Evaluate f (w/2):
f ⎜
= Solve for C 0 to obtain:
C 0
=
(2)
2π LC ( x )
1 2π
1 LC 0
κ + 1
2
2 (κ + 1) LC 0
1 ⎛ w ⎞ 2π 2 f 2 ⎜ ⎟ L(κ + 1) ⎝ 2 ⎠
Substitute numerical values and evaluate C 0: C 0
=
1 = 5.4 fF 2 ⎛ 20 cm ⎞ 2 2π (90 MHz ) ⎜ ⎟ (2.0 mH)(4.8 + 1) ⎝ 2 ⎠
2795
2796
Chapter 29
(b) Substitute for C ( x) in equation (2) to obtain:
1
f ( x ) =
⎡ ⎣
2π Lκ C 0 ⎢1 −
κ − 1 ⎤ x κ w ⎥⎦
Substitute numerical values and evaluate f ( x): 1
f ( x ) =
2
70 MHz
=
⎡ ⎤ − (2.0 mH)(4.8) (5.39 × 10 F)⎢1 − 4.8 1 x ⎥ ⎣ 4.8(0.20 m ) ⎦
1 − (4.0 m −1 ) x
−16
Driven RLC Circuits A circuit consists of an ideal ac generator, a 20-μ F capacitor and an 62 • 80-Ω resistor, all connected in series. The output of the generator has a peak emf of 20-V, and the armature of the generator rotates at 400 rad/s. Find ( a) the power factor, (b) the rms current, and ( c) the average power supplied by the generator. Picture the Problem The diagram shows the relationship between δ , X L, X C, and R. We can use this reference triangle to express the power factor for the given circuit. In (b) we can find the rms current from the rms potential difference and the impedance of the circuit. We can find the average power delivered by the source from the rms current and the resistance of the resistor.
) C 2
2
Z
=
R
X
+ (
− X
L
X L − X C
δ
R
(a) The power factor is defined to be:
cos δ =
With no inductance in the circuit:
X L
R Z
=
R R 2
+ ( X L − X C )2
=0
and R
cos δ = R
2
+ X C 2
R
= R
2
+
1 ω C 2 2
Alternating-Current Circuits Substitute numerical values and evaluate cosδ :
80 Ω
=
cos
2797
1
(80 Ω )2 +
(400 s − ) (20 F) 1 2
2
= 0.54 (b) Express the rms current in the circuit:
I rms
=
ε rms Z
ε max 2
=
R 2 + X C 2
ε max
=
2 R 2
Substitute numerical values and evaluate I rms :
I rms
+
1 2
ω C 2
20 V
= 2
(80 Ω)2 +
1
(400 s − ) (20 μ F) 1 2
2
= 95.3 mA = 95 mA (c) The average power delivered by the generator is given by:
Pav
2 = I rms R
Substitute numerical values and evaluate Pav :
Pav
= (95.3 mA)2 (80 Ω ) = 0.73 W
63
••
[SSM]
Show that the expression Pav
2 = Rε rms
Z 2 gives the correct
result for a circuit containing only an ideal ac generator and (a) a resistor, (b) a 2 2 capacitor, and (c) an inductor. In the expression Pav = Rε rms Z , Pav is the average power supplied by the generator, ε rms is the root-mean-square of the emf of the generator, R is the resistance, C is the capacitance and L is the inductance. (In Part (a), C = L = 0, in Part (b), R = L = 0 and in Part (c), R = C = 0. Picture the Problem The impedance of an ac circuit is given by Z = R 2
+ ( X L − X C )2 . We can evaluate the given expression for Pav first for
X L = X C = 0 and then for R = 0.
(a) For X = 0, Z = R and:
Pav
=
2 Rε rms
Z 2
=
2 Rε rms
R 2
=
2 ε rms
R
2798
Chapter 29
(b) and (c) If R = 0, then:
=
Pav
Rε rms 2
=
Z 2
2 (0)ε rms
( X L − X C )2
= 0
Remarks: Recall that there is no energy dissipation in an ideal inductor or capacitor.
A series RLC circuit that has an inductance of 10 mH, a capacitance of 64 •• 2.0 μ F, and a resistance of 5.0 Ω is driven by an ideal ac voltage source that has a peak emf of 100 V. Find (a) the resonant frequency and (b) the root-mean-square current at resonance. When the frequency is 8000 rad/s, find (c) the capacitive and inductive reactances, (d ) the impedance, (e) the root-mean-square current, and ( f ) the phase angle. Picture the Problem We can use ω 0
the circuit, I rms
= ε rms
=1
LC to find the resonant frequency of
R to find the rms current at resonance, the definitions of X C
and X L to find these reactances at ω = 8000 rad/s, the definitions of Z and I rms to find the impedance and rms current at ω = 8000 rad/s, and the definition of the phase angle to find δ . (a) Express the resonant frequency ω 0 in terms of L and C :
ω 0
=
Substitute numerical values and evaluate ω 0:
ω 0
=
1 LC
1
(10 mH)(2.0 μ F)
= 7.1× 103 rad/s (b) Relate the rms current at resonance to ε rms and the impedance of the circuit at resonance:
(c) Express and evaluate X C and X L at ω = 8000 rad/s:
I rms
=
ε rms R
=
ε max 2 R
100 V
=
2 (5.0 Ω )
= 14 A X C =
1 ω C
=
1
(8000 s − )(2.0 μ F) 1
= 62.50 Ω = 63 Ω and X L
= ω L = (8000 s −1 )(10 mH) = 80 Ω
Alternating-Current Circuits (d ) Express the impedance in terms of the reactances, substitute the results from (c), and evaluate Z :
2799
+ ( X L − X C )2
Z = R 2
= (5.0 Ω)2 + (80 Ω − 62.5 Ω)2 = 18.2 Ω = 18 Ω
(e) Relate the rms current at
=
I rms
ω = 8000 rad/s to ε rms and the impedance of the circuit at this frequency:
ε rms
ε max
=
Z
2 Z
=
100 V 2 (18.2 Ω )
= 3.9 A ⎛ X L − X C ⎞ ⎟ ⎝ R ⎠
( f ) δ is given by:
δ = tan −1 ⎜
⎛ 80 Ω − 62.5 Ω ⎞ ⎟⎟ = 74° = tan −1 ⎜⎜ Ω 5 . 0 ⎝ ⎠
Substitute numerical values and evaluate δ :
65 •• [SSM] Find (a) the Q factor and (b) the resonance width (in hertz) for the circuit in Problem 64. (c) What is the power factor when ω = 8000 rad/s? Picture the Problem The Q factor of the circuit is given by Q =
resonance width by Δ f = f 0 Q =
0
L R , the
0
2π Q , and the power factor by cos δ = R Z .
Because Z is frequency dependent, we’ll need to find X C and X L at ω = 8000 rad/s in order to evaluate cosδ . Using their definitions, express the Q factor and the resonance width of the circuit:
Q
=
ω 0 L
(1)
R
and
Δ f = (a) Express the resonance frequency for the circuit:
ω 0
Substituting for ω 0 in equation (1) yields:
Q=
Substitute numerical values and evaluate Q:
Q=
=
f 0 Q
=
0
(2)
2π Q
1 LC
L LC R
=
1 L R
C
1 10 mH 5.0 Ω 2.0 μ F
= 14.1 = 14
2800
Chapter 29
(b) Substitute numerical values in equation (2) and evaluate Δ f :
Δ f
=
7.07 × 103 rad/s 2 (14.1)
= 80 Hz
(c) The power factor of the circuit is given by:
cos δ =
R Z
R
=
R 2
+ ( X L − X C )2
R
= R
2
1 ⎞ + ⎛ ⎜ ω L − ⎟ ω C ⎠ ⎝
2
Substitute numerical values and evaluate cosδ : 5.0 Ω
cos δ =
⎛
(5.0 Ω )2 + ⎜⎜ (8000 s −1 )(10 mH) − ⎝
⎞ ⎟ (8000 s −1 )(2.0 μ F) ⎠⎟ 1
2
= 0.27
FM radio stations typically operate at frequencies separated by 66 •• 0.20 MHz. Thus, when your radio is tuned to a station operating at a frequency of 100.1 MHz, the resonance width of the receiver circuit should be much smaller than 0.20 MHz, so that you do not receive a signal from stations operating at adjacent frequencies. Assume your receiving circuit has a resonance width of 0.050 MHz. When tuned in to this particular station, what is the Q factor of your circuit? Picture the Problem We can use its definition, Q = f 0
Δf to find the Q factor for
the circuit. The Q factor for the circuit is given by:
Q=
Substitute numerical values and evaluate Q:
Q=
f 0
Δ f 100.1 MHz 0.050 MHz
≈ 2.0 ×103
A coil is connected to a 60-Hz ac generator with a peak emf equal to 67 •• 100 V. At this frequency, the coil has an impedance of 10 Ω and a reactance of 8.0 Ω. (a) What is the peak current in the coil? (b) What is the phase angle between the current and the applied voltage? (c) A capacitor is put in series with the coil and the generator. What capacitance is required so that the current is in phase with the generator emf? (d ) What is the peak voltage measured across this capacitor?
Alternating-Current Circuits
2801
Picture the Problem We can use I peak = ε peak Z to find the current in the coil and
the definition of the phase angle to evaluate δ . We can equate X L and X C to find the capacitance required so that the current and the voltage are in phase. Finally, we can find the voltage measured across the capacitor by usingV C = IX C . (a) Express the current in the coil in terms of the potential difference across it and its impedance:
I peak =
ε peak Z
100 V 10 Ω
Substitute numerical values and evaluate I peak :
I peak =
(b) The phase angle δ is given by:
δ = cos
Substitute numerical values and evaluate δ : (c) Express the condition on the reactances that must be satisfied if the current and voltage are to be in phase: Substitute numerical values and evaluate C :
−1
= 10 A
⎛ R ⎞ = sin −1 ⎛ X L ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ Z ⎠ ⎝ Z ⎠
⎛ 8.0 Ω ⎞ ⎟⎟ = 53° = sin −1 ⎜⎜ 10 Ω ⎝ ⎠ X L
= X C =
C =
1 1 ⇒ C = ω C ω X L
1 2π (60 s
−1
)(8.0 Ω )
=
1 2π fX L
= 332 μ F
= 0.33 mF (d ) Express the potential difference across the capacitor: Relate the peak current in the circuit to the impedance of the circuit when X L = X C : Substitute for I to obtain:
Relate the impedance of the circuit to the resistance of the coil:
V C = I peak X C
I peak
V C =
=
V peak R
V peak X C R
=
Z = R 2 + X 2
V peak
2π fCR
⇒ R =
Z 2 − X 2
2802
Chapter 29
Substituting for R yields:
V C =
V peak
2π fC Z 2 − X 2
Substitute numerical values and evaluate V C: V C =
100 V 2π (60 s
−1
= 0.13 kV
)(332 μ F) (10 Ω) − (8.0 Ω) 2
2
An ideal 0.25-H inductor and a capacitor are connected in series with 68 •• an ideal 60-Hz generator. An digital voltmeter is used to measure the rms voltages across the inductor and capacitor independently. The voltmeter reading across the capacitor is 75 V and that across the inductor is 50 V. ( a) Find the capacitance and the rms current in the circuit. ( b) What is the rms voltage across the series combination of the capacitor and the inductor? Picture the Problem We can find C using V C = I rms X C and I rms from the
potential difference across the inductor. In the absence of resistance in the circuit, the measured rms voltage across both the capacitor and inductor is V = V L − V C . (a) Relate the capacitance C to the potential difference across the capacitor: Use the potential difference across the inductor to express the rms current in the circuit: Substitute for I rms to obtain:
Substitute numerical values and evaluate C :
V C = I rms X C =
I rms
C =
C =
=
V L X L
=
2π fC
⇒ C =
I rms
2π fV C
V L
2π fL
V L
(2π f )2 LV C 50 V
[2π (60 s − )] (0.25 H)(75 V ) 1
= 19 μ F (b) Express the measured rms voltage V across both the capacitor and the inductor when R = 0:
I rms
V = V L
− V C
2
Alternating-Current Circuits
2803
Substitute numerical values and V = 50 V − 75 V = 25 V evaluate V : 69 •• [SSM] In the circuit shown in Figure 29-42 the ideal generator produces an rms voltage of 115 V when operated at 60 Hz. What is the rms voltage between points (a) A and B, (b) B and C , (c) C and D, (d ) A and C , and (e) B and D? Picture the Problem We can find the rms current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge of the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations.
(a) Express the potential difference between points A and B in terms of I rms and X L: Express I rms in terms of ε and Z :
Evaluate X L and X C to obtain:
V AB
= I rms X L
I rms
=
X L
= 2π fL = 2π 60 s −1 (137 mH) = 51.648 Ω
ε Z
=
(1)
ε R 2
+ ( X L − X C )2
and X C =
1 1 = −1 2π fC 2π (60 s )(25 μ F)
= 106.10 Ω Substitute numerical values and evaluate I rms :
I rms
=
115 V
(50 Ω)2 + (51.648 Ω − 106.10 Ω)2
= 1.5556 A = (1.5556 A )(51.648 Ω ) = 80.344 V
Substitute numerical values in equation (1) and evaluate V AB:
V AB
(b) Express the potential difference between points B and C in terms of I rms and R:
V BC = I rms R = (1.5556 A )(50 Ω)
= 80 V
= 77.780 V = 78 V
2804
Chapter 29
(c) Express the potential difference between points C and D in terms of I rms and X C :
V CD
= I rms X C = (1.5556 A )(106.10 Ω ) = 165.05 V = 0.17 kV
(d ) The voltage across the inductor leads the voltage across the resistor as shown in the phasor diagram to the right:
V AC
V AB
V BC
Use the Pythagorean theorem to find V AC :
2 V AC = V AB
2 + V BC
= (80.0 V )2 + (77.780 V )2 = 111.58 V = 0.11 kV
(e) The voltage across the capacitor lags the voltage across the resistor as shown in the phasor diagram to the right:
V BC
V CD
Use the Pythagorean theorem to find V BD:
V BD
=
V BD
2 V CD
2 + V BC
= (165.05 V )2 + (77.780 V)2 = 182.46 V = 0.18 kV
When an RLC series circuit is connected to a 120-V rms, 60-Hz line, 70 •• the rms current in the circuit is 11 A and this current leads the line voltage by 45º. (a) Find the average power supplied to the circuit. (b) What is the resistance in the circuit? (c) If the inductance in the circuit is 50 mH, find the capacitance in the circuit. (d ) Without changing the inductance, by how much should you change the capacitance to make the power factor equal to 1? (e) Without changing the capacitance, by how much should you change the inductance to make the power factor equal to 1? Picture the Problem We can use Pav
= ε rms I rms cos δ to find the power supplied
2 R to find the resistance. In Part (c) we can relate the to the circuit and Pav = I rms
capacitive reactance to the impedance, inductive reactance, and resistance of the circuit and solve for the capacitance C. We can use the condition on X L and X C at
Alternating-Current Circuits
2805
resonance to find the capacitance or inductance you would need to add to the circuit to make the power factor equal to 1.
(a) Express the power supplied to
Pav
= ε rms I rms cos δ
Pav
= (120 V )(11A )cos 45° = 933 W
the circuit in terms of ε rms, I rms , and the power factor cosδ : Substitute numerical values and evaluate Pav :
(b) Relate the power dissipated in the circuit to the resistance of the resistor:
= 0.93 kW Pav
2 = I rms R ⇒ R =
933 W
Pav 2 I rms
Substitute numerical values and evaluate R:
R =
(c) Express the capacitance of the capacitor in terms of its reactance:
C =
Relate the capacitive reactance to the impedance, inductive reactance, and resistance of the circuit:
Z 2
= R 2 + ( X L − X C )2
Z 2
=
Express the impedance of the circuit in terms of the rms emf ε and the rms current I rms : Equating these two expressions yields: Solve for X L − X C :
Note that because I leads ε , the circuit is capacitive and X C > X L. Hence:
ε 2 2 rms
I
(11A )2 1 ω X C
=
= 7.71Ω = 7.7 Ω 1 2π fX C
ε 2 2 I rms
= R 2 + ( X L − X C )2
ε 2
X L − X C
=
X L − X C
= −( X L − X C )
and
2 rms
I
− R 2
(1)
2806
Chapter 29
ε 2
X C = X L +
2 rms
I
ε 2
= 2π fL + Substitute numerical values and evaluate X C:
− R 2
2 rms
I
(
X C = 2π 60 s
−1
− R2
)(50 mH)
(120 V )2 2 Ω) ( 7 . 71 + − (11A )2 = 18.8 Ω + 7.7 Ω = 26.6 Ω
Substitute in equation (1) and evaluate C :
1
C =
2π (60 s −1 )(26.6 Ω )
= 99.9 μ F
= 0.10 mF (d ) Let C pf = 1 represent the
ΔC
= C pf = 1 − C = C pf = 1 − 99.9 F
capacitance required to make cosδ = 1. The necessary change in capacitance is given by: Relate C pf = 1 to X L:
Solving for C pf = 1 yields:
Substitute for C pf = 1 in the expression for ΔC to obtain: Substitute numerical values and evaluate ΔC :
X L
= X C =
C pf = 1
ΔC
=
ΔC
=
=
1 2π fC pf = 1
1 2π fX L
1 2π fX L
− 99.9 μ F 1
2π (60 s
−1
)(18.8 Ω )
− 99.9 μ F
= 41 μ F (e) Let L pf = 1 represent the inductance required to make cosδ = 1. The necessary change in inductance is given by:
Δ L
= L pf = 1 − L = L pf = 1 − 50 mH
Alternating-Current Circuits Relate L pf = 1 to X C :
X L
Solving for L pf = 1 yields:
= X C = 2π fL pf = 1
L pf = 1
Substitute for L pf = 1 in the expression for Δ L to obtain: Substitute numerical values and evaluate Δ L:
2807
Δ L
=
Δ L
=
=
X C
2π f
X C
2π f
− 50 mH
26.6 Ω 2π (60 s −1 )
− 50 mH = 20 mH
Plot the circuit impedance versus the angular frequency for each of the 71 •• following circuits. (a) A driven series LR circuit, (b) a driven series RC circuit, and (c) a driven series RLC circuit. Picture the Problem The impedance for the three circuits as functions of the angular frequency is shown in the three figures below. Also shown in each figure (dashed line) is the asymptotic approach for large angular frequencies.
(a)
(b)
(c)
In a driven series RLC circuit, the ideal generator has a peak emf equal 72 •• to 200 V, the resistance is 60.0 Ω and the capacitance is 8.00 μ F. The inductance can be varied from 8.00 mH to 40.0 mH by the insertion of an iron core in the solenoid. The angular frequency of the generator is 2500 rad/s. If the capacitor voltage is not to exceed 150 V, find (a) the peak current and (b) the range of inductances that is safe to use. Picture the Problem We can find the maximum current in the circuit from the maximum voltage across the capacitor and the reactance of the capacitor. To find the range of inductance that is safe to use we can express Z 2 for the circuit in 2 2 terms of ε peak and I peak and solve the resulting quadratic equation for L.
(a) Express the peak current in terms of the maximum potential difference across the capacitor and its
I peak =
V C , peak X C
= ω CV C , peak
2808
Chapter 29
reactance: Substitute numerical values and evaluate I peak :
(b) Relate the maximum current in the circuit to the emf of the source and the impedance of the circuit: Express Z 2 in terms of R, X L, and X C: Substitute to obtain:
I peak = (2500 rad/s )(8.00 F)(150 V )
= 3.00 A I peak =
Z
⇒ Z = 2
2 ε peak 2 I peak
= R 2 + ( X L − X C )2
2
Z
2 ε peak 2 peak
I
Evaluating X C yields:
ε peak
X C
=
= R 2 + ( X L − X C )2 1 ω C
=
1
(2500 rad/s)(8.00 μ F)
= 50.0 Ω Substitute numerical values to obtain:
(200 V )2 = (60.0 Ω)2 + ((2500 rad/s)L − 50.0 Ω )2 2 (3.00 A ) Solving for L yields:
Denoting the solutions as L+ and L−, find the values for the inductance: The ranges for L are:
L =
L+
50.0 Ω ± 844 Ω 2 2500 s −1
= 31.6 mH and L− = 8.38 mH
8.00 mH < L < 8.38 mH and
31.6 mH < L < 40.0 mH A certain electrical device draws an rms current of 10 A at an average 73 •• power of 720 W when connected to a 120-V rms, 60-Hz power line. (a) What is the impedance of the device? (b) What series combination of resistance and reactance would have the same impedance as this device? (c) If the current leads the emf, is the reactance inductive or capacitive?
Alternating-Current Circuits
2809
Picture the Problem We can find the impedance of the circuit from the applied 2 emf and the current drawn by the device. In Part (b) we can use Pav = I rms R to find R and the definition of the impedance of a series RLC circuit to find X = X L − X C C.
(a) Express the impedance of the device in terms of the current it draws and the emf provided by the power line:
Z =
Substitute numerical values and evaluate Z :
Z =
(b) Use the relationship between the average power supplied to the device and the rms current it draws to express R:
Pav
ε rms I rms
120 V 10 A
= 12 Ω
2 = I rms R ⇒ R =
720 W
Pav 2 I rms
Substitute numerical values and evaluate R:
R =
The impedance of a series RLC circuit is given by:
Z = R 2
+ ( X L − X C )2
or 2 2 Z = R
+ ( X L − X C )2
Solving for X L
− X C yields:
Substitute numerical values and evaluate X :
(10 A )2
= 7.20 Ω = 7.2 Ω
X = X L − X C = Z 2 − R 2 X =
(12 Ω )2 − (7.20 Ω )2 =
10 Ω
(c) If the current leads the emf, the reactance is capacitive. A method for measuring inductance is to connect the inductor in series 74 •• with a known capacitance, a known resistance, an ac ammeter, and a variablefrequency signal generator. The frequency of the signal generator is varied and the emf is kept constant until the current is maximum. ( a) If the capacitance is 10 μ F, F, the peak emf is 10 V, the resistance is 100 Ω, and the rms current in the circuit is maximum when the driving frequency is 5000 rad/s, what is the value of the inductance? (b) What is the maximum rms current? Picture the Problem (a) We can use the fact that when the current is a maximum,
2810
Chapter 29
X L = X C rms, max from C , to find the inductance of the circuit. In Part (b), we can find I rms,
ε peak and the impedance of the circuit at resonance. (a) Relate X L and X C C at resonance:
X L
= X C or ω 0 L =
Solve for L L to obtain:
L =
1 ω 02C
Substitute numerical values and evaluate L:
L =
(b) Noting that, at resonance, X = 0, express I rms, rms, max in terms of the applied emf and the impedance of the circuit at resonance: 75
1 ω 0C
1
(5000 s ) (10 μ F)
I rms, max
−1 2
=
ε rms
=
Z
=
= 4.0 mH
ε max 2 Z
10 V 2 (100 Ω )
= 71 mA
A resistor and a capacitor are connected in parallel across an ac
••
generator (Figure 29-43) that has an emf given by ε
=
ε peak cos ω t.t. (a) Show that
the current in the resistor is given by I R = (ε peak / R R) cos ω t t . (b) Show that the current in the capacitor branch is given by I C = (ε peak / X X C t + 90º). (c) Show C ) cos(ω t that the total current is given by I = I peak cos(ω t ), where tan δ = R/ X t + δ ), X C C and I peak = ε peak / Z Z . Picture the Problem Because the resistor and the capacitor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through through the resistor and the current through the capacitor. Because these two currents are not in phase, we use phasors to calculate their sum. The amplitudes of the applied voltage and the currents are equal to the magnitude of the
r
phasors. That is
ε
= ε peak ,
r
r
I = I peak , I R
(a) The ac source applies a voltage given by ε ε peak cos t . Thus, the
=
r
= I R , peak , and I C = I C , peak .
ε peak cos
t = I R R
voltage drop across both the load resistor and the capacitor is: The current in the resistor is in phase with the applied voltage:
I R = I R , peak cos t
Alternating-Current Circuits Because I R , peak =
ε peak
:
R
I R
(b) The current in the capacitor leads the applied voltage by 90°: Because I C , peak =
ε peak X C
:
Draw the phasor diagram for the circuit. The projections projections of the phasors phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, and the current in the capacitor leads the the applied voltage by 90°. The net current phasor is the sum of the r r r branch current phasors I = I C + I R .
(
cos ω t
ε peak
=
X C
cos(ω t + 90°)
I = I R + I C
I
ε
I C
max
I R δ
t
ω
)
The peak current through the parallel combination is equal to ε peak Z , where Z is the impedance of the combination:
From the phasor diagram we have:
R
I C = I C, peak cos( t + 90°)
I C
(c) The net current I is the sum of the currents through the parallel branches:
ε peak
=
2811
I = I peak cos(ω t − δ ) ,
where I peak =
ε peak Z
2 = I R2, peak + I C 2 , peak I peak 2
2
⎛ ε ⎞ ⎛ ε ⎞ = ⎜⎜ peak ⎟⎟ + ⎜⎜ peak ⎟⎟ ⎝ R ⎠ ⎝ X C ⎠ 2 1 ⎞ ε peak 2 ⎛ 1 = ε peak ⎜⎜ 2 + 2 ⎟⎟ = 2 ⎝ R X C ⎠ Z where
Solving for I I peak yields:
From the phasor diagram:
1 Z 2
I peak =
=
1 R 2
ε peak Z
+
1 X C 2
where Z −2 = R −2 + X C −2
I = I peak cos(ω t + δ )
where
2812
Chapter 29
ε peak tan δ =
I C I R
=
X C
ε peak
=
R X C
R
76 ••• Figure 29-44 shows a plot of average power Pav versus generator frequency ω for a series RLC circuit driven by an ac generator. The average power Pav is given by Equation 29-56. The full width at half-maximum, Δω , is the width of the resonance curve between the two points, where Pav is one-half its maximum value. Show that for a sharply peaked resonance, Δω ≈ R/ L and that Q ≈ ω 0/Δω in this case (Equation 29-58). Hint: The half-power points occur when the denominator of Equation 29-56 is equal to twice the value it has at resonance; that is, when L2 ω 2 − ω 02 + ω 2 R 2 ≈ 2ω 02 R 2 . Let ω 1 and ω 2 be the solutions of this
(
)
equation. Then, show that Δω = ω 2 – ω 1 ≈ R/L. Picture the Problem We can use the condition determining the half-power points to obtain a quadratic equation that we can solve for the frequencies corresponding to the half-power points. Letting ω 1 be the half-power frequency that is less than ω 0 and ω 2 be the half-power frequency that is greater than ω 0 will lead us to the result that Δω = ω 2 −ω 1 ≈ R/ L. We can then use the definition of Q to complete the proof that Q ≈ ω 0 /Δω .
Equation 29-56 is:
Pav
(
=
2 2 V app, rms Rω
(
− ω 02 ) + ω 2 R 2 2
L2 ω 2
)
2
The half-power points occur when the denominator of Equation 29-56 is twice the value near resonance; that is, when:
2 2 2 L ω − ω 0
For a sharply peaked resonance, + 0 ≈ 2 0 . Hence:
L [(ω − ω 0 )(2ω 0 )]
Let ω 1 be a solution to this equation. Noting that, for a sharply peaked resonance, 1 ≈ 0 , it follows that:
+ ω 2 R 2 ≈ 2ω 02 R 2
or 2 L2 [(ω − ω 0 )(ω + ω 0 )] + ω 2 R 2
2
2
+ ω 2 R 2 ≈ 2ω 02 R 2
or 2 4ω 02 L2 (ω − ω 0 ) + ω 2 R 2 2
4ω 02 L2 (ω 1 − ω 0 )
or, simplifying, R 2 2 (ω 1 − ω 0 ) ≈ 2 4 L
≈ 2ω 02 R 2
≈ 2ω 02 R 2
+ ω 02 R 2 ≈ 2ω 02 R 2
Alternating-Current Circuits Solving for ω 1 yields:
ω 1
≈ ω 0 −
R
2 L where we’ve used the minus sign because ω 1 < ω 0.
2813
2814
Chapter 29
Similarly for ω 2:
R
≈ ω 0 +
ω 2
2 L where we’ve used the plus sign because ω 2 > ω 0.
Evaluating
Δ =
2
−
1
yields:
Δω ≈ ω 0 + R
From the definition of Q:
=
L
Substitute in the expression for Δ to obtain:
R ⎞ − ⎛ ⎜ ω 0 − ⎟ = 2 L ⎝ 2 L ⎠ R
R L
0
Q
Δω ≈
0
Q
⇒Q ≈
ω 0
Δω
2
d Q dQ 1 77 ••• Show by direct substitution that L 2 + R + Q =0 dt dt C (Equation 29-43b) is satisfied by Q = Q0e−t τ cos ω ' t , whereτ = 2 L R ,
ω ' = 1 (LC ) − 1 τ 2 , and Q0 is the charge on the capacitor at t = 0. Picture the Problem We’ll differentiate Q = Q0e −
t τ
cos ω ' t twice and substitute
this function and both its derivatives in the differential equation of the circuit. Rewriting the resulting equation in the form Acosω ′t + Bsinω ′t = 0 will reveal that B vanishes. Requiring that Acosω ′t = 0 hold for all values of t will lead to the result that ω ' = 1 (LC ) − 1 τ 2 . Equation 29-43b is:
d 2 Q
L
dt 2
+ R
Q = Q0 e −
Assume a solution of the form:
t τ
dQ dt
+
1
Q=0 C
cos ω ' t
Differentiate Q(t ) twice to obtain: dQ dt
= Q0
d
[e−
dt
t τ
]
cos ω ' t
⎞ = Q0 ⎛ ⎜ e −t τ cos ω ' t + cos ω ' t e −t τ ⎟ dt ⎝ dt ⎠ 1 ⎞ = Q0e −t τ ⎛ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟ τ ⎝ ⎠ d
d
Alternating-Current Circuits
2815
and d 2Q dt 2
⎛ − ω ' sin ω ' t − 1 cos ω ' t ⎞⎤ ⎜ ⎟⎥ dt ⎣ τ ⎝ ⎠⎦ ⎡ 1 2ω ' ⎞ ⎤ sin ω 't ⎥ = Q0e−t τ ⎢⎛ ⎜ 2 − ω ' 2 ⎟ cos ω 't + τ ⎠ ⎦ ⎣⎝ τ = Q0
d ⎡
⎢e
− t τ
Substitute these derivatives in the differential equation and simplify to obtain: ⎡⎛ 1 2ω ' 1 ⎞ ⎤ ⎛ ⎞ sin ω 't ⎥ + RQ0e−t τ ⎜ − ω ' sin ω ' t − cos ω ' t ⎟ LQ0e−t τ ⎢⎜ 2 − ω ' 2 ⎟ cos ω 't + τ τ ⎠ ⎦ ⎝ ⎠ ⎣⎝ τ
+
1
Q0e − C
t τ
cos ω ' t = 0
Because Q0 and e −t τ are never zero, divide them out of the equation and simplify to obtain:
⎛ 1 − ω ' 2 ⎞ cos ω 't + 2 Lω ' sin ω 't − ω ' R sin ω ' t − R cos ω ' t + 1 cos ω ' t = 0 ⎟ τ τ C ⎝ τ 2 ⎠
L⎜
Rewriting this equation in the form Acosω ′t + Bsinω ′t = 0 yields:
⎡ 1 ⎞ 1 R ⎤ ( Rω ' − Rω ' ) sin ω 't + ⎢ L⎛ ⎜ 2 − ω ' 2 ⎟ + − ⎥ cos ω 't = 0 ⎠ C τ ⎦ ⎣ ⎝ τ or
⎡ ⎛ 1 1 R ⎤ 2 ⎞ ⎢ L⎜ τ 2 − ω ' ⎟ + C − τ ⎥ cos ω 't = 0 ⎠ ⎣ ⎝ ⎦ If this equation is to hold for all values of t, its coefficient must vanish:
⎛ 1 − 2 ⎞ + 1 − R = 0 ω ' ⎟ ⎝ τ 2 ⎠ C τ
L⎜
Solving for ω ′ yields:
ω ' =
1 LC
1 ⎞ − ⎛ ⎜ ⎟ ⎝ 2 L ⎠
2
,
the condition that must be satisfied if t Q = Q0e− τ cos ω 't is the solution to Equation 29-43b. 78 ••• One method for measuring the magnetic susceptibility of a sample uses an LC circuit consisting of an air-core solenoid and a capacitor. The resonant frequency of the circuit without the sample is determined and then measured
2816
Chapter 29
again with the sample inserted in the solenoid. Suppose you have a solenoid that is 4.00 cm long, 3.00 mm in diameter, and has 400 turns of fine wire. You have a sample that is inserted in the solenoid and completely fills the air space. Neglect end effects. (a) Calculate the inductance of your empty solenoid. (b) What value for the capacitance of the capacitor should you choose that the resonance frequency of the circuit without a sample is exactly 6.0000 MHz? (c) When a sample is inserted in the solenoid, you determine that the resonance frequency drops to 5.9989 MHz. Use your data to determine the sample’s susceptibility. Picture the Problem We can use L = μ 0 n 2 Al to determine the inductance of the
empty solenoid and the resonance condition to find the capacitance of the samplefree circuit when the resonance frequency of the circuit is 6.0000 MHz. By expressing L as a function of f 0 and then evaluating df 0/dL and approximating the derivative with Δ f 0/Δ L , we can evaluate χ from its definition. (a) Express the inductance of an aircore solenoid:
L = μ 0 n 2 Al
Substitute numerical values and evaluate L: 2
⎛ 400 ⎞ π ⎟⎟ (3.00 cm )2 (4.00 cm ) = 3.553 mH = 3.55 mH L = (4π × 10 N/A )⎜⎜ ⎝ 4.00 cm ⎠ 4 −7
2
(b) Express the condition for resonance in the LC circuit:
X L
Solving for C yields:
C =
Substitute numerical values and evaluate C :
= X C ⇒ 2π f 0 L =
C =
1 2π f 0C
(1)
1 4π 2 f 02 L 1 4π 2 (6.0000 MHz) (3.553 mH) 2
= 1.9803 ×10 −13 F = 0.198 pF (c) Express the sample’s susceptibility in terms of L and Δ L: Solve equation (1) for f 0:
χ =
f 0
=
Δ L
(2)
L
1 2π LC
Alternating-Current Circuits Differentiate f 0 with respect to L:
df 0 dL
=
1
d
2π C dL 1
=−
−
L 1 2
4π L LC
=−
=−
−
L 3 2 4π C
f 0
2 L
Approximate df 0/dL by Δ f 0/Δ L:
Δ f Δ L Δ f 0 f = − 0 or 0 = − f 0 2 L Δ L 2 L
Substitute in equation (2) to obtain:
χ = −2
Substitute numerical values and evaluate χ :
1
2817
Δ f 0 f 0
⎛ 5.9989 MHz − 6.0000 MHz ⎞ ⎟⎟ 6.0000 MHz 0 ⎝ ⎠ = 3.7 × 10− 4
χ = −2⎜⎜
The Transformer 79 • [SSM] A rms voltage of 24 V is required for a device whose impedance is 12 Ω. (a) What should the turns ratio of a transformer be, so that the device can be operated from a 120-V line? (b) Suppose the transformer is accidentally connected in reverse with the secondary winding across the 120-V-rms line and the 12-Ω load across the primary. How much rms current will then be in the primary winding? Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use V 2 N 1 = V 1 N 2 and N 1 I 1 = N 2 I 2 to find the turns ratio and
the primary current when the transformer connections are reversed. (a) Relate the number of primary and secondary turns to the primary and secondary voltages:
V 2, rms N 1 = V 1, rms N 2
Solve for and evaluate the ratio N 2/ N 1:
N 2
(b) Relate the current in the primary to the current in the secondary and to the turns ratio:
N 1
=
I 1, rms
V 2, rms V 1, rms
=
N 2 N 1
=
24 V 120 V
I 2, rms
(1)
=
1 5
2818
Chapter 29
Express the current in the primary winding in terms of the voltage across it and its impedance: Substitute for I 2, rms to obtain:
Substitute numerical values and evaluate I 1, rms:
I 2, rms
=
I 1, rms
=
I 1
V 2, rms Z 2
N 2 V 2, rms N 1 Z 2
5 ⎞⎛ 120 V ⎞ ⎟ = 50 A = ⎛ ⎜ ⎟⎜⎜ ⎝ 1 ⎠⎝ 12 Ω ⎠⎟
A transformer has 400 turns in the primary and 8 turns in the 80 • secondary. (a) Is this a step-up or a step-down transformer? ( b) If the primary is connected to a 120 V rms voltage source, what is the open-circuit rms voltage across the secondary? (c) If the primary rms current is 0.100 A, what is the secondary rms current, assuming negligible magnetization current and no power loss? Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can decide whether the transformer is a step-up or step-down transformer by examining the ratio of the number of turns in the secondary to the number of terms in the primary. We can relate the open-circuit rms voltage in the secondary to the primary rms voltage and the turns ratio.
(a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer. (b) Relate the open-circuit rms voltage V 2, rms in the secondary to the rms voltage V 1, rms in the primary: Substitute numerical values and evaluate V 2, rms : (c) Because there are no power losses:
Substitute numerical values and evaluate I 2, rms:
V 2, rms
=
N 2
V 2, rms
=
8
N 1
V 1, rms
400
V 1, rms I 1, rms
(120 V ) =
2.40 V
= V 2, rms I 2, rms
and I 2, rms
=
I 2, rms
=
V 1, rms V 2, rms
I 1, rms
120 V 2.40 V
(0.100 A ) =
5.00 A
Alternating-Current Circuits
2819
The primary of a step-down transformer has 250 turns and is 81 • connected to a 120-V rms line. The secondary is to supply 20 A rms at 9.0 V rms. Find (a) the rms current in the primary and ( b) the number of turns in the secondary, assuming 100 percent efficiency. Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use I 1, rmsV 1, rms = I 2, rmsV 2, rms to find the current in the primary
and V 2, rms N 1
= V 1, rms N 2 to find the number of turns in the secondary.
(a) Because we have 100 percent efficiency:
I 1, rmsV 1, rms = I 2, rmsV 2, rms
and I 1, rms = I 2, rms
Substitute numerical values and evaluate I 1, rms: (b) Relate the number of primary and secondary turns to the primary and secondary voltages: Substitute numerical values and evaluate N 2/ N 1:
I 1, rms
=
V 1, rms
= (20 A )
V 2, rms N 1
N 2
V 2, rms
9.0 V 120 V
= 1.5 A
= V 1, rms N 2 ⇒
N 2
=
V 2, rms V 1, rms
N 1
9.0 V (250) ≈ 19 120 V
An audio oscillator (ac source) that has an internal resistance of 82 •• 2000 Ω and an open-circuit rms output voltage of 12.0 V is to be used to drive a loudspeaker coil that has a resistance of 8.00 Ω. (a) What should be the ratio of primary to secondary turns of a transformer, so that maximum average power is transferred to the speaker? (b) Suppose a second identical speaker is connected in parallel with the first speaker. How much average power is then supplied to the two speakers combined? Picture the Problem Note: In a simple circuit maximum power transfer from source to load requires that the load resistance equals the internal resistance of the source. We can use Ohm’s law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the speaker(s).
(a) Express the effective loudspeaker resistance at the primary of the transformer:
Reff
=
V 1, rms I 1, rms
2820
Chapter 29
Relate V 1, rms to V 2, rms, N 1, and N 2:
Express I 1, rms in terms of I 2, rms, N 1, and N 2: Substitute for V 1, rms and I 1, rms and simplify to obtain:
= V 2, rms
V 1, rms
I 1, rms = I 2, rms
V 2, rms Reff = I 2, rms
Solve for N 1/ N 2:
N 1
Evaluate N 1/ N 2 for Reff = Rcoil:
N 1
N 2 N 1 N 1
N 2 N 2 N 1
V 2, rms
2000 Ω 8.00 Ω
=
N 2
N 2
I 2, rms Reff
=
N 2
N 1
(b) Express the power delivered to the two speakers connected in parallel:
Psp = I 12, rms Reff
Find the equivalent resistance Rsp of
1 Rsp
the two 8.00-Ω speakers in parallel:
=
=
Reff
(1)
R2
= 15.811 = 15.8 (2)
1 1 ⇒ Rsp + 8.00 Ω 8.00 Ω
Solve equation (1) for Reff to obtain:
⎛ N ⎞ Reff = R2 ⎜⎜ 1 ⎟⎟ ⎝ N 2 ⎠
Substitute numerical values and evaluate Reff :
Reff
Find the current supplied by the source:
⎛ V 2, rms ⎞ ⎛ N 1 ⎞ 2 ⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟ ⎝ I 2, rms ⎠ ⎝ N 2 ⎠
= 4.00 Ω
2
= (4.00 Ω )(15.811)2 = 1000 Ω
I 1, rms
=
V rms Rtot
=
12.0 V 2000 Ω + 1000 Ω
= 4.00 mA Substitute numerical values in equation (2) and evaluate the power delivered to the parallel speakers:
Psp
= (4.00 mA)2 (1000 Ω) = 16.0 mW
Alternating-Current Circuits
2821
General Problems The distribution circuit of a residential power line is operated at 83 • 2000 V rms. This voltage must be reduced to 240 V rms for use within residences. If the secondary side of the transformer has 400 turns, how many turns are in the primary? Picture the Problem We can relate the input and output voltages to the number of turns in the primary and secondary usingV 2, rms N 1 = V 1, rms N 2 .
Relate the output voltages V 2, rms to the input voltage V 1, rms and the number of turns in the primary N 1 and secondary N 2:
V 2, rms
Substitute numerical values and evaluate N 1:
N 1
=
N 2 N 1
V 1, rms ⇒ N 1 = N 2
V 1, rms V 2, rms
⎛ 2000 V ⎞ ⎟⎟ = 3.33 ×103 = (400)⎜⎜ ⎝ 240 V ⎠
A resistor that has a resistance R carries a current given by 84 •• (5.0 A) sin 2π ft + (7.0 A) sin 4π ft, where f = 60 Hz. (a) What is the rms current in the resistor? (b) If R =12 Ω, what is the average power delivered to the resistor? (c) What is the rms voltage across the resistor?
(I ) to relate the rms current to the current carried by the resistor and find ( I ) by integrating I 2. Picture the Problem We can use its definition, I rms
=
2
av
2
av
(a) Express the rms current in terms of the ( I 2 )av :
I rms
=
(I ) 2
av
Evaluate I 2: I 2
= [(5.0 A ) sin 2π ft + (7.0 A ) sin 4π ft ] 2 = (25 A 2 )sin 2 2π ft + (70 A 2 )sin 2π ft sin 4π ft + (49 A 2 )sin 2 4π ft
Find ( I 2 )av by integrating I 2 from t = 0 to t = T = 2π /ω and dividing by T :
( I ) 2
av
=
ω
2π
2π ω
∫ {(25 A )sin 2
2
2π ft + (70 A 2 )sin 2π ft sin 4π ft
0
+ (49 A 2 )sin 2 4π ft }dt
2822
Chapter 29
Use the trigonometric identity sin 2 x = 12 (1 − cos 2 x ) to simplify and
( I ) 2
av
= 12.5 A 2 + 0 + 24.5 A 2 = 37.0 A 2
evaluate the 1st and 3rd integrals and recognize that the middle term is of the form sin xsin2 x to obtain: Substitute for ( I 2 )av and evaluate I rms:
I rms
= 37.0 A 2 = 6.1 A
(b) Relate the power dissipated in the resistor to its resistance and the rms current in it:
2 P = I rms R
Substitute numerical values and evaluate P:
P = (6.08 A ) (12 Ω) = 0.44 kW
(c) Express the rms voltage across the resistor in terms of R and I rms :
V rms = I rms R
Substitute numerical values and evaluate V rms :
V rms
2
= (6.08 A )(12 Ω) = 73 V
85 •• [SSM] Figure 29-45 shows the voltage versus time for a squarewave voltage source. If V 0 = 12 V, (a) what is the rms voltage of this source? (b) If this alternating waveform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what is the new rms voltage? Picture the Problem The average of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT . We can use this definition to find both the average of the voltage squared, V 2 av and then use the definition
( )
of the rms voltage. (a) From the definition of V rms we
V rms
=
(V )
V rms
=
V 02
2 0 av
have: Noting that V rms :
− V 02 = V 02 , evaluate
= V 0 = 12 V
Alternating-Current Circuits
(b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval 1 ΔT : 2
V = V 0
Express the square of the voltage during this half cycle:
V 2
Calculate (V 2 )av by integrating V 2 from t = 0 to t =
1 2
ΔT and dividing
= V 02
(V ) 2
av
=
by ΔT : Substitute to obtain:
2823
V rms
=
1 ΔT 2
V 02
∫
dt =
=
V 0
ΔT 0
1 2
V 02
V 02 ΔT
=
2
1 Δ 2
[t ] 0 T = 12 V 02
12 V 2
= 8.5 V
What are the average values and rms values of current for the two 86 •• current waveforms shown in Figure 29-46? Picture the Problem The average of any quantity over a time interval ΔT is the integral of the quantity over the interval divided by ΔT . We can use this definition to find both the average current I av , and the average of the current squared,
( I ) 2
av
From the definition of I av and I rms we have: Waveform (a) Express the current during the first half cycle of time interval ΔT :
Evaluate I av, a :
I av
=
1 ΔT
∫
ΔT
Idt and I rms
0
=
(I ) 2
av
4A t ΔT where I is in A when t and T are in seconds.
I a
=
I av, a
=
1
ΔT
∫
ΔT 0
4.0 A ΔT
tdt =
4.0 A
(ΔT )2
ΔT
=
4.0 A ⎡ t 2 ⎤ (ΔT )2 ⎢⎣ 2 ⎥⎦ 0
= 2.0 A
ΔT
∫ tdt 0
2824
Chapter 29
Express the square of the current during this half cycle: Noting that the average value of the squared current is the same for each time interval ΔT , calculate ( I a2 )av by
(4.0 A)2 2 I = t (ΔT )2 T 1 (4.0 A )2 2 2 t dt ( I a )av = ∫ 2 ΔT 0 (ΔT ) 2 a
Δ
ΔT
(4.0 A )2 ⎡ t 3 ⎤ 16 = = A2 ⎥ 3 ⎢ 3 (ΔT ) ⎣ 3 ⎦ 0
integrating I a2 from t = 0 to t = ΔT and dividing by ΔT : Substitute in the expression for I rms, a to obtain: Waveform (b) Noting that the current during the second half of each cycle is zero, express the current during the first half cycle of the time interval 12 ΔT :
=
I rms, a
I b
16 2 A 3
= 2.3 A
= 4.0 A
Evaluate I av, b : I av, b
=
4.0 A
1 ΔT 2
ΔT
∫
dt =
0
4.0 A ΔT
1 ΔT 2
[t ] 0
= 2.0 A Express the square of the current during this half cycle: Calculate ( I b2 )av by integrating I b2 from t = 0 to t =
1 2
ΔT and dividing
2
I b
= (4.0 A )2
( I )
2 b av
by ΔT :
= =
Substitute in the expression for I rms, b
I rms, b
(4.0 A )2 ΔT
(4.0 A )2 ΔT
1 Δ 2
∫ dt 0
1 Δ 2
[t ] 0 T = 8.0 A 2
= 8.0 A 2 = 2.8 A
to obtain: 87
••
In the circuit shown in Figure 29-47, ε 1 = (20 V) cos 2π ft , where
f = 180 Hz; ε 2 = 18 V, and R = 36 Ω. Find the maximum, minimum, average, and rms values of the current in the resistor. Picture the Problem We can apply Kirchhoff’s loop rule to express the current in the circuit in terms of the emfs of the sources and the resistance of the resistor.
Alternating-Current Circuits
2825
We can then find I max and I min by considering the conditions under which the time-dependent factor in I will be a maximum or a minimum. Finally, we can use
=
I rms
(I ) 2
av
to derive an expression for I rms that we can use to determine its
value. Apply Kirchhoff’s loop rule to obtain: Solving for I yields:
ε 1, peak cos
t + ε 2
ε 1, peak
I =
R
− IR = 0
cos ω t +
ε 2 R
or I = A1 cos t + A2 where A1 Substitute numerical values to obtain:
ε 1, peak
=
R
and A2
=
ε 2 R
⎛ 20 V ⎞ 18 V ⎟⎟ cos(2π (180 s −1 )t ) + 36 Ω ⎝ 36 Ω ⎠ = (0.556 A ) cos(1131s −1 )t + 0.50 A
I = ⎜⎜
The current is a maximum when cos 1131s −1 t = 1 . Hence :
I max
= 0.50 A + 0.556 A = 1.06 A
Evaluate I min :
I min
= 0.50 A − 0.556 A = − 0.06 A
Because the average value of cosω t = 0:
I av
The rms current is the square root of the average of the squared current:
I rms
[ I ] 2
av
= 0.50 A
=
[I ] 2
av
is given by:
[ I ] 2
av
[
]
Because cos 2 ω t av
[cos t ]av = 0 :
= ( A1 cos ω t + A2 )2 av = [ A12 cos 2 ω t + 2 A1 A2 cos ω t + A22 ]av = [ A12 cos 2 ω t ]av + [2 A1 A2 cos ω t ]av + [ A22 ]av = A12 [cos 2 ω t ]av + 2 A1 A2 [cos ω t ]av + A22
= 12 and
[ I ] 2
av
= 12 A12 + A22
(1)
2826
Chapter 29
Substituting in equation (1) yields:
I rms
=
1 2
A1
I rms
=
1 2
⎛ ε 1 ⎞ + ⎛ ε 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ R ⎠ ⎝ R ⎠
Substitute for A1 and A2 to obtain:
Substitute numerical values and evaluate I rms:
88
••
2
+ A22 2
2
2
I rms
20 V ⎞ ⎛ 18 V ⎞ = 12 ⎛ ⎜ ⎟ +⎜ ⎟ ⎝ 36 Ω ⎠ ⎝ 36 Ω ⎠ = 0.64 A
2
Repeat Problem 87 if the resistor is replaced by a 2.0-μ F capacitor.
Picture the Problem We can apply Kirchhoff’s loop rule to obtain an expression for charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find I max and I min by considering the conditions under which the time-dependent factor in I
will be a maximum or a minimum. Finally, we can use I rms
=
(I ) 2
av
expression for I rms that we can use to determine its value. q (t )
Apply Kirchhoff’s loop rule to obtain:
ε 1 peak cos ω t + ε 2 −
Solving for q(t ) yields:
q(t ) = C ε 1 peak cos t + ε 2
C
=0
= A1 cos ω t + A2 where A1 = C ε 1 peak and A2 Differentiate this expression with respect to t to obtain the current as a function of time:
I =
dq
=
= C ε 2
d
( A1 cos ω t + A2 ) dt dt = −ω A1 sin ω t
Substituting numerical values yields:
(
)
I = −2π (180 Hz)(2.0 μ F)sin(2π (180 Hz) t ) = (− 2.26 mA) sin 1131 s −1 t
The current is a minimum when sin(1131s −1 )t = 1 . Hence:
I min
= − 2.3 mA
The current is a maximum when sin 1131s −1 t = −1 . Hence:
I max
= 2.3 mA
to derive an
Alternating-Current Circuits
2827
2828
Chapter 29
= 0
Because the dc source sees the capacitor as an open circuit and the average value of the sine function over a period is zero:
I av
The rms current is the square root of the average of the squared current:
I rms
I 2
av
=
[I ] 2
(1)
av
is given by:
[ I ] 2
av
Because cos 2 ω t av
= ( A1 cos ω t + A2 )2 av = [ A12 cos 2 ω t + 2 A1 A2 cos ω t + A22 ]av = [ A12 cos 2 ω t ]av + [2 A1 A2 cos ω t ]av + [ A22 ]av = A12 [cos 2 ω t ]av + 2 A1 A2 [cos ω t ]av + A22
= 12 and
I 2
av
= 12 A12 + A22
[cos t ]av = 0 : Substituting in equation (1) yields: Substitute for A1 and A2 to obtain:
I rms
=
1 2
A1
I rms
=
1 2
(C ε 1 )2 + (C ε 2 )2
2
+ A22
= C 12 (ε 1 )2 + (ε 2 )2 Substitute numerical values and evaluate I rms:
I rms
= (2.0 μ F)
1 2
(20 V )2 + (18 V)2
= 46 μ A
89 ••• [SSM] A circuit consists of an ac generator, a capacitor and an ideal inductor ⎯ all connected in series. The emf of the generator is given by ε peak cos t . (a) Show that the charge on the capacitor obeys the equation L
d 2 Q 2
dt
+
Q C
= ε peak cos ω t . (b) Show by direct substitution that this equation is
ε peak
satisfied by Q = Q peak cos t where Q peak = − . (c) Show that the current 2 2 L(ω − ω 0 ) can be written as I = I peak cos( t − δ ) , where I peak =
ε peak 2
L ω
− ω 02
=
ε peak X L
− X C
and
δ = –90º for ω < ω 0 and δ = 90º for ω > ω 0, where ω 0 is the resonance frequency.
Alternating-Current Circuits
2829
Picture the Problem In Part (a) we can apply Kirchhoff’s loop rule to obtain the 2nd order differential equation relating the charge on the capacitor to the time. In Part (b) we’ll assume a solution of the form Q = Q peak cos t , differentiate it twice,
and substitute for d 2Q/dt 2 and Q to show that the assumed solution satisfies the differential equation
ε peak
provided Q peak = − . In Part (c) we’ll use our L(ω 2 − ω 02 )
results from ( a) and (b) to establish the result for I peak given in the problem statement. (a) Apply Kirchhoff’s loop rule to obtain:
ε −
Substitute for ε and rearrange the differential equation to obtain:
L
Because I = dQ dt :
Q C
dI dt
− L
+
2
Q C
d Q
L
2
dt
+
dI dt
=0
= ε max cos ω t
Q C
= ε max cos ω t
(b) Assume that the solution is:
Q = Q peak cos t
Differentiate the assumed solution twice to obtain:
dQ dt
= −ω Q peak sin ω t
and d 2 Q 2
dt
Substitute for
dQ
and
d 2Q
in the
dt 2 differential equation to obtain: dt
Factor cosω t from the left-hand side of the equation:
If this equation is to hold for all values of t it must be true that:
= −ω 2Q peak cos ω t
− ω 2 LQ peak cos ω t +
Q peak
cos ω t C = ε peak cos ω t
Q ⎞ ⎛ 2 ⎜⎜ − ω LQ peak + peak ⎟⎟ cos ω t C ⎠ ⎝ = ε peak cos ω t
− ω 2 LQ peak +
Q peak C
= ε peak
