= VrmsΙrmscos φ =
Vrms R
2
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PURELY PURELY CAPA CAPA CITIVE CIRCUIT CIR CUIT: CUIT:
vs = Vm sin ωt
vs -
q =0 C
i C
1 and is called capacitive reactance. Its unit is ohm Ω. ωC From the graph of current versus time and voltage versus time , it is clear
XC =
page 5
Vm V d(Cv ) d(CVm sin ωt ) dq or i = = = =CVmωcos ωt = cos ωt = m cos ωt = Ιm cos ωt. 1 dt XC dt dt ωC
v V
T
T that current attains its peak value at a time before the time at which 4
t i
T voltage attains its peak value. Corresponding to the phase difference 4
I
t
2π T 2π π = ω∆t = = = . iC leads v C by π/2 Diagrammatically (phasor T 4 4 2
Ιm .
diagram) it is represented as Vm
Since φ =90º,
= Vrms Ιrmscos φ = 0
Ques. A capacitor acts as an infinite resistance for (A) DC (B) AC (C) DC as well as AC Ans. (A)
(D) neither AC nor DC
Ex. 10 An alternating voltage E = 200 2 sin (100 t) V is connected to a 1µF capacitor through an ac ammeter (it reads rms value). What will be the reading of the ammeter? Sol.
Comparing E = 200 2 sin (100 t) with E = E0 sin ωt we find that, E0 = 200 2 V and ω = 100 (rad/s)
1 1 = = 10 4 Ω ωC 100 × 10 −6 And as ac instruments reads rms value, the reading of ammeter will be, So,
XC =
Ι rms =
i.e.
Ι rms =
Erms = XC
200 2 2 × 10 4
E0 2 XC
E0 as Erms = 2
= 20mA
Ans
Ques. A 10 µF capacitor is connected with an ac source E = 200 rms value) . What will be the reading of the ammeter? Ans: 200 mA
2 sin (100 t) V through an ac ammeter (it reads
Ques. Find the reactance of a capacitor (C = 200 µF) when it is connected to (a) 10 Hz AC source, (b) a 50 Hz AC source and (c) a 500 Hz AC source. Ans. (a) 80 Ω for 10 Hz AC source, (b) 16 Ω for 50 Hz and (c) 1.6 Ω for 500 Hz.
9.
PURELY PURELY INDUCTIVE CIRCUIT CIR CUIT: CUIT:
v s = Vm sin ωt
Writing KVL along the circuit,
vs - L
di =0 dt
∫ Ldi = ∫ V
m
⇒ sin ωt dt
L
di =Vmsin ωt dt
Vm i=– cos ωt + C ωL
i
L
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Writing KVL along the circuit,
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com ∴ i=–
Vm cosωt ωL
⇒
C=0
⇒
Vm Ιm = X L
v
From the graph of current versus time and voltage versus time , it is clear
V
T
T before the time at which that voltage attains its peak value at a time 4
t
page 6
i T current attains its peak value. Corresponding to the phase difference = 4
I
t
2π T 2π π ω∆t = = = . Diagrammatically (phasor diagram) it is repreT 4 4 2 Vm
. i lags behind v L by π/2. Ιm L Since φ = 90º,
= VrmsIrmscos φ = 0 sented as
Summary :
φ
AC source connected with
Z
Pure Resistor
0
v R is in same phase with iR
Pure Inductor
π/2
v L leads iL
XL
Pure Capacitor
π/2
v C leads iC
XC
R
Phasor Diagram
Vm
Im
Vm
Ιm Ιm Vm
Ques. An inductor (L = 200 mH) is connected to an AC source of peak current. What is the intantaneous voltage of the source when the current is at its peak value? Ans. zero
10.
RC SERIES CIRCUIT WITH AN AC SOURCE : Let i = Ιm sin (ωt + φ)
v C= Ιm XCsin (ωt + φ − or
Vm =
π ) 2
⇒
v R=iR= ΙmR sin (ωt+φ)
⇒
v S=v R + v C
vS= Vm sinωt
C R
π Vm sin (ωt+φ)= ΙmR sin (ωt + φ) + Ιm XCsin (ωt + φ − ) 2
(Ι mR )2 + (Ι m XC )2 + 2(ΙmR)(Ιm XC) cos π
2
OR Ι m =
Vm 2
⇒
tan φ=
Ι m XC XC = . Ι mR R
Ιm
Z= R 2 + XC 2
R + XC Using phasor diagram also we can find the above result. 2
ΙmR
ΙmXC
φ
Vm
Ιm Vm
Ques. An AC source producing emf ξ = ξ0[cos(100 π s -1)t + cos(500 π s -1)t] is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to
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=0
Ans.
be i = i1 cos[(100 π s -1)t + ϕ1]+i2 cos[(500 π s -1)t+ ϕ1] (A) i1 > i2 (B) i1 = i2 (C) i1 < i2 (D) the information is insufficient to find the relation between i1 and i2 (C)
220V,50HZ
Ex. 11 In an RC series circuit, the rms voltage of source is 200V and its frequency 100 µF, find π Impedance of the circuit (ii) Power factor (iv) Maximum current (vi) voltage across C (viii) max voltage across C (x) < PR > (xii)
(i) (iii) (v) (vii) (ix) (xi)
Sol.
XC=
Power factor angle Current voltage across R max voltage across R
< PC >
C R
10 6 =100 Ω 100 ( 2π50) π
(i)
Z= R 2 + XC 2 = 100 2 + (100) 2 =100 2 Ω
(iii)
Power factor= cosφ =
(v)
Maximum current =Ιrms 2 =2Α
(vi)
voltage across R=VR,rms=ΙrmsR= 2 ×100 Volt
(vii)
voltage across C=VC,rms=ΙrmsXC= 2 ×100 Volt
(viii)
max voltage across R= 2 VR,rms= 200 Volt
(ix)
max voltage across C= 2 VC,rms= 200 Volt
(x)
=VrmsΙrmscosφ =200 × 2 ×
(xi)
tan φ=
(iv)
Current Ιrms=
(x)
1 2
1 2
XC =1 R
(ii)
∴ φ=45º
200 Vrms = = 2Α 100 2 Z
=200 Watt
Ex. 12 In the above question if v s(t) = 220 2 sin (2π 50 t), find (a) i (t), (b) v R and (c) v C (t) Sol.
(a)
i(t) = Ιm sin (ωt + φ)
=
2 sin (2π 50 t + 45º)
(b)
v R = iR . R
=
2 × 100 sin (100 πt + 45º)
= i(t) R
(c) v C (t) = iCXC (with a phase lag of 90º) = 2 ×100 sin (100 πt + 45 – 90) Ex. 13 An ac source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is Ι. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω. Sol. According to given problem, Ι=
V V = Z [R 2 + (1/ Cω)2 ]1/ 2
... (1)
Ι V = ...(2) 2 [R 2 + (3 / Cω)2 ]1/ 2 Substituting the value of Ι from Equation (1) in (2),
and,
1 9 1 3 2 4 R 2 + 2 2 = R 2 + 2 2 . i.e., R 2 2 = 5 C ω C ω C ω
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is 50 Hz.If R =100 Ω and C=
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Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com 3 5
Ans.
LR SERIES CIRCUIT WITH AN AC SOURCE :
vS= Vsinωt
ΙXL
V
V
φΙ
ΙR L
Ι
R
From the phasor diagram V=
(ΙR )2 + (ΙXL )2
tan φ =
2 2 = Ι (R ) + (XL ) =
ΙZ
ΙXL XL = ΙR R
9 H inductor and a 12 ohm resistance are connected in series to a 225 V, 50 Hz ac source. Calculate 100 π the current in the circuit and the phase angle between the current and the source voltage.
Ex. 14 A
Sol.
Here XL = ωL = 2π f L = 2π × 50 × So,
Z=
So (a) I =
R 2 + XL
2
=
9 =9Ω 100 π
12 2 + 9 2 = 15 Ω
225 V = = 15 A 15 Z
Ans
XL 9 and (b) φ = tan–1 R = tan–1 = tan–1 3/4 = 37º 12 i.e., the current will lag the applied voltage by 37º in phase. Ans
Ex. 15 When an inductor coil is connected to an ideal battery of emf 10 V, a constant current 2.5 A flows. When the same inductor coil is connected to an AC source of 10 V and 50 Hz then the current is 2A. Find out inductance of the coil . Sol. When the coil is connected to dc source, the final current is decided by the resistance of the coil . 10 = 4 Ω 2.5 When the coil is connected to ac source, the final current is decided by the impedance of the coil .
∴
r=
∴
Z=
But ∴ ∴
10 = 5Ω 2
Z= (r ) + (XL ) XL= 3 Ω ωL = 2 π fL = 3 L = 3/100π Henry 2
2
XL2 = 52– 42 = 9 ∴
2 π 50 L = 3
Ex. 16 A bulb is rated at 100 V,100 W , it can be treated as a resistor .Find out the inductance of an inductor (called choke coil ) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 Hz. 2
Sol:
page 8
11.
=
From the rating of the bulb , the resistance of the bulb is R=
Vrms =100 Ω P
200V ,50 Hz 1A
L R=100Ω
For the bulb to be operated at its rated value the rms current through it should be 1Α Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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1/ 2
3 2 R X (1/ Cω) 5 So that, = = R R R
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com 200
Vrms Z
∴
1=
L=
2
100 + ( 2π50L ) 2
3 H π
Ex. 17 A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The arc lamp has an effective resistance of 5 Ω when running of 10 A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160 V (dc), what additional resistance is required? Compare the power losses in both cases. Sol. As for lamp VR = ΙR = 10 × 5 = 50 V, so when it is connected to 160 V ac source through a choke in series, V2 = VR2 + VL2,
VL =
160 2 − 50 2 = 152 V VL = ΙXL = ΙωL= 2πfLI
and as,
Ark lamp
R L VL 152 = = 4.84 × 10–2 H Ans. 2πfΙ 2 × π × 50 × 10 V V Now the lamp is to be operated at 160 V dc; instead of choke if additional resistance r is put in series with it, V = Ι(R + r), i.e., 160 = 10(5 + r) i.e., r = 11 Ω Ans. ~ In case of ac, as choke has no resistance, power loss in the choke V = V sin ωt will be zero while the bulb will consume, P = Ι2 R = 102 × 5 = 500 W However, in case of dc as resistance r is to be used instead of choke, the power loss in the resistance r will be. PL = 102 × 11 = 1100 W while the bulb will still consume 500 W, i.e., when the lamp is run on resistance r instead of choke more than double the power consumed by the lamp is wasted by the resistance r.
So,
L=
L
R
0
Ques. An alternating voltage of 220 volt r.m.s. at a frequency of 40 cycles/sec is supplied to a circuit containing a pure inductance of 0.01 H and a pure resistance of 6 ohms in series. Calculate (i) the current, (ii) potential difference across the resistance, (iii) potential difference across the inductance, (iv) the time lag, (v) power factor. Ans. (i) 33.83 amp. (ii) 202.98 volts (iii) 96.83 volts (iv) 0.01579 sec (v) 0.92
12.
LC SERIES CIRCUIT WITH AN AC SOURCE :
ΙXL
vS= Vsinωt
V
V
90 0
Ι
Ι L C
From the phasor diagram V= Ι (XL − XC ) =
ΙZ
φ = 90º
Ques. Which of the following plots may represnet the reactance of a series LC combination ?
Ans.
D
13.
RLC SERIES CIRCUIT WITH AN AC SOURCE : Ι(XL – XL)
ΙXL V
vS= Vsinωt
V
φ ΙXR
ΙR
Ι
L
R
C
ΙXC
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page 9
Ιrms=
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Also,
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com 2 2 (ΙR)2 + (ΙXL − ΙXC)2 = Ι (R) + (XL − XC)
tan φ =
=
Z= (R ) + (XL − XC )
ΙZ
2
2
Ι(XL − XC ) (XL − XC ) = ΙR R
Ques. A series AC circuit has resistance of 4 Ω and a reactance of 3 Ω. the impedance of the circuit is (A) 5 Ω (B) 7 Ω (C) 12/7 Ω (D) 7/12 Ω Ans. (A)
13.1 Resonance : Amplitude of current (and therefore Irms also) in an RLC series circuit is maximum for a given value of Vm and R , if the impedance of the circuit is minimum, which will be when XL-XC =0. This condition is called resonance. So at resonance: XL-XC =0. or
ωL=
or
ω=
1 ωC
1 LC
. Let us denote this ω as ωr.
Ιm Ιm/max
Z Zm in
ωr
ωr Ex. 18 In the circuit shown in the figure , find (a) the reactance of the circuit . (b) impedance of the circuit (c) the current (d) readings of the ideal AC voltmeters (these are hot wire instruments and read rms values). Sol:
(a)
2 XL = 2 π f L = 2π × 50 × = 200 Ω π XC =
200V,50Hz 2Η
100 µF
π
100Ω
π
V1
V2
V3
V4
V5
1 = 100 Ω 100 2π50 × 10 − 6 π
∴
The reactance of the circuit X = XL–XC = 200-100 = 100 Ω Since XL > XC ,the circuit is called inductive.
(b)
impedance of the circuit Z = R 2 + X 2 = 100 2 + 100 2 =100 2 Ω
(c)
the current Irms=
(d)
readings of the ideal voltmeter
200 Vrms = = 2A 100 2 Z
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V=
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From the phasor diagram
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
V2: ΙrmsR = 100 2 Volt V3: ΙrmsXc =100 2 Volt
V5: ΙrmsZ = 200 Volt ,which also happens to be the voltage of source.
13.1 Q VAL VALUE AL UE (QUALIT (QU ALIT Y FA FA CTOR) CTOR) OF LCR LCR SERIES CIRCUIT CIR CUIT (NOT (NO T IN IIT SYLLA S YLLAYLLA BUS) : Q value is defined as
XL where XL is the inductive reactance of the circuit, at resonance. R
More Q value implies more sharpness of Ι VS ω curve more Q value less Q value
i ωr
ω
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2 V4: Ιrms R 2 + XL = 100 10 Volt
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V1: ΙrmsXL = 200 2 Volt
EXERCISE-1 SECTION (A) : AVERAGE, PEAK AND RMS VALUE r.m.s. value of current i = 3 + 4 sin ( t + /3) is: (A) 5 A
(B)
17 A
(C)
5 2
A
(D)
7 2
A
A 2.
A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage V = 10 sin (100 t). The peak current in the circuit will be : (A) 2 amp (B) 1 amp (C) 10 amp (D) 20 amp
A 3.
The peak value of an alternating e.m.f E given by E = E 0 cos t is 10 volt and frequency is 50 Hz. At time t = (1/600) sec, the instantaneous value of e.m.f is : (A) 10 volt
A* 4.
(B) 5 3 volt
(C) 5 volt
(D) 1 volt
The voltage of an AC source varies with time according to the equation, V = 100 sin 100 t cos 100 t. Where t is in second and V is in volt. Then : (A) the peak voltage of the source is 100 volt (B) the peak voltage of the source is (100/ 2 ) volt (C) the peak voltage of the source is 50 volt (D) the frequency of the source is 50 Hz
A 5.
An alternating voltage is given by : e = e1 sint + e2 cost. Then the root mean square value of voltage is given by : (A)
A 6.
e12 e 22
(B)
(C)
e1 e 2
e1 e 2 2
(D)
e12 e 22 2
An AC voltage is given by : 2 t T Then the mean value of voltage calculated over time interval of T/2 seconds : (A) is always zero (B) is never zero (C) is (2e0/) always (D) may be zero
E = E 0 sin
A 7. A 8.
A 9.
Average value of A.C. current in a half time perios may be : (A) positive (B) negative (C) zero
(D) none
If the frequency of the source e.m.f. in an AC circuit is n, the power varies with a frequency : (A) n (B) 2 n (C) n/2 (D) zero An AC voltage of V = 220 2 sin 250 t is applied across a DC voltmeter, its reading will be: 2 (A) 220 2 V (B) 2 V (C) 220 V (D) zero
A 10.
The current in a discharging LR circuit is given by i = o e–t/T where is the time constant of the circuit calculate the rms current for the period t = 0 to t =
A 11.
Find the rms value for the saw-tooth voltage of peak value V0 from t=0 to t=2T as shown in figure. +V0
+
+
2T
V 0 –
T 2
T
–
3 T 2
t
–V0
SECTION (B) : POWER CONSUMED IN AN AC CIRCUIT B 1.
The average power delivered to a series AC circuit is given by (symbols have their usual meaning) : (A) Erms rms (B) Erms rms cos (C) Erms rms sin (D) zero
B 2.
Energy dissipates in LCR circuit in :
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page 12
A 1.
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Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com (D) all of these
B 4.
B 5.
A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two identical resistances. The ratio of heat produced in the two resistances in the same time interval will be (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 4 : 1
B 6.
A resistor and an inductor are connected to an AC supply of 120 volt and 50 Hz. The current in the circuit is 3 ampere. If the power consumed in the circuit is 108 watt, then the resistance in the circuit is : (A) 12 ohm
(B) 40 ohm
(C)
(52 28) ohm
(D) 360 ohm
B 7.
What is the rms value of an alternating current which when passed through a resistor produces heat, which is thrice that produced by a current of 2 ampere in the same resistor in the same time interval? (A) 6 ampere (B) 2 ampere (C) 2 3 ampere (D) 0.65 ampere
B 8.
A resistor and a capacitor are connected to an AC supply of 200 volt, 50 Hz in series. The current in the circuit is 2 ampere. If the power consumed in the circuit is 100 watt, then the resistance in the circuit is : (A) 100 ohm (B) 25 ohm (C) 125 75 ohm (D) 400 ohm
B 9.
The impedance of a series circuit consists of 3 ohm resistance and 4 ohm reactance. The power factor of the circuit is : (A) 0.4 (B) 0.6 (C) 0.8 (D) 1.0
B 10.
An electric bulb and a capacitor are connected in series with an AC source. On increasing the frequency of the source, the brightness of the bulb : (A) increase (B) decreases (C) remains unchanged (D) sometimes increases and sometimes decreases
B 11.
By what percentage the impedance in an AC series circuit should be increased so that the power factor changes from (1/2) to (1/4) (when R is cosntant) ? (A) 200% (B) 100% (C) 50% (D) 400%
B 12*. Average power consumed in an A.C. series circuit is given by (symbols have their usual meaning) : 2
B 13.
2
0 | z | cos 2(| z |) 2 An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness. What would be the peak voltage of the source ?
(A) Erms rms cos
(B) (rms)2 R
(C)
E0 R
2
(D)
B 14.
A resistor of resistance 100 is connected to an AC source = (12V) sin (250 s – 1)t. Find the power consumed by the bulb.
B 15.
In an ac circuit the instantaneous values of current and applied voltage are respectively i = 2(Amp) sin (250 s – 1)t and = (10V) sin ((250 s – 1)t + at t =
B 16.
). Find the instantaneous power drawn from the source 3
2 ms and its average value. 3
A 2000 Hz, 20 volt source is connected to a resistance of 20 ohm, an inductance of 0.125/ H and a capacitance of 500/ nF all in series. Calculate the time in which the resistance (thermal capacity = 100 joule/ºC) will get heated by 10º C
SECTION (C) : AC SOURCE WITH R, L, C CONNECTED IN SERIES
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page 13
(C) R only
The potential difference V across and the current flowing through an instrument in an AC circuit are given by : V = (5 cos t) volt = (2 sin t) A The power dissipated in the instrument is : (A) zero (B) 5 watt (C) 10 watt (D) 2.5 watt 2.2 A coil has an inductance of H and is joined in series with a resistance of 220 . When an alternating e.m.f. of 220 V at 50 cps is applied to it, then the wattless component of the rms current in the circuit is (A) 5 ampere (B) 0.5 ampere (C) 0.7 ampere (D) 7 ampere
B 3.
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(B) C only
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(A) L only
C 2.
An LCR series circuit with 100 resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Then the current and power dissipated in LCR circuit are respectively (A) 1A, 200 watt. (B) 1A, 400 watt. (C) 2A, 200 watt. (D) 2A, 400 watt.
C 3.
In an AC series circuit, the instantaneous current is zero when the instantaneous source voltage is maximum. Connected to the source may be a (A) pure inductor (B) pure capacitor (C) pure resistor (D) combination of an inductor and a capacitor
C 4.
A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same AC supply also gives the same value of peak current but the current lags behind by 90°. If the series combination of X and Y is connected to the same supply, what will be the rms value of current ? (A)
10 2
amp
(B)
5 2
amp
(C)
5 amp 2
(D) 5 amp
C 5.
In an AC circuit the potential differences across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is (A) 20 V (B) 25.6 V (C) 31.9 V (D) 53.5 V
C 6.
An AC voltage source V = 200 2 sin 100 t is connected across a circuit containing an AC ammeter(it reads RMS value) and capacitor of capacity 1 F. The reading of ammeter is : (A) 10 mA (B) 20 mA (C) 40 mA (D) 80 mA
C 7.
When 100 V DC is applied across a solenoid, a steady current of 1 A flows in it. When 100 V AC is applied across the same solenoid, the current drops to 0.5 amp. If the frequency of the AC source is 150 3 / Hz, the impedance and inductance of the solenoid are : (A) 200 and 1/3 H (B) 100 and 1/16 H (C) 200 and 1.0 H (D) 1100 and 3/117 H
C 8.
If in a series LCR AC circuit, the rms voltage across L, C and R are V 1, V2 and V3 respectively, then the voltage of the source is always : (A) equal to V 1 + V2 + V3 (B) equal to V 1 – V2 + V3 (C) more than V1 + V2 + V3 (D) none of these is true
C 9.
A resistor R, an inductor L, a capacitor c and voltmeters V 1, V2 and V3 are connected to an oscillator in the circuit as shown in the adjoining diagram. When the frequency of the oscillator is increased, upto resonance frequency, the voltmeter reading (at resonance frequency) is zero in the case of : (A) voltmeter V1 (B) voltmeter V2 (C) voltmeter V3 (D) all the three voltmeters
C 10.
In the series LCR circuit, the voltmeter and ammeter readings are :
(A) V = 100 volt, = 2 amp (C) V = 1000 volt, = 2 amp C 11.
C 12.
(B) V = 100 volt, = 5 amp (D) V = 300 volt, = 1 amp
An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage ? In a series LCR circuit with an AC source R = 300 C = 20 µF, L = 1.0 henry, E rms = 50 V and = 50/
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A 0.21-H inductor and a 88- resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. Use = 22/7. (A) 2 A, tan–1 3/4 (B) 14.4 A, tan–1 7/8 (C) 14.4 A, tan–1 8/7 (D) 3.28 A, tan–1 2/11
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Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com C 1.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
C 13.
Consider the situation of the previous problem find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil .
C 14.
A circuit has a coil of resistance 50 ohms and inductance
40 F and AC supply voltage of 200 V and 50 cycles/sec. Calculate the impedance of the circuit, the p.d. across inductance coil and condenser.
denser of (i) (ii) C 15.
An inductor 2/ Henry, a capacitor 100/ µF and a resistor 75 are connected in series across a source of EMF V = 10 sin 100 t. (a) find the impedance of the circuit.(b) find the energy dissipated in the circuit in 20 minutes.
SECTION (D) : RESONANCE D 1*. D 2.
Power factor may be equal to 1 for : (A) pure inductor (B) pure capacitor
(C) pure resistor
The value of power factor cos in series LCR circuit at resonance is : (A) zero (B) 1 (C) 1/2
(D) An LCR circuit (D) 1/2 ohm
D 3.
A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency 4 × 10 3 rad s –1 . At resonance, t he v o l tage across r esi stance and i nductance are 60V and 40 V respectively. The values of L and C are respectively : (A) 20 mH, 25/8 F (B) 2mH, 1/35 F (C) 20 mH, 1/40 F (D) 2mH, 25/8 nF
D 4.
In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in resonance ? (A) 4 times (B) 1/4 times (C) 8 times (D) 2 times
D 5.
A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency n. If the resonant frequency is n r, then the current lags behind voltage, when : (A) n = 0 (B) n < nr (C) n = rr (D) n > nr
D 6.
A series circuit consists of a resistance, inductance and capacitance. The applied voltage and the current at any instant are given by E = 141.4 cos (5000 t – 10º) and = 5 cos (5000 t – 370º) The inductance is 0.01 henry. Calculate the value of capacitance and resistance.
D 7.
An inductance of 2.0 H, a capacitance of 18µ F and a resistance of 10k are connected to an AC source of 20 V with adjustable frequency (a) W hat frequency should be chosen to maximise the current(RMS) in the circuit ? (b) What is the value of this maximum current (RMS) ?
D 8.
An inductor-coil a capacitor and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied a maximum rms current of 6.0 A is observed. If this inductor coil is connected to a battery of emf 12 V and internal resistance 4.0 what will be the steady current ?
D 9.
A wave of wavelength 300 metre can be transmitted by a transmission centre. A condenser of capacity 2.5 F is available. Calculate the inductance of the required coil for a resonant circuit.Use 2=10.
SECTION (E) : TRANSFORMER E 1.
A power (step up) transformer with an 1 : 8 turn ratio has 60 Hz, 120 V across the primary; the load in the secondary is 104 . The current in the secondary is (A) 96 A (B) 0.96 A (C) 9.6 A (D) 96 mA
E 3.
A transformer is used to light a 140 watt, 24 volt lamp from 240 V AC mains. The current in the main cable is 0.7 amp. The efficiency of the transformer is : (A) 48% (B) 63.8% (C) 83.3% (D) 90%
E 4.
In a step-up transformer the voltage in the primary is 220 V and the current is 5A. The secondary voltage is found to be 22000 V. The current in the secondary (neglect losses) is (A) 5 A (B) 50 A (C) 500 A (D) 0.05 A The core of a transformer is laminated to reduce
E 5.
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page 15
3 henry. It is connected in series with a con
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Hz. Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com (C) copper loss
(D) magnetic loss
E1
A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary ?
E 2.
In a transformer ratio of secondary turns (N2) and primary turns (N1) i.e.
N2 4 . If the voltage applied in N1
primary is 200 V, 50 Hz, find (a) voltage induced in secondary (b) If current in primary is 1A, find the current in secondary if the transformer is (i) ideal and (ii) 80% efficient and there is no air loss.
SECTION (F) : MISCELLANEOUS * Marked are more than one correct options. F 1.
F 2.
A capacitor is a perfect insulator for : (A) direct current (C) direct as well as alternating current
(B) alternating current (D) None of the above
A choke coil sould have : (A) high inductance and high resistance (C) high inductance and low resistance
(B) low inductance and low resistance (D) low inductance and high resistance
F 3.
A choke coil is preferred to a rheostat in AC circuit as : (A) it consumes almost zero power (B) it increases current (C) it increases power (D) it increases voltage
F 4.
With increase in frequency of an AC supply, the inductive reactance : (A) decreases (B) increases directly proportional to frequency (C) increases as square of frequency (D) decreases inversely with frequency
F 5.
With increase in frequency of an AC supply, the capacitive reactance : (A) varies inversely with frequency (B) varies directly with frequency (C) varies directly as square of frequency (D) remains constant
F 6.
An AC ammeter is used to measure current in a circuit. When a given direct constant current passes through the circuit, the AC ammeter reads 3 ampere. When another alternating current passes through the circuit, the AC ammeter reads 4 ampere. Then the reading of this ammeter if DC and AC flow through the circuit simultaneously, is : (A) 3 ampere (B) 4 ampere (C) 7 ampere (D) 5 ampere
F 7.
In an a.c. circuit consisting of resistance R and inductance L, the voltage across R is 60 volt and that across L is 80 volt. The total voltage across the combination is (A) 140 V (B) 20 V (C) 100 V (D) 70 V
F 8.
In the AC circuit shown below, the supply voltage has constant rms value V but variable frequency f. At resonance, the circuit : V R (B) has a resonance frequency 500 Hz (C) has a voltage across the capacitor which is 180 0 out of phase with that across the inductor
(A) has a current given by =
(D) has a current given by =
1 F
1 H
V,f
~
V 1 1 R2
F 9.
R
2
In the circuit shown in figure, if both the bulbs B 1 and B2 are identical : (A) their brightness will be the same (B) B2 will be brighter than B 1
(C) as frequency of supply voltage is increased the brightness of bulb B 1 will increase and that of B 2 will decrease. (D) only B2 will glow because the capacitor has infinite impedance F 10*. An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power
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(B) hysteresis loss
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(A) eddy current loss
delivered by the source : (A) must be 1000 W (C) may be greater than 1000 W
An inductor coil having some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle ? (A) current (B) induced emf in the inductor (C) joule heat (D) magnetic energy stored in the inductor
F 12*. A town situated 20 km away from a power house at 440 V, requires 600 KW of electric power at 220 V. The resistance of line source carrying power is 0.4 per km. The town gets power from the line through a 3000 V–220 V step-down transformer at a substitution in the town. Which of the following is/are correct (A) The loss in the form of heat is 640 kW (B) The loss in the form of heat is 1240 kW (C) Plant should supply 1240 kW (D) Plant should supply 640 kW F 13.
11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22000 V. Which of the following is correct (A) first mode of transmission consumes less power (B) second mode of transmission consumes less power (C) first mode of transmission draws less current (D) second mode of transmission draws less current
F 14.
In a series LCR circuit with an AC source(E rms = 50 V and = 50/ Hz), R = 300 , C = 0.02 mF, L = 1.0 H, Which of the following is correct (A) the rms current in the circuit is 0.1 A (B) the rms potential difference across the capacitor is 50 V (C) the rms potential difference across the capacitor is 14.1 V (D) the rms current in the circuit is 0.14 A
F 15.
A circuit is set up by connecting L = 100 mH, C = 5 F and R =100 in series. An alternating emf of 500 Hz is applied across this series combination. Which of the following is correct (A) the impedance of the circuit is 141.4 (B) the average power dissipated across resistance 225 W (C) the average power dissipated across inductor is zero. (D) the average power dissipated across capacitor is zero.
(150 2 ) volt,
F 16*. In a series RC circuit with an AC source( peak voltage E 0 = 50 V and f = 50 /Hz), R = 300 ,C = 25 F. Then : (A) the peak current is 0.1 A (B) the peak current is 0.7 A (C) the average power dissipated is 1.5 W (D) the average power dissipated is 3 W F 17.
A coil of inductance 5.0 mH and negligible resistance is connected to an oscillator giving an output voltage E = (10V) sin tWhich of the following is correct (A) for = 100 s–1 current is 20 A (B) for = 500 s–1 current is 4 A –1 (C) for = 1000 s current is 2 A (D) for = 1000 s–1 current is 4 A
F 18.
A pure inductance of 1 henry is connected across a 110 V, 70Hz source. Then correct option are (Use = 22/7): (A) reactance of the circuit is 440 (B) current of the circuit is 0.25 A (C) reactance of the circuit is 880 (D) current of the circuit is 0.5 A
EXERCISE-2 1.
An LCR series circuit with 100 resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Calculate the current and power dissipated in LCR circuit. [REE - 90]
2.
A 100 volt AC source of frequency 500 hertz is connected to a LCR circuit with L = 8.1 millihenry, C = 12.5 microfarad and R = 10 ohm, all connected in series. Find the potential difference across the resistance. [REE - 91] The current in a coil of self inductance 2.0 Henry is increasing according to i = 2 sin t 2 ampere. Find the amount of energy spend during the period when the current changes from zero to 2 ampere.[REE- 91]
3.
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page 17
F 11.
(B) may be 1000 W (D) may be less than 1000 W
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Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com voltage of frequency
by.. 2
1 (A) tan–1 CR
(B) tan–1 (CR)
[REE - 91] 1 (C) tan–1 R
(D) cos–1 (CR)
An alternating potential V = V 0 sin t is applied across a circuit. As a result the current = 0 sin t flows in it. The power consumed in the circuit per cycle is [REE - 92] 2 (A) zero (B) 0.5 V00 (C) 0.707 V00 (D) 1.414 V00
6.
In a purely resistive AC circuit, the current (A) Lags behind the EMF in phase (B) Is in phase with the e.mf. (C) Leads the EMF in phase (D) Leads the EMF in half the cycle and lags behind it in the other half.
7.
A current of 4 A flows in a coil when connected to a 12 V d.c. source. If the same coil is connected to a 12 V, 50 rad/s, AC source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a 2500 µF condenser is connected in series with coil. [REE - 93]
8.
The current and voltage in an AC circuit are respectively given by = 0 cos t, V = V 0 sin t. The power consumed in the circuit is [REE - 93] (A) zero
(B)
V0 0 2
(C)
V0 0 2
[REE - 92]
(D)
2 V0 0
9.
If a resistance of 30, a capacitor reactance 20 , and an inductor of inductive reactance 60 are connected in series to a 100 V, 50 Hz power source, then [REE - 94] (A) A current of 2.0 A flows (B) A current of 3.33 A flows (C) Power factor of the circuit is zero (D) Power factor of the circuit is 3/5
10.
A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency 4 × 10 5 rad s–1. At resonance, the voltage across resistance and inductance are 60V and 40 V respectively. Find the values of L and C. At what frequency the current in the circuit lags the voltage by 45º? [REE - 95]
11.
In a circuit, an inductance of 0.1 Henry and a resistance of 1 are connected in series with an AC source of voltage V = 5 sin 10 t. The phase difference between the current and applied voltage will be (A) (B) 2 (C) /4 (D) 0 [REE - 96]
12.
An inductive reactance, X L = 100 , a capacitive reactance, X C = 100 , and a resistance R = 100 , are connected in series with a source of 100 sin (50 t) volts. Which of the following statements are correct? [REE - 96] (A) The maximum voltage across the capacitor is 100 V. (B) The net impedance of the circuit is 100 . (C) The maximum voltage across the inductance is 100 V. (D) The maximum voltage across the series is 100 V.
13.
A series LCR circuit is operated at resonance. Then [REE - 97] (A) Voltage across R is minimum (B) Impedance is minimum (C) Power transferred is maximum (D) Current amplitude is minimum
14.
A box P and a coil Q are connected in series with an AC source of variable frequency. The EMF. of source is constant at 10 V. Box P contains a capacitance of 1 µF in series with a resistance of 32 . Coil Q has a self inductance 4.9 mH and a resistance of 68 . The frequency is adjusted so that the maximum current flows in P and Q. Find the impedance of P and Q at this frequency. Also find the voltage across P and Q respectively. [REE - 98] An inductor 20 × 10–3 Henry, a capacitor 100 µF and a resistor 50 are connected in series across a source of EMF V = 10 sin 314 t. Find the energy dissipated in the circuit in 20 minutes. If resistance is
15.
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page 18
5.
The current in a circuit containing a capacitance C and a resistance R in series leads over the applied
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4.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com The electric current in an AC circuit is given by i = i 0 sin t. What is the time taken by the current to change from its maximum value to the rms value? [REE - 99]
17.
An alternating EMF of angular frequency is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency. [REE - 99] (A)
4
(B)
2
(C)
(D) 2
18.
A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then. [REE - 99] (A) bulb will give more intense light (B) bulb will give less intense light (C) bulb will give light of same intensity as before (D) bulb will stop radiating light
20.
In an AC circuit, the power factor (A) is zero when the circuit contains an ideal resistance only (B) is unity when the circuit contains an ideal resistance only (C) is zero when the circuit contains an ideal inductance only (D) is unity when the circuit contains an ideal inductance only
21.
When an AC source of emf e = E 0 sin (100 t) is connected across a circuit, the phase difference
[REE - 2000]
, as shown in the diagram. If the 4 circuit consists possibly only of R-C or R-L or L-C series, find the relationship between the two elements. [JEE 2003]
between the emd e and the current i in the circuit is observed to be
(A) R = 1k, C = 10 F (C) R = 1k, L = 10 H
(B) R = 1k, C = 1 F (D) R = 1k, L = 1H
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page 19
16.
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removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit. [REE - 99]
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EXERCISE 1 SECTION (A) :
SECTION (E) : E 1.
D
E 2.
C
E5
zero
E 3.
D
B
A 2.
D
A 3.
B
E 4.
A
A* 4.
C
A 5.
D
A 6.
D
E 6.
(a) 800 V (b) (i) 0.25 A (ii) 0.2 A.
A 7.
ABC
A 8.
B
A 9.
D
C
F 3.
A
A 10.
o (e 2 1) / 2 e
SECTION (F) : F 1. A F 2. F 4.
B
A
F 6.
D
F 7.
C
A 11.
V0 3
SECTION (B) :
F 5.
page 20
A 1.
* Marked are more than one correct options.
B 1.
B
B 2.
C
B 3.
A
B 4.
B
B 5.
C
B 6.
A
F8
ABC
F 9.
B 7.
C
B 8.
B
B 9.
B
F 11.
AB
F 12*. AC
B 10.
A
B 11.
B
B 12*. ABCD
F 14.
AB
F 15.
ABCD F 16*. AC
B 13. B 15.
12 2 volts 10 W, 5 W
F 17.
ABC
F 18.
AB
B 14. B 16.
0.72 W 50 sec
SECTION (C) :
BC
F 10*. BD F 13.
BD
EXERCISE 2 1.
2A; 400 watt.
C 1.
A
C 2.
D
C 3.
ABD
3.
4 joule
4.
A
C 4.
C
C 5.
B
C 6.
B
5.
A
6.
B
C 7.
A
C 8.
D
C 9.
B
7.
08 H; 17.28 watt A, D
A
C 11.
2.2 3 =1.2 H
9.
C 10.
10.
2 × 10–4 Henry;
11.
C
13.
B, C
14.
P=76.96 ,Q=97.59 P = 7.6 V; Q = 9.8 V, impedance = 100
15.
952 J; 0.52 cos 314 t
16.
T/8 or
18.
A
20.
A
C 12.
(a) 0.1 A (b) 50 V, 30 V, 10 V (Note that the sum of the rms potential differences across the three elements is greater that the rms voltage of the source.)
C 13. C 14.
25 mJ, 5mJ Z = 50 2 ohm, V C = 500 2 v olt and VL= 600 2 volt
C 15.
125 , 288 J
SECTION (D) : D 1*.
CD
D 2.
B
D 3.
A
D 4.
A
D 5.
D
D 6.
4 F
D 7.
(a)
D 8.
1.5 A
2.
8.
A
1 µF; 8 × 105 rad/s 32
12.
4
100 volt
ABCD
17.
D
19.
B, C
250 Hz (b) 2 mA 3
D 9.
1×10–8 henry
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ANSWER
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ALTERNATING CURRENT Some questions (Assertion–Reason type) are given below. Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice : Choices are : (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1. (C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True. 535.
STATEMENT – 1 The alternating current cannot be used to conduct electrolysis. STATEMENT – 2
The ions due to their inertia, cannot follow the changing E . 536.
STATEMENT – 1 In a series LCR circuit at resonance, the voltage across the capacitor or inductor may be more than the applied voltage. STATEMENT – 2 At resonance in a series LCR circuit, the voltages across inductor and capacitor are out of phase.
537.
STATEMENT – 1 By only knowing the power factor for a given LCR circuit it is not possible to tell whether the applied alternating emf leads or lags the current. STATEMENT – 2 cos θ = cos (–θ)
538.
STATEMENT – 1 In the purely resistive element of a series LCR, AC circuit the maximum value of rms current increases with increase in the angular frequency of the applied emf. STATEMENT – 2 2
Imax 539.
ε 1 = max , z = R 2 + ωL − , where Imax is the peak current in a cycle. z ωC
STATEMENT – 1 AC source is connected across a circuit. Power dissipated in circuit is P. The power is dissipated only across resistance. STATEMENT – 2 Inductor and capacitor will not consume any power in AC circuit.
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STATEMENT – 1 : In series RLC circuit potential drop across inductive reactance will be same as capacitive reactance at resonance. STATEMENT – 2 : At frequency less than resonance frequency for series RLC nature of circuit will be capacitive, frequency more than resonance nature of overall circuit will be inductive.
541.
STATEMENT – 1 : For series RLC network, power factor of circuit in region (1) is positive and in region (2) is negative. STATEMENT – 2 : Overall nature of circuit in region (1) is inductive while in region (2) is capacitive. I (1) (2)
fr 542.
f
STATEMENT – 1 : In a series LCR circuit, at resonance condition power consumed by circuit is maximum. STATEMENT – 2 : At resonance condition effective resistance of circuit is maximum.
543.
STATEMENT – 1 : In series L–R circuit voltage leads the current. STATEMENT – 2 : In series L–C circuit current leads the voltage.
544.
STATEMENT – 1 : Average value of a.c. over a complete cycle is always zero. STATEMENT – 2 : Average value of a.c. is always defined over half cycle.
545.
STATEMENT – 1 : In series LCR circuit resonance can take place. STATEMENT – 2 : Resonance takes if inductance and capacitive reactance are equal.
546.
STATEMENT – 1 : KVL rule is also being applied in AC circuit shown below. STATEMENT – 2 :
8V
~ V=10V VC in the circuit = 2V. 547.
STATEMENT – 1 : AC generators are based upon EMI principle. STATEMENT – 2 : Resistance offered by capacitor for alternating current is zero.
548.
STATEMENT – 1 : For sinusoidal a.c. ( I = I0 sin ωt )
I rms =
I0 . 2
STATEMENT – 2 : The r.m.s. value of alternating current is defined as the square root of the average of I2 during a complete cycle. 549.
STATEMENT – 1 Rate of heat generated when resistance is connected with AC source depends on time. STATEMENT – 2 RMS voltage may be greater than maximum AC voltage.
550.
An inductor, capacitor and resistance connected in series. The combination is connected across AC source. STATEMENT – 1 : Peak current through each remains same.
Download FREE Study Package from www.TekoClasses.com & Learn on Video www.MathsBySuhag.com Phone : 0 903 903 7779, 98930 58881 Alternating Current Page: 23 STATEMENT – 2 : Average power delivered by source is equal to average power developed across resistance. 551.
STATEMENT – 1 : In alternating current direction of motion of free electrons changes periodically. STATEMENT – 2 : Alternating current changes its direction after a certain time interval.
552.
STATEMENT – 1 : When frequency is greater than resonance frequency in a series LCR circuit, it will be an inductive circuit. STATEMENT – 2 : Resultant voltage will lead the current.
553.
STATEMENT – 1 : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads the current. STATEMENT – 2 : The phase angle is the angle between the alternating e.m.f. and alternating current of the circuit.
554.
STATEMENT – 1 : An alternating current shows magnetic effect. STATEMENT – 2 : Alternating current varies with time.
Hint & Solution 535. 537. 541. 545. 549. 553. 537.
538.
542.
(A) 536. (B) (A) 538. (D) 539. (C) 542. (C) 543. (A) 546. (C) 547. (C) 550. (B) 551. (C) 554. (B) For a certain values of cos θ (power factor) two values of negative. Accordingly the applied emf may lead or lag. The maximum value of rms current =
Pav =
(A) (B) (C) (B)
540. 544. 548. 552.
θ are possible. One is positive the other is much
ε rms ε rms . It does not depend upon ω. = z R
VI cos φ 2
At resonance condition cos φ = 1 But Z = R Which is minimum. 543.
L–R circuit
V
R
WL N
H C–R circuit. I YwC R
544.
For half cycle Imean = 0.636I0 or Emean = 0.636 E0
(B) (B) (A) (A)
V
Download FREE Study Package from www.TekoClasses.com & Learn on Video www.MathsBySuhag.com Phone : 0 903 903 7779, 98930 58881 Alternating Current Page: 24 Average value is always defined over a half cycle cause in next half cycle it will be opposite in direction. Hence for one complete cycle, average value will be zero. 545.
At resonant frequency XL = XC ∴ Z = R (minimum) Therefore current in the circuit is maximum.
546.
Voltage will be added vectorially.
547.
XC =
1 ωC
ω ≠ O area for AC.
1/ 2
T 2 ∫ I dt Irms 0T ∫ dt 0
1/ 2
2π/ ω 2 2 ∫ I0 sin ωt dt = 0 2π/ ω dt ∫ 0
=
I0
.
2
549.
Rate of heat generated depends on time.
550.
Average power consumed by capacitor or inductor is zero.
551.
Motion of electron is random with drift velocity opposite to the direction of current.
552.
VL V
VL–VC
φ
I VR
VC
553.
X − XC tan φ = L = R
ωL −
1 ωC
R
When X L > XC then tan φ is positive i.e. φ is positive (between 0 and π/2). Hence e.m.f. leads the current.
554.
Like direct current, alternating current also produces magnetic field. But the magnitude and direction of the field goes on changing continuously with time.