CURRENT ELECTRICITY
Preface In the past chapters we studied the interactions of electric charge at rest ; Now we are zready to study charges in motion. An electric electric current consists of charges in motion from one region to another. another. When this motion m otion takes place within the conducting path that forms a closed loop, the path is called the electric circuit. In this chapter we will study the basic properties of electric currents. To To understand the behavior of currents in electric circuits, we will describe the properties of conductors and how they depends on temperature. If you look inside insi de TV, TV, your computer or your stereo ste reo receiver receive r or under the hood of a car, ca r, you you will find circuits of much greater complexity than the the simple circuits. In this chapter we study general methods for analyzing such networks, including how to find unknown voltages, currents and properties of circuits element. We will determine the equivalent for several resistors connected connec ted in series or in parallel. For more general networks we need two rule called kirchoff's rule. our principal concern in this chapter is with direct current circuits in which directionof current does not change with time. This book consists of theoritical & practical explanations of all the concepts involved in the chapter. Each article followed foll owed by a ladder of illustration. ill ustration. At the end of the theory part, there are miscellaneous solved s olved examples which involve the application of multiple multip le concepts of this chapter. Students are advised to go through all these solved examples in order to develope bettter understanding of the chapter and to have better grasping level in the class.
Total number of Questions Quest ions in this chapter are a re : (i) In chapte chapterr Exam Example pless ............... ....................... ........ 50 (ii) (ii) Solv Solve ed Exam Exampl ples es
.... ....... ..... .... .... .... ..... ..... ..... ... 38
Total otal no. no. of que quest stion ions s .............. ...................... .......... 88
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CURRENT ELECTRICITY
70
(k) Electric Electric field outside outside a current current carrying carrying conductor is zero, but it is non zero inside v the conductor and is given by e = – l
1. ELECTRIC CURRENT
(a) When a charge charge flows flows in a conducto conductorr from one place to the other, then the rate of flow of charge is called electric current(i) (b) The electric electric current current in in measured measured by 'rate 'rate of flow of charge'. or (c) (c) Charge Charge flowing flowing per per second second from any any cross cross section of the conductor is called electric current, Current i = i =
q t
Ch arg e dq = , if flow is uniform Time dt
(d) (d) Unit : Ampere (A) 1 ampere = 1 coulomb/second. i.e. if 1 coulomb of charge flows per second then 1 ampere of current is said to be flowing. (e) Dimen Dimensio sion n : (M0L0T0 A1) (f) If n electr electrons ons pass pass throug through h any cross cross section section ne in every t seconds then i = t –19 where e = 1.6 × 10 coulomb. (g) 1 ampere ampere of current current means means the flow flow of 18 6.25 × 10 electrons per second through any cross section of conductor (h) Directio Direction n of of flow flow of of current current is taken taken to be be opposite to the direction of flow of electrons. (i) (i) Value of of the curren currentt is same same throug throughout hout the conductor, irrespective of the cross section of conductor at different points. (j) (j) Net charg charge e in a curre current nt carry carrying ing conduct conductor or is zero at any instant of time. Note : A curre current nt carrying carrying conduct conductor or cannot cannot said said to be charged, because in conductor the current is caused by electron (free electron). The no. of electron (negative charge) and proton (positive charge) in a conductor is same. Hence the net charge in a current carrying conductor is zero.
Note : The electric field inside charged conductor is zero, but it is non zero inside a current carrying conductor
(l) (l) Electric Electric curren currentt is a scalar scalar quality quality Although Although in diagrams, we represent current in a wire by an arrow but the arrow simply indicate the direction of flow of positive charges in the wire. Note : Though electri electric c current current needs needs direction direction for its representation, yet it is scalar quantity. It is because, the current can be added algebraically. Only scalar quantities can be added algabraically not the vector quantities.
Electric current Ex.1
Sol
If a charge of 1.6 × 10 –19 coulomb flows per second through any cross section of any conductor, the current constitute will be– (A) 2.56 × 10 –19 A (B) 6.25 × 10 –19 A (C) 1.6 × 10 –19 A (D) 3.2 × 10 –19 A (C) (C) q t Here q = 1.6 x 10 –19 C and t = 1 sec From definition of current i =
1 10 –19 = 1.6 × 10 –19 ampere 1 Thus if dq charge flows in dt seconds, then dq electric current, i is given by i = dt If charge flowing through a conductor is given by q = 1.5 t 2 + t. The current caused in t = 2 second will be– (A) 4 ampere (B) 5 ampere (C) 6 ampere (D) 7 ampere (D) (D)
Ex.2
Sol
E
dq dt 2 Here q = 1.5 t + t
We know i =
+
– F
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i =
dq = 3t + 1 dt
dq Now dt at t = 2s = 3 x 2 + 1 = 7
Current caused in 2 seconds = 7 ampere CURRENT ELECTRICITY
71
(k) Electric Electric field outside outside a current current carrying carrying conductor is zero, but it is non zero inside v the conductor and is given by e = – l
1. ELECTRIC CURRENT
(a) When a charge charge flows flows in a conducto conductorr from one place to the other, then the rate of flow of charge is called electric current(i) (b) The electric electric current current in in measured measured by 'rate 'rate of flow of charge'. or (c) (c) Charge Charge flowing flowing per per second second from any any cross cross section of the conductor is called electric current, Current i = i =
q t
Ch arg e dq = , if flow is uniform Time dt
(d) (d) Unit : Ampere (A) 1 ampere = 1 coulomb/second. i.e. if 1 coulomb of charge flows per second then 1 ampere of current is said to be flowing. (e) Dimen Dimensio sion n : (M0L0T0 A1) (f) If n electr electrons ons pass pass throug through h any cross cross section section ne in every t seconds then i = t –19 where e = 1.6 × 10 coulomb. (g) 1 ampere ampere of current current means means the flow flow of 18 6.25 × 10 electrons per second through any cross section of conductor (h) Directio Direction n of of flow flow of of current current is taken taken to be be opposite to the direction of flow of electrons. (i) (i) Value of of the curren currentt is same same throug throughout hout the conductor, irrespective of the cross section of conductor at different points. (j) (j) Net charg charge e in a curre current nt carry carrying ing conduct conductor or is zero at any instant of time. Note : A curre current nt carrying carrying conduct conductor or cannot cannot said said to be charged, because in conductor the current is caused by electron (free electron). The no. of electron (negative charge) and proton (positive charge) in a conductor is same. Hence the net charge in a current carrying conductor is zero.
Note : The electric field inside charged conductor is zero, but it is non zero inside a current carrying conductor
(l) (l) Electric Electric curren currentt is a scalar scalar quality quality Although Although in diagrams, we represent current in a wire by an arrow but the arrow simply indicate the direction of flow of positive charges in the wire. Note : Though electri electric c current current needs needs direction direction for its representation, yet it is scalar quantity. It is because, the current can be added algebraically. Only scalar quantities can be added algabraically not the vector quantities.
Electric current Ex.1
Sol
If a charge of 1.6 × 10 –19 coulomb flows per second through any cross section of any conductor, the current constitute will be– (A) 2.56 × 10 –19 A (B) 6.25 × 10 –19 A (C) 1.6 × 10 –19 A (D) 3.2 × 10 –19 A (C) (C) q t Here q = 1.6 x 10 –19 C and t = 1 sec From definition of current i =
1 10 –19 = 1.6 × 10 –19 ampere 1 Thus if dq charge flows in dt seconds, then dq electric current, i is given by i = dt If charge flowing through a conductor is given by q = 1.5 t 2 + t. The current caused in t = 2 second will be– (A) 4 ampere (B) 5 ampere (C) 6 ampere (D) 7 ampere (D) (D)
Ex.2
Sol
E
dq dt 2 Here q = 1.5 t + t
We know i =
+
– F
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i =
dq = 3t + 1 dt
dq Now dt at t = 2s = 3 x 2 + 1 = 7
Current caused in 2 seconds = 7 ampere CURRENT ELECTRICITY
71
Ex.3
Sol
The no. of electrons flowing per second through any cross section of wire, if it carries a current of one ampere, will be – (A) 2.5 × 10 18 (B) 6.25 × 10 18 (C) 12.5 × 10 18 (D) 5 × 10 18 (B)
Ex.6
q ne = [ q = ne, from quantization t t rule of charge]
i=
Ex.4
Sol
Sol
It
e
=
1 1
18 19 = 6.25 × 10 1.6 10
The no. of electron passing through a heater wire in one minute, if it carries a current of 8 ampere, will be– (A) 2 × 1020 (B) 2 × 1021 (C) 3 × 10 20 (D) 3 × 10 21 (D) (D) It
8 60
= 3 × 10 21 e 1.6 10 19 In hydrogen atom, the electron moves in an orbit of radius 5 × 10 –11 m with a speed of 2.2 × 106 m/sec. the equivalent current will be – (A) 1.12 mA (B) 4.32 mA (C) 3.32 mA (D) 7.12 mA (A) Time taken by the electron in 1 revolution is 2r T = v where r is the radius of orbit and v is the speed. Therefore, the no. of revolutions in 1 second is n =
Ex.5
n =
=
2.2 10 6 v 1 n = = = 22 2r T 2 ( 5 10 11) 7
F IJ G HK
1015 /sec
= 7 × In one revolution, 1.6 × 10 –19 coulomb of charge flows through any point of the orbit. Hence Hence the total total charge charge flown in 1 sec. is q = charge in 1 revolution × no. of revolution in 1 sec. = (1.6 × 10 –19) × (7.0 × 10 15) = 1.12 × 10 –3 coulomb
Current
I=
q t
=
= 1.12 ×
11.2 10 3 1 10 –3 amp
Sol Ex.7
Sol
Ex.8
Sol
A conductor conduc tor of non–uniform non–unifor m cross–sectio cro ss–sectional nal area, has cross–sectional area at three points as A1 = 2cm 2, A2 = 4cm 2, A3 = 6cm2. If a current of 5 ampere is passed through A 1, the current will give values, when passed through A2 and A3 respectively as– (A) 10 ampere, 15 ampere (B) 20 ampere, 30 ampere (C) 2.5 ampere, 1.66 ampere (D) 5 ampere, 5 ampere Current will remain same. A steady current is flowing in a cylindrical conductor. Is there any electric field within the conductor ? [IIT–82] Yes, The reason is that the current in a conductor flows only when the electric field established in the conductor applies a force on each free electron. An electr ele ctron on moves mov es in a circle cir cle of radius rad ius 10 cm with a constant speed of 4 × 10 6 m/s find find the electric current at a point on on the circle. Consider a point A on the circle. T he electron crosses this point once in every revolution. The number of revolutions made by electron in one second is – A
4 10 6
2 = × 107 2 10 10 2
Hence, the charge crossing 'A' every second 2 3.2 is × 107 × 1.6 × 10 –19 C = × 10 –12 C
By the definition of current, this is the current at this point
1
× 10 –12 A
2. CURRENT DENSITY
(a) The current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross–section of the conductor of that point. i A i = Electric current A = Area of cross cross section. section.
(b) It is denoted by j i.e. j =
Note : Area Area 'A' 'A' is normal normal to curre current nt 'I'. If A is not
= 1.12 mA
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normal to I, but makes an angle the normal to current, then CURRENT ELECTRICITY
with
72
Current density
I
P
Ex.9
A I
I
j =
A normal = A cos
I = j A cos
J =
= j . A
6.25
4
22 10 6 Sol
2
× 106 A/m2 200 A/cm 2 (c) It is a VECTOR quantity It's direction is the direction of motion of the positive charges at that point. (d) Units : ampere / meter 2 (A/m 2) (e) Dimension : [M0L –2T0 A] (f) If, n = number of free electrons per unit volume of conductor. A = cross sectional area of conductor vd = Drift. velocity. then I = neA v d and J = ne vd (g) (g) Drift velocity : An applied potential difference does not give an accelerated motion to electrons but simply gives them a small constant velocity ( 10 –4 m/s) along the length of wire towards the end at higher potential. This is called Drift velocity of the electrons. Note : The speed speed of random random motion of electron electrons s is determined by temperature and is given by
1 3 mv2 = kT 2 2
v =
J = I/A =
For
Ex.11
Sol
The relations between relaxation time () and drift velocity (v d) is given v d
r 2 6.25
2.5 2 10 6
= 127 × 106 A/m 2 = 127 A/cm 2 and for Ex.10 (i) A current of 5 amp. is passing through metallic wire of cross–sectional area 4 × 10 –6 / m 2, the drift drift speed speed of electro electrons ns will be (in m/sec) (density of electron per unit [Roorkee–91] volume is 5 × 1026) –2 (A) 1.56 × 10 (B) 1.6 × 10 –19 (C) 1.56 × 10 2 m/se m/secc (D) (D) 1. 1.6 × 1019 Sol (A) We know I = neAvd
(h) Electrons collide with the ions of metal while moving. The average time–interval between two two successive collisions is called relaxation– relaxation– time, denoted by .
A, J =
I
4
3kT m
where m is mass of electron, T is absolute temp. and k is Boltz man's constant.
One end of an aluminium wire, whose diameter is 2.5 mm is welded to one end of a copper wire whose diameter is 2.0 mm. The composite wire carries a steady current 6.25 A. The current densities in Al and Cu will be respectively – (A) 127 A/cm 2, 200 A/cm 2 (B) 126 A/cm 2, 180 A/cm 2 (C) 125 A/cm 2, 160 A/cm 2 (D) 125 A/cm 2, 180 A/cm 2 (A)
= – e E m
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5
vd =
5 10 26 1.6 10 19 4 10 6 = 1.56 × 10 –2 m/sec The diameter of a copper wire is 2mm. If a steady current of 6.25 A is caused by 8.5 × 10 28 /m –3 electrons flowing through it. The drift velocity of conduction electrons will be – (A) 0.25 mm/s (B) 0.35 mm/s (C) 0.15 mm/s (D) 0.45 mm/s (C) J =
6.25 i = A 2.5 10 6 4 = 200×104 A/m 2
J 200 10 4 vd = = ne 8.5 10 25 1.6 10 19
= 14.7 × 10 –5 m/s
0.15
mm/s
CURRENT ELECTRICITY
73
Ex.12 A silver wire 1mm diameter carries a charge
Sol
of 90 coulombs in 1 hour and 15 minutes. Silver contains 5.8 × 1028 free electrons per cm3. The current (in amp.) in wire and drift velocity of the electron will be respectively– (A) 0.02, 2.69 × 10 –7 (B) 0.03, 3.69 × 10 –7 (C) 3.2, 2.69 × 10 –7 (D) 2.3, 3.69 × 10 –7 (A) i =
J =
where, R is a constant. This is called 'Electrical resistance' of the conductor. (b) This is true for metallic conductors only which have free electrons. (c) The law is not applicable for ionized gases, transistors, semi–conductors etc. I
90 coulombs q = 4500 sec s = 0.02 ampere t
(a)
i 0.02 amp i = = A r 2 (0.05 )2 metre 2
V
= 2.55 × 104 amp/m2
I
2.55 10 4 J vd = = ne (5.8 10 28 )(1.6 10 19 )
Diode
(b)
= 2.69 × 10 –7 m/sec. Ex.13 The total momentum of electrons in a straight wire of length l = 1000m carrying a current I = 70A, will be – (in N.s) (A) 0.40 × 10 –6 (B) 0.20 × 10 –6 (C) 0.80 × 10 –6 (D) 0.16 × 10 –6 Sol (A)
V
I Semi
conductor
(c)
We know I = neAvd where vd drift velocity n
V
no. of density of electron. I
Total no. of electron N = nA l Total momentum (P) of electron = Nmv d or P = (nA l m)
I
neA
=
Torch
(d)
Il m
Bulb
e V
P =
70 1000 9.3 10
31
1.6 10 19
= 0.40 N.s 3. OHM'S LAW
(a) If there is no change in the physical state of a conductor (Such as temperature) then the ratio of the potential difference applied at it's ends and the current flowing through it is constant i.e.
(d) Units of resistance : ohm( ) 1 ohm = 1 volt / 1 ampere. (e) Dimensions of resistance : [M1L2T –3 A –2] (f) If, L = length of conductor R = resistance of conductor A = cross sectional area of conductor perpendicular to current Then,
R
L
, R
1 A
L A This constant of proportionality, is called 'Resistivity' or 'Specific resistance'.
I
R=
V
V
I
or
V = I R ;
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CURRENT ELECTRICITY
74
Note : Effect of stretching a wire on its resistance
Effect of temperature on resistance :
(i) If the length of wire is changed, then
(a)
l 12 = 2 l2
R1 R2
Rt
(ii) If the radius of wire is changed, then
R0
l 24 = 4 f 1
R1 R2
O
(g) Units of = Ohm–meter Dimensions of
Rt = R0 (1 + t) where, Rt = Resistance at t 0 C. R0 = Resistance at 00 C t = change in temperature.
= [M1 L3 T –3 A2]
(h) If T = Temperature in kelvin R = R 0 (1 + (T – T 0) where R 0 = Resistance at temp. T 0 and = thermal coefficient of resistance so, as T increases R increases.
= Temperature coefficient of resistance at 0º C = +ve for metals. = –ve for semi conductors and insulators. = 0 for alloys.
(i) Resistivity is also defined as the ratio of the intensity of the electric field E at any point within the conductor and the current density j at that point
E = j or
j
(b) R2 = R1 [1 + (t2 – t1)]. This formula gives an approximate value. (c) Resistance of the conductor decreases linearly with temperature and becomes zero at a specific temperature. This temperature is called critical or transition temperature, conductor becomes a super conductor at this temperature. (d) There is no loss of energy in a circuit formed by super conductors. Current passed in loop formed by superconductor will continue flowing for infinite time if there is no resistance in the loop.
E
(j) Resistivity is' characteristic property of the material of the conductor. It does not depend upon length area etc. of the conductor. Although it depends on temperatu re. It increases with increase in temperature (k) Value of resistivity is least for conductors and most for insulators. (l) Inverse of resistivity is called conductance of wire denoted by
=
1
Ohm's law
(m) Units of conductance : Mho
Ex.14
Important points :
(a) If a conductor is stretched to n times of it's original length, it's new resistance will be n 2 times older one (b) if x% of change is brought in length of a wire, it's resistance will change by 2x%. This is true for x < 5 only. (c) If a conductor is stretched such that it's radius is reduced to 1/nth of it's original values, then resistance will increases n4 times similarly resistance will decrease n 4 times if radius is increased n times by contraction –
t0C
Sol
In a wire of length 4m and diameter 6mm, a current of 120 ampere is passed. The potential difference across the wire is found to be 80 volt. The resistance of wire will be– (A) 0.15 ohm (B) 0.25 ohm (C) 6.660 ohm (D) none (A) From he definition of resistance
180 = 0.15 ohm 120 I Ex.15 Is the formula V = iR true for non–ohmic resistance also ? Sol Yes, this formula defines resistance and not ohm's law.
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R =
V
=
CURRENT ELECTRICITY
75
Ex.16
Sol
The resistance between two rectangular faces of a block of dimensions 4cm × 4cm × 10cm of maganin ( = 48 × 10 –8 ohm, m), will be (A) 4.8 (B) 3.8 (C) 30 (D) 3 (A) From R =
l , we have A
4.8 10 8 4 10 2
( 2.46) 31.4 (0.016) 2 RA = = l 1
31.4 (0.016)2
= 1.59 × 10 –6 ohm. cm Note :
resistance R is streched uniformly to a wire of radius (r/2). The new resistances will be – (A) 2R (B) 4R (C) 8R (D) 16R (D) The volume of given wire remains unchanged, hence
10 –6
1 volume = Area = 31.4 (0.006 )2
will be same for any shape of wire formed
A l = A' l ' R=
R =
l
(2l )
2
'
2 volt (i) E= V/d =2volt/50cm = 0.5 metre = 4 V/m (ii) J = i/A = 0.25 amp/10 3×10 –4 m 2 = 2.5 × 10 6 A/m 2
Sol
(iii) = J/E = (2.5 × 10 6 amp/m 2)/4volts /metre) = 6.25 × 10 5 mho/m. Ex.21 A copper wire is stretched to make, 0.1%
longer, the percentage change in its resistance will be – (A) 0.2 (B) 0.4 (C) 0.8 (D) 1.6 (A)
Sol
The resistance of a wire of length l , cross– sectional area A and resistivity is given by
2 l A R' = l A ' R
Further A l = A' (2 l ) [Volume remains conserved] A/A' = 2
2
2
and R' = , A A' = specific resistance.
l ' l and R' = A A'
r 2 r ' 2 R' r = 2 = 16 2 [ r = ] R 2 r ' r R' = 16R Ex.20 The potential difference across a wire of 10 –3 cm 2 cross–sectional area and 50cm length is 2 volt, when a current of 0.25amp exists in wire. Calculate– (i) field strength in the wire (ii) the current density, and (iii) the conductivity of the metal
Ex.18 A given piece of wire length l , cross sectional
Sol
(A'/A) = ( l / l ')
or
A R' A = = R A ' ' A '
by metal. area A and resistance R is stretched uniformly to a wire of length 2 l . The new resistance will be– (A) 2R (B) 4R (C) R/2 (D) Remains unchanged (B)
R' = 4 or R' = 4R R
Ex.19 A given piece of wire of length l , radius r and
Sol
= 4.8 × 4 10 10 4 Ex.17 If resistance of a wire formed by 1.cc of copper be 2.46. The diameter of wire is 0.32mm, then the specific resistance of wire will be– (A) 1.59 × 10 –6 ohm.cm (B) 2.32 × 10 –6 ohm.cm (C) 3.59 × 10 –6 ohm.cm (4)1.59 × 10 –8 ohm. cm Sol (A) length of wire R =
R =
l
...................... (A) A If d is density and m the mass of wire, then A l d = m or Area, A =
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m ld
CURRENT ELECTRICITY
76
Substituting this value in (A), we get
l
R =
But
(m / l d)
d m
=
d m
T1
l 2
= constant for a given wire, V
d Taking log, log R = log m + log l 2
d
+ 2 log l m Differentiating = log
T2
I
dR 2dl = R l
Sol
(A) T 1 = T2 (B) T 1 > T2 (C) T 1 < T2 (D) None (C) The slope of I – V curve I
R
or
I
R
i.e. slop e
I
R The slope of graph at temperature T 1 is greater than that at temperature T 2.
dl dR × 100 = 2 l 100 % R dl × 100 = 0.1% l Percentage (%) change in resistance = 2 × 0.1 = 0.2% i.e. the resistance increases by 0.2%. Ex.22 There are two wires of the same length and of the same material and radii r and 2r. The ratio of their specific resistance is– (A) 1 : 2 (B) 1 : 1 (C) 1 : 4 (D) 4 : 1 Sol (B) Specific resistance is a characteristic property of metal and doesn't depend upon dimensions of the wire used.
=
Resistance
at T 2 > Resistance at T 1 For metallic wire the resistance R increases with increase of temperature. Hence T2 > T1.
Given
Ex.25
The resistance of a tungsten filament at 150º C is 133 ohm. Its resistance at 500º C will be– (The temperature coefficient of resistance of tungsten is 0.0045 per ºC)
(C) 50 (A) 257 Sol
The resistance of wire is 50 then the graph between log v and log I is– (A) straight line passing through origin (B) parabola (C) hyperbola (D) none of the above. Sol (D) V = IR log V = log I + log R This is a straight line but not passing through origin. Ex.24 The current voltage graph for a given metallic wire at two different temperatures T 1 and T2 are shown in fig. Which is true– Ex.23
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(B) 79
(D) none of these
(A) If the resistance of a wire at 0º C be R 0 and at t0 C be Rt, then R t = (1 + t) or R 0 Rt = where is the temperature coefficient 1 t of resistance. The resistance of the filament at 150º C is 133 ohm. Therefore, its resistance at 0º C will be given by R 0 =
133 = 79.0 ohm 1 (0.0045) 150
Now, the resistance of the filament at 500º C will be R500 = R0 (1 + t500) = 79.0 [1+ (0.0045) x 500] = 257 ohm
CURRENT ELECTRICITY
77
Ex.26
Sol
The resistance of a conductor at 20º C is 3.15 ohm and at 100º C is 3.75 ohm. Determine the temperature coefficient of resistance of the conductor. The resistance of the conductor at 0º C will be– (A) 2 (B) 3 (C) 6 (D) 5 (C) If the resistance of the conductor at t 1 ºC be R1 and at t 2 ºC be R 2, then R1 = R0 (1 + t1), and R2 = R0 (1 + t2) R1 1 t1 On dividing : R = 1 t or R1 (1 + 2 2
t2)
R 2 R1 = R t R t 12 2 1 Here t1 = 20º C, R1 = 3.15 ohm, t 2 = 1000º C and R2 = 3.75 ohm 3.75 315 = (3.15 100) (3.75 20) 0.60 0.60 = = = 0.0025 per ºC] 315 75 240 Now from the formula R t = R0 (1 + t), we have = R2 (1 +
t1)
R0 = Rt/(1 +
or
t)
Note : If n resistance (each R) are connected in series there resultant will be nR
(iv) For a series combination of two resistances R1
V (A) equivalent resistance (B) I = V / (R 1 + R2)
R2 V (D) V2 (voltage across R 2) = IR 2 = R R 1 2 (b) PARALLEL COMBINATION :
A
V1
V2
V I
i2 i3
R2
B
(i) There is same drop of potential across each resistance. (ii) Current in each resistance is inversely proportional to the value of resistance i.e.
R3
V
V
V
i1 = R , i2 = R , i3 = R etc. 3 1 2
V3
V (i) Same current passes through each resistance. (ii) Voltage across each resistance is directly proportional to it's value. V1 = IR 1, V2 = IR2 (iii) Sum of the voltages across resistances is equal to the voltage applied across the circuit i.e. V = V1 + V2 + V3 + .............. V = IR1 + IR 2 + IR 3 + ..............
R1
V
(a) Series Combination
I
i1
R3
3.15 1 (0.0025) ( 20) = 3.0 ohm =
R2
R = R1 + R2
R1V (C) V1 (voltage across R 1) = IR 1 = R R 1 2
4. COMBINATION OF RESISTANCES
R1
R2
(iii) Current flowing in the circuit is sum of the currents in individual resistances i.e. i = i1 + i2 + i3 , V V V i = R + R + R + ........ 1 2 3
1 1 1 i 1 = = R + R + R + .... V R 1 2 3
where R = equivalent resistance. Note : (a) You are asked to find R and not
= R1 + R2 + R3 + ..................
= RWhere, R = equivalent resistance. Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159
1 in R
the question, so be careful. (b) The equivalent resistance of parallel combination is lower than the value of lowest resistance in the combination. CURRENT ELECTRICITY
78
(c) For a parallel combination of two resistances ....... R1 i1 i R2 i2
Sol
(A) Suppose the resultant resistance of the given resistance be R', then R' = R + (R + 1) + (R + 2) + ........ (R + n) = =
V(R1 R2 ) R1R2
(ii) If n resistance are connected in series and parallel respectively the ratio of their resultant will be nR : R/n = n2 .
Combination of resistances
Sol
=
1 2
=
r 22 2
×
Ex.29
Sol
Two wires of the same material having lengths in the ratio of 1 : 2 and diameters in the ratio 2 : 1 are connected with a cell of 6 volt and internal resistance 1 . The ratio of the potential difference across the two wires will be– (A) 1 : 2 (B) 2 : 1 (C) 1 : 8 (D) 8 : 1 (C) Since in series, current flowing will be same, IR1 V1 R1 thus V = IR = R 2 2 2
Ex.30
Sol
l
r 12
8
Resistance R (R + 1), (R + 2) ......... (R + n) are connected in series, their resultant resistance will be – n (A) (n + 1) R 2
n (B) (n – 1) R 2
(C) n (R + n)
(D) n (R – n)
Resistance R, 2R, 4R, 8R ..................... are connected in parallel. Their resultant resistance will be – (A) R (B) R/2 (C) 0 (D) (B) Let R' be the resultant resistance then 1 1 1 1 = + + + ...................... R' R 2R 4R
1 1 1 [1 + + + ............................. R 2 4
1 1 = 2 R' = R = 1 R 1 R 2 2
1
or V1:V2 = 1 : 8 Ex.28
The Resistance 4R, 16R, 64R, ............ are connected in series, their resultant will be– (A) 0 (B) (C) 4/3 R (D) 3/4 R (B) Let the resultant resistance of the given combination be R', then
=
r 12
2 1 1 = 2 × = 2
2
n [2R + n] = (n + 1) R 2
R' = 4R + 16R + 64R + .................. =
r 22
×
n 1
[2R + (n + 1) – 1]
d is common difference In the given question total terms are (n + 1)]
Note : (i) If n resistances (each R) are connected in parallel, their resultant will be R/n
Ex.27
2
[ Sum of n terms in arithmetic series is n Sn = [2a + (n – 1) d] 2 Where a is first term
V (i) i = i1 + i2 =
n 1
Ex.31
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The resistance of two conductors in series is 40 and their resistance becomes 7.5 when connected in parallel. The resistance are– (A) R1 = 30, R2 = 40 . (B) R1 = 20, R2 = 30 . (C) R1 = 30 , R2 = 10 . (D) R1 = 20 , R2 = 20 . CURRENT ELECTRICITY
79
(C)
Sol
(A) Let x the resultant resistance. If in the following small combination
Sol
In Series R1 + R2 = 40 ................ (A) R1R2 In Parallel R R = 7.5 1 2
1
Since (R1 – R2)2 = (R1 + R2)2 – 4R1R2 = 402 – 1200 = 400 R1 – R2 = 20 ..................... (B)
1
R
A
B up to n
(b)
C
R
x = 1 + R R
1
(a) is added, the value of x will remain unaffected. hence the resultant circuit will be as shown in fig b From fig (b) the resultant resistance
R R
1 1
1
1 (a)
Solving (A) and (B), we get R 1 = 30 and (B), we get R 1 = 30 and R2 = 10 Ex.32 In the following figure the resultant. Resistance between A and C will be– R
1
up to n
x
x 1.x + 1 = 2 + 1 x 1 x
= 1 +
3
Hence current (A) R
n 1 n
n 1 (B) R n
n2 1 (C) R n
n 1 (D) R n
2
I =
Sol
R R' = R1 + R2 = nR + n Ex.33 In the following fig, the current drawn by the battery of 12V supply (in amp) will be –
12V
3 1 A A
complete circle. It's resistance between two diametrically opposite points will be (in )–
n 2 1 = R n
+A
3) = 6
Ex.34 A wire of resistance 10 is bent to form a
(A) The resistance are connected in series between the points A and B and those between B and C are in parallel. Let R 1 and R 2 be the resultant of these two combinations, then R1 = nR and R2 = R/n
Sol
V = 12/(1 + R
(A) 3.5 (C) 2.5 (C)
(B) 5 (D) 1.5 5
A
C 5
Let's find the resistance between A and C. Then the configuration is similar to D .
C
A
B –B
Hence, equivalent resistance
(A) 6 ( 3 1)
(B) 6 ( 3 1)
(C) 12 ( 3 1)
(D) 12 ( 3 1)
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55 = 2.5 55
CURRENT ELECTRICITY
80
Ex. 35
Ex36
Find the equivalent resistance and hence current supplied by the battery in the diagram shown – 4 4 4 g c e a 3
6V
3
Find effective resistance between points A and B in figure. 3 D E 3 3 6
6
2
6
3 b
d
Sol
(1)
Solving the circuit one by one as follows. Hence current supplied by 6V battery = = 1A. 6 4 4 g c e a 3
6V b
3
d c
Sol
Reducing the figure as follows in order from 1 to 8 3 (1) D E 3 6
6
6
e
(2)
d
(3)
e
D
b
6
6
6
4
D
c 3
(4) 2
6V
a
3 A
B
3
c
C
(5) 6V
3
6
(4)
d
b
3 B
3
A
f
d
3 C
6
3
3
B
3
f
c
6V
a
C
(3)
4
a
3
6
A b
D
2
3
B
6
3 6V
3
3
3
E
(2)
C
6
A h
4
B
A
6
f 4
a
3
h
f
3
6
(5)
b d 6V Hence current supplied by the battery = 6 = 1A Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159
6 6
A
3
3
B
CURRENT ELECTRICITY
81
(6)
C
3
A
6
(7)
3
B
(8)
B
3
3
A
B
A
5. COMPARATIVE STUDY OF COMBINATION OF RESISTANCES AND CAPACITORS Resistance
S.
Capacitors
No.
In series (i)
I
In parallel
R1
R2
R3
V3
V2
V1
+ – E
In series +q
C1 V1
+ –
+q
C2 V2
+q
In parallel +q1
C3 V3
V
C1
-q1 +q2 C2
Earth A
C3
+q3 -q3 V
E
B
-q2
(ii)
R = R 1 + R2 + R 3
1 1 1 1 R R1 R 2 R 3
1 1 1 1 C C1 C 2 C 3
(iii)
V = V1 + V2 + V3
V 1 = V 2 = V3 = V
V = V1 + V2 + V3
V1 = V2 = V
Earth
C = C 1 + C2 + C3
(iv) Current is same in all the resistances
Currents through different but p.d. across
Charge is equal in all capacitors
Different capacitor diff. charge.
(v)
If n resistances, each of value R are
all of them is the same If n resistances each of value R are
If n cap. are in series then C' = C/n
If n capacitor are connected in parallel
connected in series then R' = nR
connected in parallel, then R' = R/n
(vi) Effective resistance is greater than the highest resistance in the combination
Effective resistance is less than the smallest resistance in the combination
C' = nC Effective capacitance is Effective capacitance less than smallest is greater than the capacitor in the combination
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largest capacitor in the combination
CURRENT ELECTRICITY
82
6. KIRCHOFF'S LAWS
(i) Let i be current following through 2 resistance. This current is divided into two parts i 1 and i2 at C as shown in fig. The effective resistance of 6 and 9 resistance in series = 6 + 9 = 15 . The p.d. between CD, V = i1 × 15 = i 2 × 5
Sol
Kirchoff in 1842 gave two laws for solving complicated electrical circuits. These laws are as follows– (a)
First law :
In an electrical circuit, the algebraic sum of the current meeting at any junction in the circuit is zero. OR Sum of the currents entering the junction is equal to sum of the currents leaving the junction
i
i
i = i1 + i2 1 or i1 = 4
Kirchoff's laws In the following figure the 5 ohm–resistance develops heat 10.24 cal's due to a current flowing through it. Calculate – (i) heat developed per second in the 2 resistance and (ii) the potential difference across 6 resistance.
D
B
or i2 = 3i1
i1 = i2 – i – 3i1
3i 4 Heat developed in 5 resistance per
i2 = 3i1 =
2
3 i 5 2 i1 R 4 second H = cal = 4.2
E
(a) In applying this law, when we traverse in the direction of current then the product of the currrent and the corresponding resistance is taken as positive, and the emf is taken as positive when we traverse from the negative to the positive electrode of the cell through the electrolyte. (b) This law is based on 'law of conservation of energy'.
i1 5 1 = = i2 15 3
As
4. 2
(3 i / 4)2 5 = 10.24 cals –1 4 .2 Solving, we get i = 3.92 A. Heat developed in 2 resistor per sec But given H=
Important notes
Ex.36
5
This law is based on law of conservation of charge. In other words, when a steady current flows in a circuit then their is neither accumulation of charge at point in the circuit nor any charge is removed from there. Second law : In a 'closed' mesh of a circuit the algebric sum of the products of the current and the resistance in each part of the mesh is equal to the algebric sum of the e.m.f.'s in that mesh. i.e. =
C
i2
Note :
iR
A
= 0
i1 – i2 – i3 – i4 + i5 = 0 or i1 + i5 = i2 + i3 + i4
(b)
2
9
6
i1
i2 R (3.92)2 (2) cal = =7.31 cal. 4.2 4.2 Potential difference across 9 resistor sec = current × resistance =
(ii)
1 = i1 × 6 = × 6 = 5.86 volt. 4 Ex.37 The value of current i in the following circuit is– 4A 2A B A i 3A C 1.3A
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(A) 2.7 A (C) 3 A
(B) 3.7 A (D) 4A CURRENT ELECTRICITY
83
Sol
(B) i = ((4 + 3) – 2 – 1.3) = 3.7 A
Ex.41
Using kirchoff's first law at A, B and C. Ex.38
Write KCL (first law) and KVL (second law) for the circuit shown below : E1
1 A
Ex.39
i2
R2
E2
2
2 i1
i1
R1
Sol
R3
p B
a
Ex.40
2.8
KCL at point A : i 1 + i2 + i3 = 0 KVL for loop 1 : i 1 R1 – i2 R2 = E1 – E2 KVL for loop 2 : i 2 R2 – i3 R3 = E2
6V
Sol
(C) Mass and charge. (D) Charge and mass (A)
a i2
i1 E1
i1 b
i3
E 6 = = 1.5A R 4 Potential difference across pq = iR' = 1.5 × 1.2 = 1.8V Current drawn from cell, i =
i3
R2= 3.5 E2 E2= 6V
R1
Sol
R1
V 18 Current in 2 resistor, i1 = = = 0.9A 2 2
R1
1 4
(A) 1.4V
(B) 4.6V
(C) 3.6V (B)
(D) 5.6V
(A) 1.5 A (B) 0.9 A (C) 1.2 A (D) 1.3 A (B) In steady state the branch containing capacitance acts as the open circuit since capacitance offers infinite resistance to d.c. The capacitance simply collects charge. The effective resistance of 2 and 3 resistors connected in parallel is R1R 2 23 6 R' = R R = = = 1.2 23 5 1 2
The potential difference between points a and b in the circuit of figure is – (i1 = 1.0 A, i 3 = 0.6 A) R1
4
C
i3
Kirchoff's first and second laws respectively show the conservation of
q
3 i2
i
(A) Charge and energy (B) Energy and charge
Sol
The steady state current in a 2 resistor shown in fig will be– (The internal resistance of the battery is negligible and the capacitance of the condenser C is 0.2 F)
Ex.42
In going from a (potential V a) to b (potential Vb) we have Va – I2R2 – E2 = Vb Va – Vb = E2 + I2 R2 = 6 + (– 0.4) × (3.5) = 6 – 1.4 volt = 4.6 volt (from above example I 2 = – 0.4 Amp.)
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From the fig. determine (i) potential at A, (ii) potential at C, and (iii) reading of the voltmeter connected across the 10V bettery –
i
A
10V 1
B
C
16V 0.5
0.5
4 F
G
D
CURRENT ELECTRICITY
84
The current in circuit is (consider loop (CBAFGDC)
Sol
I =
E 2 E1 16 10 = r 1 r 2 R1 R 2 1 0.5 4 0.5
6 = 1A 6 (i) V A – VF = IR = 4 volt Because VF= 0(grounded), therefore V A=4 volt (ii) VD – VC = 1 × 0.5 = 0.5 volt
Combinations of cells :
E1
E2
r 1
0 (grounded), So VC = – 0.5 volt (iii) The 10V battery is being charged therefore V = E + Ir = 10 + 1 × 1 = 11 volt 7. CELLS (a) Electro Motive Force (EMF) : The potential
difference across the terminals of a cell when it is not giving any current is called EMF of the cell. or The energy given by the cell in the flow of unit charge in the whole circuit (including the cell) is called the EMF of the cell. W E = Q (b)
Terminal voltage :
(i) The resistance offered by the electrolyte of the cell to the flow of current through it is called internal resistance of the cell. (ii) When current is drawn through the cell or current is supplied to cell then, the potential difference across its terminals is called terminal voltage. (iii) When i current is drawn from cell, then terminal voltage is less than it's emf E. V = E – i r E r
r 3
i
=
VD =
E3
r 2
R (a) Series Combination :
(i) Equivalent emf E = E 1 + E2 + E3 ......... Note : Direction of emf is taken into consideration. (ii) Equivalent internal resistance r is given by r = r 1 + r 2 + r 3 ............. (iii) Current, i =
Ei E = r R r R i
Imp :
(iv) For maximum current, R = r i.e. The load resistance must be equal to the equivalent internal resistance. (v) If all emf are equal (E), then for series nE combinations of n such cells, I = R nr E Cases : (a) if nr >> R, I = r nE (b) If nr << R, I = R (c) Cells are employed in series only when internal resistance is less than the load resistance. (b) Parallel Combination :
(i) Equivalent internal resistance, r is
I I I + r + r + ......... r 1 2
(ii) Equivalent emf E1
i E3
Where V = terminal voltage, r = internal resistance of battery (iv) When current is supplied to the cell, the terminal voltage is greater than the emf E i.e. V = E + i r (v) Units of both emf and terminal voltage are volt.
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E2
E4
r 1
r 2 r 3 r 4
R
E1 E2 E3 E ........... i r 1 r 2 r 3 r i E = 1 1 1 = 1 ............... r1 r2 r 3 r i CURRENT ELECTRICITY
85
(iii) Current, i =
(B)
Sol
E R r r
E
R (iv) When all 'n' cells with emf E and internal resistance r each, are connected in parallel, then equivalent emf = E, equivalent internal r resistance = n E nE (v) In this (5) case I = R r = nR r n
5A
Potential drop across internal resistance = 0.1 × 5 = 0.5V Hence, potential difference across terminals will be 2 + 0.5 = 2.5V 8. WHEAT STONE BRIDGE
B
(c) Mixed combination :
mnE , For maximum current mR nr Internal resistance = External resistance i=
i.e. R =
nr m E
i1 – i g Q ig
p
i1 A
i2
C
G S i2+ i g
R
i = i 1+ i2
i
D
Cases :
E (a) If r << nR, I = R nE (b) If r >> nR, I = r (c) This combination is used only when load resistance is lower than internal resistance.
0.1
2V
E + – (a) The configuration in the adjacent figure is called wheat stone bridge. (b) If ig = 0 i.e. current in galvanometer is zero, then bridge is said to be balanced. (c) For ig = 0 (i) VD = VB
(ii)
P R = Q S
(d) Equivalent resistance in balanced condition (P Q) (R S) = PQRS P R (d) If < then VB > VD and current will flow Q S from B to D. P R > , the VB < V D and current will flow Q S from D to B. (f) Meter bridge and post office box work on this principle.
(e) If
r i
Wheat stone bridge Ex.44
R
In the following figures, the resistance A and B will respectively be– C
Ex.43 A ba tt er y of em f 2 vo lt s an d in te rn al
resistance 0.1 is being charged with a current of 5A. The potential difference between terminal of the battery is? (A) 1.5V (B) 2.5V (C) 3.5V (D) 4.5V Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159
R
R
A
B
R
R D CURRENT ELECTRICITY
86
R
R/2 (4,5)
1
a
R
A
Sol
R
b
R
c
d
a (A) R/2, R (C) R, R (C) A
B
R
R
This is a balanced bridge (P/Q = R/S) For this bridge, we can ignore the resistor between b and d. hence both the above fig can be put as fig (iii). Obviously, the resistance between A and B will be R (D) Equivalent resistance in balanced (P Q)(R S) condition = PQRS Ex.45 Figure shows a cube made of 12 resistances, each of resistance R. The equivalent resistance across a and b will be – 8
1
a
4 7
6
b
3
2
(A) (7/12) R (B) (12/7) R (C) (9/5) R (D) (5/9) R (A) The given combination of resistance can be reduced as follows From the consideration a
1
R
R
A
Sol
7R 5
(B)
2
R
(3, 6)
(C)
6R 4
(D)
4R 6
(A) This is Wheatstone bridge but is unbalanced. To find equivalent resistance, we imagine, that a cell of emf E is connected between points A and B. Then the combination look as following figure. For the loop ACDA 2i = 3i1 + i2 ......(A) For the loop BDCB i = 3i 1 – 4i2 ......(B)
B (i-i1+i2) 2R (i1-
i)
R R
i2
2
D
i1
¯ E +
2R
R
R
R
5R 7
R
R
D R
A b
R
R
R R
R
C
(A)
B R
R
8
R
2 R/2 (3, 6)
R
R
(4, 5)
2R
The equivalent resistance between points A and B in the following circuit will be–
Ex.46
5
R/2
of symmetry alone, we notice that points 4 and 5 must be at the same potential, so must be points 3 and 6. This implies that the circuit can be redrawn with point 4 and 5 connected, and 6 and 3 connnected, as shown in. This figure further reduces to the combination shown in fig. The equivalent resistance between a and b is = (7/12) R
R
b
(B) 2R, R (D) 3R, R R
Sol
R
B
(i - i1)
i
7
R
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CURRENT ELECTRICITY
87
Solving considering loop ADBEA the above two equations, we get i 1 = (3/5) i, i 2 = (1/5) i E = (i – i2) 2R + (i – i 1 + i2) R = (4R/5) i + (3R/5) i E = (7R/5) i Therefore the effective resistance is Req = E/i = 7R/5
units of power = joule/sec, watt, horse power 1 watt = 1 joule/sec, 1 HP = 746 watt unit of electrical energy = watt second, kilowatt hour 1 kilowatt hour (kwh) = 36 × 10 5 Joule Combination of electrical instruments–
(a)
9. ELECTRIC ENERGY AND POWER Electric energy :
When a potential difference is applied across a wire, current starts flowing in it. The free electrons collide with the positive ions of the metal and lose energy. Thus energy taken from the battery is dissipated. The battery constantly provide energy to continue the motion of electron and hence electric current in the circuit. This energy is given to ions of the metal during collision and thus temperature of wire rises. Thus, energy taken from the battery gets transferred in to heat. This energy is called electrical energy. This effect is also called 'Heating Effect of Current'. If R = Resistance of wire I = Current in wire V = Potential difference across wire. Flow of charge in 'dt' time = Idt.
given by (b)
V2 ( 220) 2 = ohm P 40
Series combination
V2,P2
V3,P3
E (i)
If total power dissipatted if P, then
(ii)
(c) 2
V dW = VIdt = I 2Rdt = dt = Vdq R This energy is equal to work done by battery or heat produced in the wire.
R =
V1,P1
Energy dissipated dW = Vdq = VIdt, V = IR,
If 220V and 40W is written on an electrical instrument then this is called it's standard Ratings. It means that if 220V is applied across this instrument then 40W of power will be generated. Thus the resistance will be
1 1 1 1 = P + P + P , P 1 2 3
Where P' 1 s are standard powers of instrument In this combination, the bulb with least power will glow most and bulb with highest power will glow least or we can say that bulb with highest R will glow brightest and bulb with least R will glow least. Parallel combination
V1,P1 V2,P2
If energy is to be written in calorie–
V3,P3
dW cal = 24 dW cal 4.2 When dW is energy in Joules. Then dW =
E
Electrical power :
The rate of loss of energy in an electrical circuit is called electrical power. It is denoted by'P' P =
dW V2 = I2 R = IV = dt R
(i) Net power dissipation P = P1 + P2 + P3 (ii) Bulb with least power will glow least or the bulb in which maximum current is flowing will glow brightest and vice–versa.
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CURRENT ELECTRICITY
88
Note :
high temperature. On connecting with a
(a)
These formulae are applicable only if the voltage ratings of all the instruments are equal along with the power source. If voltage ratings are different then circuit is solved by considering equivalent resistances of the instruments as follows. (b) Replace the instrument by its's equivalent resistance. If standard rating is (V/P) then it resistance is R = V2/P (c) Find the currents and voltages in different branches using kirchoff's first and second laws. (d) If rating of a bulb is changed form V 1/P1 to V2/P2 then V12 V22 = = R P1 P2
or
P2 =
V22 V12
2–volt battery, the filament will be at ordinary temperature and now its resistance will be less than the resistance determined. Ex.48
The resistance of each of the three wires joined as in the fig 2 ohm, and each one can have maximum power of 18 watt (otherwise it will melt). The maximum power the whole circuit can take is – 2 2 2
P1 Sol
Electric Energy and Power
maximum current in any wire of the circuit
Ex.47 A 60–watt lamp is operated at 220–volt power
line. What will be resistance when the lamp is lighting ? How much current will flow in it? Estimate, how much current will flow in the lamp when it is connected to a 2–volt battery? If the resistance of the lamp is R ohm and it. Lights at V volt, then the electric power V2 consumed (in watt) is given by P = R Here P = 60 watt and V = 220 volt
Sol
( 220 ) 2 V2 R= = = 806.7 ohm 60 R According to Ohm's law, the current flowing in the lamp is
i =
P = R
18 watt = 3 ampere. 2ohm
This circuit has two 2– wires in parallel and a third 2– wire in series with this parallel combination. Hence the equivalent resistance of the whole 2 2 circuit is R' = + 2 = 3 . 2 2 maximum power expended in the whole circuit is Pmax = i2 R' = (3) 2 × 3 = 27 watt Ex.49 A heater is designed to operate with a power
220 V = = 0.27 ampere 806.7 R On connecting the lamp with a 2–volt battery, the estimated current is
i=
i =
(A) 27 W (B) 9 W (C) 81 W (D) 18 W (A) Electric power expended in a wire of resistance R is P = i 2R
of 1000 watt in a 100–volt line. It is connected to two resistance of 10ohm and R ohm as shown in the fig. If the heater is now giving a power of 62.5 watt. The value of the resistance R, will be – 1000 watt 10 HEATER
2 V = 806.7 R = 0.0025 ampere.
Note : This is only an estimation, not the actual current. The acual current will be slightly larger. The reason is that we have determined the resistance of lamp at very Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159
R
100 volt (A) 5 (C) 2.5
(B) 10 (D) 1.25
CURRENT ELECTRICITY
89
(A) Suppose the resistance of the heater is r.
Sol
1796 .6 5 30 = Rs. 269.19 1000 cost of electricity = 269.19 × 0.40 = Rs. 107.68 Your stage : Now you may solve the questions from Q. No. 43 to 54 of Ex.# 1
= Rs.
V2 (100)2 or 1000= or r = 10 ohm r r Suppose, on connecting as shown in the fig, the potential difference between the ends of Then P =
the heater is V'. Then 62.5 = V' = 25 volts
V2 r
or
POINTS TO REMEMBER
potential difference across 10 = 100 – 25
1.
75 volt = 75 volt and current in 10 , i = 10 ohm
In liquids, the charge carriers are positive and negative ions.
2.
In gases, the charge carriers are positive ions and free electrons.
3.
In semiconductors, the charge carriers are holes and free electrons. The conventional direction of flow of current is opposite to the direction of flow of electrons.
4.
Current is a scalar quantity. Current density is a
= 7.5 ampere. A part of this current goes into the heater and the rest goes into R. Now, current in the heater, i = =
25
= 2.5 ampere
current in R = 7.5 – 2.5 = 5.0 ampere
R = 5 .0 ampere = 5 ohm
Ex.50
Sol
62.5
P V'
vector quantity. Direction of j is the same as
25 volt
the of E .
In a house there are 12 bulbs each of resistance 440 ohm, one motor of 1/10 horse– power, and 4 fans each of 100 watt. These are operated 5 hours per day. The expenditure of electricity in a month of 30 days will be– (1 H.P. = 746 watt, cost of electricity is 40 paisa per unit and voltage of power is 220 volt. (A) Rs. 269.19 (B) Rs. 107.68 (C) Rs. 1794.6 (D) Rs. 179.46 (B) Electric power expended in each bulb 2
5.
Reciprocal fo slope of V – I graph is equal to resistance.
6.
The resistance of conductor depends on the material of the conductor, shape and size of the conductor as well as on the physical state (Temperature) of the conductor.
7.
Reciprocal of resistance is called conductance, = I/R.
8.
Fu se W ire : Fuse wire is used in a circuit to
control the maximum current flowing in a circuit. It is a thin wire having high resistance and is made up of a material with low melting point.
2
(200 ) V = 440 R electric power expended in 12 bulbs = 12 x 110 = 1320 watt Electric power expended in 4 fans = 4 × 100 = 400 watt
=
Electric power expended in a motor = H.P. =
1 10
1 × 746 = 74.6 watt 10
While calculating the resistance of a wire by the formula R = ( l /A) it should be remembered that 'A' is the area, normal to the direction of current flow and ' l ' is the length in the direction of current flow.
10. Free electron density in a metal is given by n =
total
power expended = 1320 + 400 + 74.6 = 1794.6 watt Units consumed in 30 days =
9.
watt hour 1000
(Nx.d/A) where N Avogadro number, x number of free electron per atom, d = density of metal, A Atomic weight of the metal, n is of the order of 10 23 per cm 3.
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CURRENT ELECTRICITY
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11. For carbon, India rubber, mica, electrolytes and
22. In series resistance circuit, it should be
insulators the value of negative. It means that their resistance decreases with increase in temperature.
remembered that : (i) The current is same in every part of the series circuit. (ii) The total resistance in the circuit is the sum of the individual resistances including internal resistance of the cell (if any) (iii) Voltage across any part of a circuit is proportional to the resistance of that part. (iv) Current in the circuit is independent of the relative positions of the various resistance in the series. (v) The effective resistance of n equal resistances each of resistance r in series is R = nr.
12. Resistance of a conductor increases with
decreases in density or when it is subjected to mechanical stress. 13. The direction of current density is the direction of
motion of positive charge at that point. 14. House wiring circuits are in parallel therefore the
voltage across each bulb is constant. The power of the bulb is given by the formula : P = V 2/R. For constant voltage P (1/R) therefore, the greater the resistance, the smaller is the power. Hence, if we take two bulbs of 60W and 100W, then the resistance of 60W bulb will be more than the resistance of 100 watt bulb. 15. The filament of 60W bulb is thinner than the
filament of 100 watt bulb. 16. If the two bulbs, one of 60W and the other of 100
W are connected in parallel, then 100 W bulb will glow more. 17. If two bulbs, one of 60W and the other of 100W
are connected in series, then the 60 watt bulb will glow more, because in series combination current through both the bulbs is the same and the power consumed P = I 2R or P R. Now, since the resistance of 60W bulb is more than the resistance of 100 watt bulb, 60W bulb will glow more.
(vi) If 1
then = E. Hence, sp. resistance is the electric field per unit current density. Greater the current density, greater is the field. 20. If two bulbs of power P 1 and P2 are connected
in parallel and the rated voltage is applied, then the total power consumed is P = P 1 + P2. 21. If two bulbs of power P 1 and P2 are connected in series and the rated voltage is developed across each bulb, then the total power consumed is 1/P = (1/P 1) + (1/P2)
1
1
1
23. In parallel combination of resistance it should be
remembered that : (i) Total current through the combination is the sum of individual currents through the various branches. (ii) The potential difference across all the resistance is the same. (iii) The current through each branch is inversely proportional to the resistance of that branch. (iv) The reciprocal of the total resistance of the combination is equal to the sum of the reciprocals of the individual resistance. (v) If two resistances R 1 and R2 are in parallel, then the currents I 1 and I2 in them will be R2I R1I distributed as I1 = R R , I2 = R R 1 2 1 2
and maximum for insulators.
= E/J, if j = 1amp/m 2,
is the effective conductance then
= + + ................... 1 2 3
18. Specific Resistance is minimum for conductors
19. The specific resistance
where, I is the net current that flows through the circuit. (vi) The total resistance of n equal resistances each of resistance r, in parallel is : Rseries (vii) For n equal resistances R = n2. parallel
is the effective = + 2 + 3
(viii) If
conductance then
24. Using 'n' conductors of equal resistances the
number of combinations one can have, using all at a time is 2 n–1.
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CURRENT ELECTRICITY
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25. If the resistance of 'n' conductors are entirely
different, then the number of possible combinations are 2n. 26. Resistivity of a conductor changes with impurity.
The impurity increases the resistivity, resistivity of an alloly is greater than the resistivity of its constituents.
38. For some materials
is nearly zero. There is no
change in resistance with temperature. 39. The temperature coefficient of thermistors is negative and it is high. 40. The direction of flow of current in the cell i s from negative electrode to the positive electrode, while outside the cell it is from positive electrode to the negative electrode.
27. Resistivity increases with temperature. 41. Open circuit means, that no current is being drawn 28. Mechanical stress increases the resistivity.
from the cell.
29. Resistivity of most of the metals increases on
heating.
42. Closed circuit means that current is being drawn
from the cell.
30. Resistivity of Antimony and Bismuth and semi–
conductors decreases with increase of temperature.
43. Energy is consumed inside the cell due to its
internal resistance. 44. Due to the presence of internal resistance in the
31. When the charge is stationary, the energy stored
is calculated by the formula
VQ while if the 2
charge is flowing then the energy dissipated is calculated by VQ. 32. When a wire is drawn to n times the new length,
then its Resistance becomes n 2R where n is the original resistances.
cell, there is a potential drop inside the cell. 45. E.m.f. (E) is the characteristic of each cell and
its value remains constant for the cell, while the potential difference (V) goes on decreasing on taking more and more current from the cell. 46. While charging a cell the positive terminal of the
charger should be connected to the positive terminal of the cell and the negative terminal of the charger should be connected to the negative terminal of the cell.
33. When wires are drawn from same volume but 47. While charging a cell, we send a current in the
with difference area of cross–section, then Resistance Resistance
1 ( Area of cross sec tion)2
or
1 (radius) 4
34. When a wire is folded n times on its own length
to 1/nth length then the new resistance becomes R/n2 (R is the initial resistance of the wire) 35. Temperature coefficient of materials is given by
=
R t R0 R0 t . Its units are per ºC.
36. If
is
37. If
positive the resistance increase with increase in temperature. is negative the resistance decrease with increase in temperature.
cell by some other electric source (as a battery), called charger then the direction of current inside the cell will be from the positive plate to the negative plate. In this case the potential difference V between the plates of the cell will be greater than the emf of the cell. V = E + ir. 48. Greater the length of the potentiometer wire,
smaller is the potential gradient (k = V/I) and more is the balancing length hence, more is the accuracy of observations. 49. Potential difference across the terminals of the
cell when current is drawn from it is V = E – Ir. 50. Potential difference across the terminals of the
cell when the cell is charged V = E + Ir.
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CURRENT ELECTRICITY
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51. Internal resistance of the cell r =
E V R, V
where E = emf of the cell, V = Potential difference across the terminals of the cell, R = external resistance. 52. Energy supplied by the cell in t seconds = Eit
(if r = 0), = Vit (if r is finite). 53. Energy dissipated in the internal circuit of
resistance R' =
i2
rt.
54. Energy dissipated in the external circuit of
resistance R' = i2Rt. 55. If the diameter of the potentiometer wire is not
uniform, then the value of potential gradient will not be uniform throughout the wire. 56. The current flowing in the circuit if a cell of emf
E and internal resistance r is connected to an E external resistance R is I = r R 57. If two cells of emf E 1 and E 2 and internal
(e) The common terminal voltage V =
E1r 2 E 2r 1 r 1 r 2
62. When an external resistance R is connected to
a battery of emf E and internal resistance r, the power transferred from the cell to the resistance is maximum when r = R. 63. Nichrome is used as electrical heater element
because of its high specific resistance, high melting point and low temperature coefficient of resistance. (i) Nichrome is an alloy of Nickel, chromium and copper. (ii) Manganin is an alloy of copper, manganese and Nickel. (iii) Constantan is an alloy of copper and nickel. (iv) Fuse wire is an alloy of lead and tin. 64. Heater wire has high resistance and high melting
point. 65. Heater wire is always connected parallel to the
mains.
resistance r 1 and r 2 are connected so as to support each other, then the current in a resistor
66. Fuse wire is always connected in series in the
E1 E 2 R is I = r r R 1 2
67. In resistance boxes Manganin wires are used as
58. If the two cells oppose each other, then I =
resistances used to measure the resistance of a conductor.
59. In a battery of N cells each of emf E, if n cells
are wrongly connected, then the net emf of the battery is E' = NE – 2nE as one wrongly connected cell destroys one more cell. emf of ch arg er emf of cell total resis tan ce of the circuit
61. When cells of emf E1 and E2 are connected in
parallel (a) net emf = E 1 – E2 if E1 > E2. (b) Total internal resistance is (r 1 + r 2).
E1 E 2 r 1 r 2 (d) As E 1 > E2, E1 is discharging while E2 is charging. (c) Current i =
their temperature coefficient is negligibly small. 68. Wheatstone bridge is the arrangement of
E1 ~ E 2 r 1 r 2 R
60. Charging current =
circuit.
69. Bridge is balanced if P/Q = R/S. 70. At balance, the potential difference across the
galvanometer arm is zero and the current through the galvanometer is also zero (ig = 0). 71. The bridge is most sensitive when the current in
all the four branches of the bridge is of same order. 72. The balance is not effected on interchanging the
position of battery and galvanometer, 73. Post office box works on the principle of
Wheatstone bridge and it is used to measure the resistance of a wire, but its sensitivity is small.
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CURRENT ELECTRICITY
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74. Metre bridge is an instrument to measure the
80. The resistance of potentiometer can be
resistance of a wire accurately. It works on the principle or W heatstone's bridge.
considered as infinity while measuring the emf. 81. The potentiometer and vaccum tube voltmeter
75. Meter bridge is more sensitive than a post office
(VTVM) are ideal voltmeters.
box. 82. The potentiometer of VTVM do not draw any 76. The drawback of a meter bridge is the appearance
of end resistances. 77. The effect of end resistances is reduced by
interchanging the gaps. 78. Complete elimination of end resistances is
possible in Carey Foster Bridge which has four gaps instead of two. 79. Potentiometer is an instrument used to measure
accurate potential differences.
current from the points across which the potential diff. is measured. 83. If
1 and 2 are
the balancing lengths for two
cells of emf E1 and E2, then E1 / E2 =
1 2
.
84. If 1 and 2 are the lengths in open and closed
circuit across a cell and R is the external resistance then the internal resistance of the cell
2
1 is. r = R 1 .
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CURRENT ELECTRICITY
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SOLVED
EXAMPLES b
Ex. 1 A cylindrical wire is stretched to increase its
Sol
length by 10%. The percentage increase in the resistance of the wire will be– (A) 20% (B) 21% (C) 22% (D) 24% (A) Let l 1 be the initial length of the wire. Then the new length will be
R a
d
2 A 1
10 10 10 = × = 11 11 11
R
R
1 1 1 1 3 i.e. R = + + = eq 2R 2R 2R 2R
11 A1 l 1 = A2 l 2 or A1 /A2 = l 2/ l 1 = 10 (where A1 and A2 are initial and final area of cross–section of the wire). If R1 and R2 are the initial and final resistances, then 1 A 2
c
R
110 11 l = l = l 2 100 1 10 1 Since, the volume remains constant
R1 R2 =
Req = Ex. 3
2 R 3
In the adjoining network of resistors, each is of resistance r ohm, the equivalent resistance between pointsbA and B is–
2
r
r
or
r A B
a r
2 R2 11 R1 = 10
d
r
c
Now, percentage change in resistance is
11 2 R 2 R1 1 × R1 × 100 = R1 ×100 = 10
(A) 5r (C) r
R
(B) 2r/3 (D) r/2.
Sol (C)
b
100 = 21% Ex. 2
The equivalent resistance between A and B is– R A
R
R
R
R
r
r A
a
r
B
r
r
B
c
d
R R
Sol.
R
(A) 2R/3 (B) R/3 (C) R (D) 3R (A) The circuit is equivalent to Fig. It is a balanced wheatstone bridge between abcd, and then in parallel (2R) resistances. Thus ignoring resistance between bd arm. The circuit is equivalent to three (2R) resistances in parallel (abc, adc, aRRc).
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Imagine, A being pulled on the left side, then abcd becomes a balanced wheatstone bridge Fig. The arm bd can be ignored. Then resistance between A, B becomes = r. 1 1 1 1 i.e. R = + = eq 2r 2r r
Req = r
CURRENT ELECTRICITY
95
Ex. 4
In the following fig. the ratio of current in 3 and 1 resistances is– 3
Ex. 5
A
x
P R = Here P = 100 , R = 200 , S Q S = 40, Q = ? 40 S Q= .P = × 100 = 20. 200 R' That is in arm BC, the net resistance should be 20, but the arm contains a combination of resistances 100 and R in parallel, therefore, we have
Z
Y 6 2 (B) 3 (D) 2
The current in 1 resistance is 3A. The current R2 in 3 resistance is I1 = R R I 1 2 6 = × 3 = 2A. 36 2 Therefore the ratio is . 3 The resultant resistance between the points A and B in the following diagram Fig. will be – 1 1 1 1
2 B
If P, Q, R, S are resistance of Wheatstone's Bridge, then in balanced position.
3A 1
1 (A) 3 (C) 1 Sol (B)
Sol (A)
2
(A) 4 (C) 6
2
1 1 1 1 1 1 = + or = – 20 100 20 100 R R
=
R=
Sol
Sol (D)
(R1 R 2 ) 1 + [(R1 + R2)2 + 4 R3 2 2 (R1 + R2)]1/2 .....(A) R1 = 1, R2 = 0. R 3 = 2. ......(B) From eqs. (A) and (B) 1 1 R = + [1 + 4 x 2 x 1]1/2 2 2 1 = [1 + 3] = 2. 2 Fig. represents a balanced Wheatstone's Bridge. The value of resistance R will be– B R 100 100 A C R=
Ex. 6
40
200 D
(A) 25 (C) 100
+ –
100 4
= 25.
Ex. 7 An electric current of 5 amp. is divided in
1
(B) 8 (D) 2
5 1 4 = 20 100
(B) 30 (D) 200
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three branches forming a parallel combination. The length of the wire in the three branches are in the ratio, 2, 3 and 4 ; their diameters are in the ratio 3, 4 and 5. Find the currents in each branch if the wire are of the same material. Let the length of three wires be 2 l , 3 l and 4 l respectively and their radii be 3r, 4r and 5r respectively. If S be the specific resistance, then 2l 3l R1 = S R2 = S 2 , (3r ) ( 4r )2 4l 2l and R3 = S , 2 or R1 = S (5r ) 9 2 3l 4l R2 = S 2 and R3 = S 16 252 2 3 4 R1 : R2 : R3 = : : 9 16 25 The ratio of their currents must be inverse of the above, i.e. 16 9 25 i1 : i2 : i3 = : : 3 2 4 or i1 : i2 : i3 = 54 : 64 : 75 or 5 54 i1 = = 1.40 amp., 193 5 64 i2 = = 1.66 amp., 193 3 75 i3 = = 1.94 amp. 193 CURRENT ELECTRICITY
96
Ex. 8 How will reading in the ammeter A of the fig.
Sol. (C)
be affected if an other identical bulb Q is connected in parallel to P as shown. The voltage in the mains is maintained at constant value Switch
Let the current, in upper branch is I 1 and in lower branch I2. The current in central resis tance will be I 1 + I2. Using Kirchhoff's laws. 2 = I1 (2) + (I1 + I2) (2) upper branch 2 = I2 (2) + (I 1 + I2) (2) lower branch adding 4 = 2(I1 + I2) + 4(I1 + I2) or I1 + I2 = 4/6 = 2/3 ampere. 0.67 ampere
Q
P
Mains
Ex. 11 The value of steady current in 2 resistance
in the following circuit diagram will be – 2 i1
A (A) the reading will be reduced to one half. (B) the reading will be double of previous one. (C) the reading will not be affected. (D) the reading will increase four fold.
i i2
+ 12 V
f (A) 10 (C) 200
e
After full charging, the steady current in the condenser is zero, hence no current will flow in 4 resistance. I=
E R R'
=
d (B) 100 (D) 500
Ex. 12 The potential different between the points X
and Y in the adjoining diagram Fig. will be– 100
Ex. 10 The reading in the ammeter is –
+ – 2V 2
2 I2
(A) 1 A (C) 0.67 A
100 200V
I1
A
6 6 = =1.5 A 28 12 2 3 28 2 3
Let current flowing in 2 resistance is I1 2 and 3 resistance are connected in parallel 2I1 = (1.5 – I 1) x 3 5I1 = 4.5 I1 = 0.9 amp.
Sol (B)
Since there is no current in edcb part, the p.d. across b, e should be 2V. Let current in 500 is I, then same current flows through X (think). Therefore, for loop abefa, 12 = I(500) + IX or 12 = I(500) + 2( IX = 2 volt) Thus I = (1/50) A or from IX = 2, X = 2 × 50 = 100 .
(B) 1.8 A (D) 0.36 A
Sol (A)
2V +
X
2.8
+ – 6V (A) 0.9 A (C) 2.7 A
Since Q is connected in parallel the net resistance becomes R/2, so the current I = 2V/R, double the value. zero. If batteries have negligible internal resistances, the value of resistance X wil be – 500 b G c a
4
0.2F
Sol (B)
Ex. 9 In a circuit shown, the galvanometer G reads
3
2
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y
x 100 100
+ – 2V (B) 2 A (D) 1.5 A
V
(A) zero (C) 10 V
(B) 50 V (D) 100 V CURRENT ELECTRICITY
97
Sol (A)
Ex.14
b 100 a
100 c
G
The value of i in the following circuit diagram will be – 8V ()
100
100
2
d
2 4 i 200V Equivalent circuit can be reduced as follows P R Because = Q S VX = VY VX – VY = 0 the reading of voltmeter will be zero. Ex.13
4 2 3 A 2 1 (C) A 2 (A)
Sol.
(B)
3 A 4
(D) 1 A
(D) Resultant resistance 1 5 8 1 1 1 R= = + + = 8 8 5 R 4 4 85 E net current I = = = 5 amp. 8 R
In the adjoining diagram R1 = 10, R2 = 20, R3 = 40, R4 = 80 and V A = 5V, V B = 10V, VC = 20V, VD = 15 V. The current in the resistance R1 will be– B
i=
i2
R2
2
1 × 5 = 1amp. 5
Ex.15 A galvanometer together with an unknown
R1 A
R3
O
i3
i1
C
R4 (A) 0.4 (B) 0.4 (C) 0.6 (D) 0.6
A A A A
towards O away from O towards O away from O.
i4
D
Sol (A)
i1 + i2 + i3 + i4 = 0 VO V A VO VB + R1 R2 +
Sol
+
VO VC R3
VO VD = 0 R4
VO 5 VO 10 VO 20 VO 15 + + + = 0 10 20 40 80 or VO = 9 volt 9 5 i1 = = 0.4 A away from O ......(B) 10
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resistance in series is connected across two identical batteries of each 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1 A and when the batteries are connected in parallel, the current is 0.6 A. The internal resistance of the battery will be– 1 1 (A) (B) 3 2 1 1 (C) (D) 5 4 (B) Let R be the combined resistance of galvanometer and an unknown resistance and r the internal resistance of each battery. When the batteries, each of e.m.f. E are connected in series, the net e.m.f. = 2E and net internal resistance = 2r 2E 2 15 Current i 1 = or 1.0 = R 2r R 2r R + 2r = 3.0. When the batteries are connected in parallel, the e.m.f. remains E and net internal resistance becomes r/2. therefore CURRENT ELECTRICITY
98
Ex.17
E
2
2E Current i2 = R r = 2R r 2 2E
2R + r = i = 2
2 15 = 5.0 0.6
3 C = 0.2F
Solving (A) and (B), we get r = 1/3 . Ex.16 A potential difference of 220 V is maintained
across a 12000 rheostat as shown in fig. The voltmeter V has resistance of 6000 and the point C is at one fourth of the distance from a to b. The reading of voltmeter will be b 220V
c V a
Sol
(A) 30V (C) 50V (B) Given ac =
(B) 40V (D) 60V
+ –
2 . 8 V=6V (a) Calculate the steady–state current in the 2 resistor shown in the circuit in fig. The internal resistance of the battery is negligible and the capacitance of the condenser is 0.2 F (b) The resistors, 400 and 800 are connected in series with a 6 volt battery. It is desired to measure the current in the circuit. An ammeter of 10 ohms resistance is used for this purpose. What will be the reading in the ammeter ? Similarly, if a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400 resistor, what will be the reading in the voltmeter ? Sol.
(a) Let R be the equivalent resistance of two resistances 2 and 3 connected in parallel.
1 ab and resistance of rheostat 4
1 3 2 5 1 1 = + = = 3 6 6 R 2 6 or R = = 1.2 . 5 400 800
= 12000. Resistance of part ac, R ac =
1 × 12000 = 3000. 4
Resistance of part bc, Rbc = 12000 – 3000 = 9000. Resistance of voltmeter, R V = 6000. The equivalent resistance of voltmeter and part ac connected in parallel is R ac R V 3000 6000 R' = R R = = 2000. 3000 6000 ac V Resistances R' and Rbc are in series, therefore net resistance in circuit, R = 9000 + 2000 = 11000 . Current in circuit, i =
220 V = = 0.02 11000 R
amp. Reading of voltmeter = p.d. across ac = iR' = 0.02 × 2000 = 40 Volt.
4
10
A
+ – 6V The branch having a capacitor C and 4 resistance acts as open circuited because D.C. will neither pass through condenser nor 4 resistance. The total effective resistance of the circuit = 1.2 + 2.8 = 4. 6V current in the circuit = = 1.5 amp. 4 Now current through 2 resistance 3 = 1.2 × = 0.9 amp. 5 ( current in parallel resistances are in ratio of the reciprocals of their resistances)
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CURRENT ELECTRICITY
99
(b) The circuit arrangement is shown in fig. 10000
3 2
V
A
4
C
+ – 2V
+ – 6V From fig.(a) total resistance of the circuit = 400 + 800 + 10 = 1210 . Current i in the circuit =
B
6
800
400
4 8
Sol
The diagram can be redrawn as shown in fig. 3
6 volt =0.00496 amp 1210
2 X
A
4 C
So ammeter will read 0.00496 amp. 4
From fig,(b) resistance 400 and 10,000
8
are in parallel. Let the effective resistance be
6
R. Hence 1 25 1 26 1 1 = + = = 10,000 10,000 10,000 400 R 10,000 26 Total resistance of the circuit 10,000 30,800 = + 800 = 26 26
+ – 2V
Potential difference across voltmeter V = current × resistance =
156 10,000 × 30,800 26
= 1.96 volt.
Ex.18
For the fig. Calculate the current through 3 ohm resistor and power dissipated in the entire
1 R AC
1 1 3 3 R AC = + = ohm 2 4 4 4
=
The effective resistance RCB between C and B 3 1 1 RCB = + = 8 8 4
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RCB =
Now, R ACB = R AC + RCB =
8 ohm. 3
4 8 + = 4ohm. 3 3
Corresponding to points X and Y, the resistances 3 ohm, 4 ohm and 6 ohm are in parallel, hence effective resistance RXY is 1 R XY
=
1 1 1 432 9 + + = = 3 6 4 12 12
circuit. The emf of battery is 2 volt and its internal resistance is 2/3 ohm.
2/3
The effective resistance R AC between A and C
R=
6 26 156 Now current in the circuit = = 30,800 30,800
Y
B
RXY =
12 4 = ohm. 9 3
CURRENT ELECTRICITY
100
Total =
resistance
R
of
the
500i – 200 i1 = 110 Solving eqs. (1) and (2), we get 3 i = amp 10 1 and i1 = amp. 5
circuit
4 2 + = 2 . 3 3
Current in the circuit =
1 amp.
1 amp. 5 1 Current in 200 ohm resistance i – i 1 = 10 3 Current in 300 ohm resistance i = amp. 10 Potential difference between A and C = Potential difference across 100 ohm resistance or potential difference across 200 ohm resistance V A – VC = current × resistance 1 = i1 × 100 = × 100 = 20 volt. 5 Potential difference between C and B is given by
Current in 100 ohm resistance i 1 =
Power dissipated in the circuit = i 2 R = 1 × 2 = 2 watts Potential difference between X and 4 4 Y = i × R XY = 1 × = V. 3 3 Potential difference across 3 ohm resistor 4 V. 3
=
Current in 3 ohm resistor =
4/3 4 = 3 9
= 0.44 amp. Two points A and B are maintained at a constant potential difference of 110 volt. A third point is connected to A by two resistances of 100 and 200 ohm in parallel, and to B by a single resistance of 300 ohm. Find the current in each resistance and the potential difference between A and C and between C and B. The circuit with current distribution is shown in fig. C
Ex.19
Sol
F 100 E i1 D i
D
VC – VB = i x 300 = Ex.20
i G H
300
200 i - i1
.....(2)
Sol
A
i B 110V Applying Kirchoff's second law to the loop DEFGHID, we have i1 × 100 – (i – i 1) × 200 = 0 300 i1 – 200 i = 0 ......(1) Now applying Kirchoff's second law to loop ADIHGCBA, we have. (i – i 1) 200 + i × 300 = 110
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3 × 300 = 90 volt. 10
In the circuit E shown in fig. + – B E, F, G and A H are cells of emf 2, 1, – 3 and 1 F+ H 2 – + volts and their internal – + resistances D C G are 2, 1, 3 and 1 ohm respectively. Calculate – (i) the potential difference between B and D and (ii) the potential difference across the terminals of each of the cells G and H. Fig. shows the current distribution. 2
A
2V
i
1V
1
2
1
i1 1V D
B
3 i2 3V
C
CURRENT ELECTRICITY
101
Applying Kirchoff's first law at point D, we have i = i1 + i2 ......(A) Applying Kirchoff's second law to mesh ADBA, we have 2i + i + 2i1 = 2 – 1 = 1 or 3i + 2i1 = 1 ......(B) Applying Kirchoff's second law to mesh DCBD, we get 3i2 + i2 – 2i1 = 3 – 1 or 4i2 – 2i1 = 2 ......(C) Solving eqs. (A), (B) and (C), we get 1 i1 = amp., 13 6 5 amp. and i = amp. 13 13 (i) Potential difference between B and D
or –3i1 + i2 = – i and in mesh BEFCD E
(i - i1- i2) (i - i1- i2) B
F i i2 (i1+ i2)
(i - i1) i A i1
C
i2 =
i1 D 2(i – i1 – i2) – i1 – i2 – i2 = 0 or –3i1 – 4i2 = – 2i .......(B) From eqs. (A) and (B).
1 2 = 2i1 = 2 13 = volt. 13 (ii) Potential difference across G = E – i 2 63 R = 3 – = 1.61 V. 13 Potential difference across H 6 = 1 – 13 (A) = 1.46 V. Ex.21
......(A)
In the adjoining circuit diagram each resistance is of 10. The current in the arm AD will be– E
i1 =
2i i , i2 = 5 5
AD i =
2i 5
What is the equivalent resistance between the terminal points A and B in the network shown in fig. Assume that the resistance of each resistor is 10. I C I1 R D I2 R F A
Ex.22
I- I1
I1- I2
R
R
R
F i B
i
G
A
(A)
2i 5
(C)
4i 5
D
(B)
3i 5
(D)
i 5
Sol (A)
Applying Kirchoff's law in mesh ABCDA 10 (i – i1) + 10i2 – 20i1 = 0
H I- I R 2
K
I
B
E
Sol
C
R
Let a battery of e.m.f. E and no internal resistance be connected across A and B. Let I current enter the corner C and leaves the corner B. The distribution of currents, according to Kirchoff's Ist law is shown in fig. applying Kirchoff's IInd law to closed loops CDHGC and DFKHD successively, we get. –RI1 – R (I 1 – I2) + 2R (I – I 1) = 0 or 4I1 – I2 = 2I ......(A) and –2RI2 + R(I – I 2) + R (I 1 – I2) = 0 or –I1 + 4I2 = I. ......(B)
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or
Solving (A) and (B), we get CURRENT ELECTRICITY
102
3 2 I and I2 = i .......(C) 5 5 Now applying krichoff’s II law to mesh ACGHKBEA, we get –2 (I –I1,) R – (I – I 2) R + E = 0 or E = 2(I – I1) R + (I – I 2)R
–
I1 =
5 IR ......(A) 6 If R AB is equivalent resistance between comers A and B, then from Ohm’s law comparing (A) and (B), we get
or
3 2 = 2 1 5 I R + 1 5 I R 7 IR. 5 If R AB is the equivalent resistance between A and B, then according to Ohm’s law E = IR AB Comparing (D) and (5), we get 7 IR AB = IR 5 7 or R AB = R. 5 Here R = 10 7 R AB = × 10 = 14 5
or
E=
1 1 1 R– R – R + E = 0 3 6 3
E =
IR AB = Ex. 24
Twelve equal wires each of resistance R are joined to form a skeleton cube. Find the resistance between two corners on the same edge of the cube. Sol.
Sol
I = x + 2y
y
I/3
I/3
I/6 I/3
I/6
I/3
I/3
I/3 I/6
I
E Let ABCDEFGH be skeleton cube formed of twelve equal wires each of resistance R. Let a battery of e.m.f. E be connected across A and G. Let the total current entering at the corner A and leaving the diagonally opposite corner G be I. By symmetry the distribution of currents in wires of cube, according to Kirchoff's Ist law is shown in fig. Applying Kirchoff's IInd law to mesh ADCGEA, we get
z (y – z)
y E
F (y – z)
x + 2y B
y
C
Let I = x + 2y current enter at point A, when a battery of e.m.f. E and no internal resistance is connected across edge AB. The edges AD and AH are symmetrically connected to A, therefore they will carry equal currents. The distribution of currents according to Kirchoff's Ist law is shown in fig.
I/6 I/3
2 (y–z)
D
x
E
G (y – z)
z
A y
I/6 I
(y – z)
H
Ex. 23
Twelve equal wires, each of resistance R ohm are connected so as to form a skeleton cube. An electric current enters this cube from one corner and leaves out the diagonally opposite corner. Calculate the total resistance of this assembly.
5 IR 6
If R AB is equivalent resistance, then from Ohm's law, E = R AB I = R AB (x + 2y) ....(A) and from Kirchoff's law applied to mesh containing AB and cell E is R x = E .....(B) (since R is resistance of each wire) Applying Kirchoff's II law to mesh AHEB yR + zR + yR – xR = 0 or x – 2y – z = 0.....(C) Applying Kirchoff's II law to mesh DGFC (y – z) R + 2 (y – z) R – zR = 0 or 4(y – z) –z = 0 or 4y = 5 z .....(D) i.e. z = (4/5) y .....(5) Substituting this value in (C), we get
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CURRENT ELECTRICITY
103
x – 2y –
4 y = 0 5
1 2 2 I1 + (I – 2I1) – I1 = 0
or
14 5 y = x i.e. y = x 5 14 Substituting value of y in (A), we get
2 2I – 5I1 = 0 or I1 = I ...... (B) 5 Applying Kirchoff's IInd law to external circuit AHEBE', we get
or
or
10 E = R AB x x 14
1 1 R + I1 R + R = E1 2 2
24 E = R AB x = R . x 14
IR +
2 IR = E' 5
[Using (B)]
R AB =
24 R 14
or
R AB =
7 R. 12
Comparing (A) and (C), wet get R AB I =
Ex. 25
Eleven equal wires each of resistance 2 form the edges of an incomplete skeleton cube. Find the total resistance between points A and B of the vacant edge.
i.e. R AB =
(I/2-I1)
I
I/2
E B
I
Sol
I1
E
I-2I1
D
y I/2-I1
F
I/2
7 7 R = × 2 = 2.4 5 5
4V
0.1
4V
0.2
I/2-I1 I/2
7 IR 5
Three 4V batteries, internal resistances 0.1, 0.2 and 0.3 are connected in parallel and in series with a 2.045 ohm resistor. Find (a) equivalent resistance of the circuit (b) equivalent voltage (c) current in the circuit (d) the terminal voltage for equivalent cells (e) the terminal voltage of each cell.
G I/2-I1
I1
A
......(C)
Ex. 26
Sol
H
7 IR = E’ 5
C
Let a battery of e.m.f. E' is applied between points A and B. Let a current I, enter through point A. If R AB is equivalent resistance between points A and B, then from Ohm's law R AB I = E' The distribution of currents, keeping in mind symmetry condition, is shown in fig. Let R (= 2) be the resistance of each wire. Applying Kirchoff's II law to mesh DGFC, we get 1 I1 R + (I – 2I1) 2
1 I1 R – I1 R = 0 2
R+
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4V
A
0.3 B
2.045 The curcuit arrangement is shown in fig. As the batteries are connected in parallel, hence total emf of the circuit = 4V. The effective resistance R AB between A and B is given by 1 R AB
=
1 1 1 110 + + = 0.1 0.2 0 .3 6
6 = 0.055 ohm. 110 (a) Equivalent resistance of the circuit R = R AB + 2.045 = 0.055 + 2.045 = 2.1 ohm. (b) Equivalent voltage = 4 volt.
R AB =
CURRENT ELECTRICITY
104
(i) When the two heating coils are in series, the effective resistance is
4 = 1.9 amp. 2 .1 (d) Terminal voltage of equivalent cell = 4 – i R AB=4–1.9 × 0.55 = 4– 0.1045 =3.8955 V. (e) Batteries are in parallel hence terminal voltage for each cell is 3.8955 V. (c) Current in the circuit =
4 7 R' = R 1 + R2 = R1 + . R1 = R . 3 3 1 with two coils in series, let the kettle take t' time to boil. The. V 2 t' V 2 t' Q = = 7 JR' 4.2 R1 3
Ex. 27
In the fig below the bulbs are identical, which bulb(s), light(s) most brightly ? (A) 1 only (B) 4 only (C) 2 and 3 (D) 1 and 5.
Comparing (A) and (D), we get
Sol (D)
Since all bulbs are identical they have the same resistances. The current I flowing through 1 branches at A. So current in 2 and 3, as well as in 4 will be less than I. The current through 5 is also I. Thus and 5 glow equally brightly and more than 2, 3 or 4. An electric tea kettle has two heating coils. When one the coils is switched on, the kettle begins to boil in 6 minutes. When the other is switched on, the boiling begins in 8 minutes. In what time will the boiling begin if both coils are switched on simultaneously (i) in series (ii) in parallel ? (A) 14 min, 3.43 min (B) 12 min, 3.43min (C) 3.43 min, 12 min (D) 3.43 min, 14 min. Sol. (A) Let R1 and R2 be the resistances of the coils, V the supply voltage, Q the heat required to boil the water. Heat produced by first coil of resistance R 1 in time t1 (= 6 min)
V 2 t1 V 2 6 60 = Q = = JR 2 4.2R 2 Equating (A) and (B), we get
.....(B)
R2 6 8 8 4 = i.e. R2 R2 R1 = 6 = 3 4 or R2 = R 3 1
.....(C)
t' 7 3
= 6 × 60 7 or t' = × 6 × 60 sec = 14 min. 3 (ii) when the two heating coils are in parallel, the effective resistance is,
Ex. 28
V 2 t1 V 2 6 60 =Q= = cal ......(A) JR1 4.2R1 Heat produced in second coil of resistance R 2 in time t2 (= 8 min)
.....(D)
4 R1 R1 3 R1 R 2 4 R" = R R = R 4 = R1 R1 7 1 1 2 3 In parallel arrangement of heating coils, let t" be the time taken by kettle to boil, so V 2 t' ' V t" 4 Q = = 4.2 R1 JR" 7 2
.....(5)
Comparing (A) and (5), we get t" 4 = 6 × 60 or t" 7 4 = × 6 × 60 sec = 3.43 min. 7 Ex. 29
A 10 m long nichrom e wire having 80 resistance, has current carrying capacity of 5 A. What is the power which can be obtained as heat by the wire from a 200 V mains supply ? If the wires are cut in two equal parts and connected in such a way that it gives maximum power. What is the arrangement to obtain maximum power ? (Rohilkhand Univ.) (A) 50 W, 200 W (B) 500 W, 2000 W (C) 50 W, 100 W (D) 500 W, 1000 W
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CURRENT ELECTRICITY
105
Sol (B)
If the wire is connected as such across the battery, then current in wire, 200 V I= = = 2.5 A and power obtained, 80 R 200 200 V2 P = = = 500 watts. 80 R The wire can carry maximum current of 5 A, therefore to double the current, the resistance should be halved. Thus if we divide the wire in two parts and the two parts are connected in parallel across 200 V mains supply, the resistance of each part = 40, therefore current 200 in each wire = = 5A. 40
i1 6 2 i
Sol (A)
Let current in 5 is i2 then P = i22 R2
45 = i22 × 5 or i2 = 3 amp R2 i1 5 1 Since = R = = i2 15 3 1 i1 = 1 amp The total current through 2 resistor is i = i1 + i2 = 3 + 1 = 4 amp. The rate of heat generation in 2 resistor is = i2R = 42 × 2 = 32 joule/sec. Potential difference across 6 resistor is V = i1 × 6 = 1 × 6 = 6 volt. Ex. 32
A 220 volt 100 watt bulb is connected to a 110 volt source. The power consume by the bulb will be – (A) 25 W (B) 20 W (C) 484 W (D) 120 W Sol (A)
220 220 V2 Resistance of the bulb R = = 100 P The new power for the voltage of 110 volt is
Sol (A)
80 × 2000. t = 4.2 ×1000 × 1 x× (100– 100
4) t =
42 1000 96 1000 = 252 sec. 2000 80
Ex. 31
In the following figure the rate of heat generated in 2 ohm resistor and potential differents across 6 ohm resister will be respectively–
B
(A) 32 J/s, 6V (B) 16 J/s, 3V (C) 8 J/s, 1V (D) 64 J/s, 12V (Heat generated in 5 ohm resistor due to current flowing in it is 45 J/s)
Ex. 30
D
i2
V2 and new power obtained, P max = R' 200 200 = = 2000 Watts. 20 Thus maximum power is 2000 watts and this is obtained when wire is cut in two halves and they are connected in parallel across the given supply.
We know that the relation between work and heat produced is W = JH P.t = J. ms
A 5
R1R 2 40 40 Net resistance, R' = R R = = 20 40 40 1 2
A heating–coil of 2000 watt is immersed in an electric kettle. The time taken in raising the temperature of 1 litre of water from 4ºC to 100ºC will be– (Only 80% part of the thermal energy produced is used in raising the temperature of water.) (A) 252 s (B) 250 s (C) 245 s (D) 247 s
9
110 110 V'2 = = 25 watt. 484 R Ex.33 An electric motor whose resistance is 2 ohm is started with a supply of 110 volt. It takes 10 ampere current at its full speed. The electric power consumed and part of the power used in mechanical work will be respectively– (A) 900 W, 82% (B) 800 W, 80% (C) 200 W, 62% (D) None of the above.
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P' =
CURRENT ELECTRICITY
106
(A) Power of the motor = VI= 110 ×10= 1100 watt Heat loss in the motor = i2R = (10)2 × 2 = 200 watt Power converted to mechanical work = (1100 – 200) watt = 900 watt Percentage of total power consumed in mechanical work = (900/1100) × 100 = 82% (approx). Ex. 34 A copper wire having cross–sectional area 0.5 mm2 and a length of 0.1 m is initially at 25º C and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire, (i) Find the time in which the wire will start melting. The change of resistance with temperature of the wire may be neglected. (ii) What will be the time taken if the length of the wire is doubled ? Giving for copper, density = 9 × 10 3 kg/m3 , specific heat = 9 × 10 –2 Kilocal kg –1 (ºC) –1, melting point = 1075 ºC, specific resistance 1.6×10 –8 ohm–metere. Sol. (i) The resistance of copper wire of length l = 0.1 m and cross–sectional area A = 0.5 mm 2 = 0.5 × 10 –6 m 2 is given by
(ii) From equation (C), it is obvious that the time required is independent of length; therefore by doubling the length of wire, the time remains the same = 588.14 sec.
Sol
R=
Ex.35 A person decide to use his bath–tub water to
generate electric power to run a 40 W bulb. The bath tub is located at a height of 10 m from the ground and it holds 20 litres of water. He installs a water driven wheel generator on the ground. The rate at which he should drain the water from the bath tub to light the bulb and the time he keeps the bulb on will be respectively– (The efficiency of the generator is 90%) (g = 9.8 m/s 2) (A) 0.345 kg/s, 441s (B) 40 kg/s, 100 s (C) 0.454 kg/s, 441s (D) None of these (C) Rate of change of Potential Energy = Electric power generated.
Sol
dm g h × 90 = 40 watt 100 dt = 40 joule/sec.
A If Q is the heat required to melt the copper wire, then i2RT Q = R t joule = kilocal ......(A) J Also Q = ms = (mass) × specific heat × rise of temperature, mass = volume × density = (area × length) d = (A l ) d and rise of temperature = – = 1075 – 25 = 1050ºC Q = A l d. s Kilocal .....(B)
40Js 1 100 = 90 9.8N / kg 10m 2 = = 0.454 kg/sec. 4.41 mass of 200 litre water Time = dm / dt 200kg = ( 2 / 4.41)kg / sec = 441 second
i2
Ex.36 A wire of resistance 0.1 ohm cm –1 bent to
2
Equating (A) and (B)
i RT = A l d.s. J
JA ds JA ds = 2 i R i2 A JA ds
Time t= =
i2 Substituting given values t =
...... (C)
4 . 2 10 3 (0 . 5 10 6 )2 (9 10 3 ) (9 10 2 ) x 1050 (10 )2 1. 6 10 8
40 watt 100 dm = 90 g h dt
Sol
form a square ABCD of side 10 cm. A similar wire is connected between the corners B and D to form the diagonal BD. Find the effective resistance of this combination between corners A and C. If a 2 V battery of negligible internal resistance is connected across A and C calculate the total power dissipated. In fig (a). A square of 10 cm side is shown. The resistance of each side is 10 × 0.1 = 1 ohm. The corners B and D also connected by the same wire. The square forms a Wheastone's bridge because the condition P/Q = R/S is satisfied. Now no current will flow through BD.
= 588.14 sec Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159
CURRENT ELECTRICITY
107
The fig has the form as shown in fig (b). 1
B
C
Two bulbs rated at 25 watts, 110 volts of 100 watts, 110 volts are connected in series to 220 volts electric supply. Perform the necessary calculations to find out which of the two bulbs, if any, will fuse. What would happen if the two bulbs were connected in parallel to the same supply.
Sol
Let i1 and i2 be the currents which can flow through the two lamps safely, then
1
1 1
1
A
B
Ex.38
C 1
D A D 1 1 Resistance of ABC part = 1 + 1 = 2 ohm (They are in series Resistance of ADC part = 1 + 1 = 2 ohm. Now the two parts are in parallel corresponding to points A and C, hence effective resistance R is given by –
25 100 = 0.227 amp. and i 2 = = 0.909 110 110 amp. The resistance of two bulbs are given by
i1 =
E 110 110 R1 = i = ohm and R 2 = ohm 0.227 0.909 1 When the two bulbs are connected in series, their total resistance
1 1 1 R = 1 ohm. = + R 2 2 When 2V battery is connected between A and C, the current. E 2 i = = = 2 amp. R 1 Power dissipated P = E i = 2 x 2 = 4 watt Ex.37
Sol
Three equal resistors connected in series across a source of emf together dissipate 10 watt of power. What would be the power dissipated if the same resistors are connected in parallel across the same source of emf Let R be the resistance of each resistor. When they are connected in series, the total resistance = R + R + R = 3R ohm. Power dissipated W 1 = E2/3R, where E = emf of the source. When the resistors are connected in parallel, their effective resistance is given by 1 1 1 1 3 = + + = R' R R R R or
R' =
R 3
E2 3E 2 Power dissipated W 2 = = R/3 R
Now or
W1 3R 3E 2 = × = 9 W2 E2 R W 2 = 9W 1 = 9 × 10 = 90 watt (W 1 = 10 watt).
110 110 R = R1 + R2 = + = 605 ohm. 0.227 0.909 When these two lamps are connected in series to 220 volts, the current passing through them is given by 220 i = = 0.363 amp. 605
Thus the first bulb will fuse because the current passing through it i.e., 0.363 is more than i1 (0.227).
Note : When the two bulbs are connected in parallel, the effective resistance R' is given by
1 1 0.227 0.909 1.136 1 = + = + = R1 R2 110 110 110 R' 110 , R' = ohm. 1.136 Current flowing through circuit i' 220 220 1.136 = = amp. 110 R' Let i3 and i4 be the currents passing through the two bulbs as shown in fig. Now the potential difference across the two bulbs is the same. Hence i3 R1 = i4 R2 i3.
110 110 = i4. 0.227 0.909
Again i3 + i4 = i' =
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or 4i3 = i4.
220 1.136 110
.....(A) .....(B)
Solving eqs. (A) and (B), get i3 = 0.454 amp. and i 4 = 1.816 amp. Thus, both the bulbs will fuse. CURRENT ELECTRICITY
108