Class 12-science H C VERMA Solutions Physics Chapter 10 - Electric Current in Conductors
Electric Current in Conductors Exercise 198
Solution 1
(a) [Q]=[A][T]2=[B][T]=[C]
[I][T]=[A][T]2=[B][T]=[C]
[A]=[IT-1]
[B]=[I]
[C]=[IT]
(b) 
I=2At+B
I=2(5)(5)+3
I=53A
Solution 2
Solution 3
Solution 4
Solution 5
Now,
Solution 6
Ω.m
Solution 7
Ω
Solution 8
m/s
Now,
Solution 9
Solution 10
Resistance of the small strip
-(i)
Now,
Differentiate with respect to x
Now,
Solution 11
(a) 
(b) 
Solution 12
Solution 13
Solution 14
Resistance at temperature T is
given by 
Let, at temperature T
Solution 15
Voltmeter reading=zero error + potential due to current
14.4=V+(1.75)R (i)
22.4=V+(2.75)R (ii)
Subtract (ii) from (i)
8=R
Put value in (i)
14.4=V+(1.75)8
V=0.4 volt
Solution 16
Initially, no current flows in circuit when switch is open.
So, emf of cell=voltmeter reading
ε=1.52V
Now, when switch is closed, cell is discharging
So, TPD=ε-ir
1.45=1.52-(1)(r)
r=0.07Ω
Solution 17
For battery
TPD=ε-ir
5.8=6-i(1)
i=0.2A
for external resistor
V=IR
5.8=(0.2)(R)
R=29Ω
Electric Current in Conductors Exercise 199
Solution 18
TPD=ε+ir
7.2=6+(2)(r)
r=0.6Ω
Solution 19
(a) Net emf across battery=9-6=3V,
Req=10Ω
current=3/10=0.3A
(b) Net emf across battery =9-6=3V
Resistance = 1Ω
Current=3/1=3A
Solution 20
Cells are in series combination
So, 
cells are in parallel combination
Divide (i) and (ii)
(a) 
(b) 
(c) 
Solution 21
For I to be maximum,
will be minimum
To minimize,
Differentiate wrt
Solution 22
Emf=100V
Req= 10000+R
Current 
For R=1Ω
For R=100Ω
So up to R=100Ω the current does not change up to 2 significant figures.
Solution 23
Voltage drop access A1=voltage drop access A2
I1R1=I2R2
(2.4)(20)=I2(30)
I2=1.6A
Reading in A3=reading in A1+reading in A2
=(2.4)+(1.6)
=4A
Solution 24
Emf=5.5V
for minimum current
Now, for maximum current 
Solution 25
For each bulb
Voltage applied=60V
Resistance=180Ω
Current I=
(a) All bulbs switched on
IN=I1+I2+I3
IN=
(b) Two bulbs switched on
IN=I1+I2
=
=0.67A
(c) Only one bulb switched on
IN=
=0.33A
Solution 26
For maximum resistance all will be in series combination
Req=20+50+100=170Ω
For minimum resistance all will be in parallel combination
Solution 27
When bulbs are used in parallel, the equivalent resistance will be less than their individual resistance
∴ two resistances are 45Ω and 22.5Ω
Solution 28
Let I be the current in 20kΩ
So(12-I) will be current in 10kΩ
Now, V20kΩ=V10kΩ(parallel combination)
I(20)=(12-I)(10)
I=4mA
So,
I20kΩ=4mA
I10kΩ=8mA
I100kΩ=12mA
Equivalent resistance between AB
VAB=I
=(12×10-3)(111.67×103)
VAB=1340V
Solution 29
Initially,
V=IR
V=5R -(i)
Now, when 10Ω is connected in parallel to R
V=IReq
From (i) and (ii)
R=2Ω
Solution 30
Solution 31
Resistance of each side of hexagon=15/6=2.5Ω
(a)
RAB=
(b)
RAC=
(c)
RAD=
Solution 32
(a) Req=10+20=30Ω
V=3 volt
V=IR
3=I(30)
I=0.1A
(b) Req=10Ω
V=IR
3=I(10)
I=0.3A
Electric Current in Conductors Exercise 200
Solution 33
Emfeq=4-2=2V
Req=4+6=10Ω
Current
Solution 34
Potential at X
Solution 35
(a) For resistor a,
For resistor b,
(b)
For resistor a,
For resistor b,
Solution 36
(a) Both cells are in parallel combination
Internal resistance of cell will be
Equivalent Resistance of circuit
So, current
Voltage difference across ab
(b) The circuit can be redrawn as shown in figure
It is similar to above circuit in (a).
Hence same answer.
Solution 37
KVL for loop 1
3-I1(1)+I2(1)-2=0
I1-I2=1 -(i)
KVL for loop 2
2-I2(1)-( I1+I2)(1)-1=0
I1-2I2=1 -(ii)
Solving (i) and (ii)
I1=1A and I2=0A
Now,
VAB=TPD for cell of 2V
= ε-ir
= 2-(0)(1)
VAB = 2V
Solution 39
Potential difference across each resistance is 0
So, current =0
Solution 38
Potential at point of X volt
Solving, X=3V
Current in 10Ω resistance=
Solution 40
The circuit is in wheat stone symmetry
So, for all values of R, current=0.
Solution 41
(a)
(b)
Solution 42
The 50R resistors are in wheat stone symmetry, so no current flows in them
Electric Current in Conductors Exercise 201
Solution 43
(a) (Emf)eq = 12+6 = 18V
Req = 10+5 = 15Ω
Current I = 
(b) Now,
(c) 
(d) Circuit in figure (b) is same as in figure (a)
Solution 44
Apply KVL for abcfa
Solution 45
(a)
(b)
(c)
(d)
(e)
Solution 46
(a)
Let equivalent resistance between AB be x
(b)
I=V/R=6/2=3A
Current in 2Ω
I2Ω=3/2=1.5A
Solution 47
(a)
Current, 
(Ammeter reading)
Now,50Ω and 200Ω are in parallel combination
So, current is divided in inverse Ratio
current in voltmeter=
= 
=0.02A
Voltmeter reading =(0.02)(200)=4V
(b)
Current, 
(Ammeter reading)
TPD across cell=voltmeter reading
Voltmeter reading= ε-Ir=4.3-0.08(1)=4.2V
Solution 48
(a) 
Since, 100Ω and 400Ω are in parallel combinationso, current is divided in inverse Ratio of resistances.
current in voltmeter Iv=
Voltmeter reading 
=
=24V
(b)
So,
V100Ω=IR=
Solution 49
I24Ω=
I50Ω=
I24Ω= I50Ω+Iv [by KCL]
0.5=0.36+Iv
Iv=0.14A
For voltmeter,
Vv=IvRv
18=0.14R
Rv≃130Ω
Solution 50
Req=25+575=600Ω;I=10mA
V=IR
V=
V=6 volt
Solution 51
IgRg=(I-Ig)Rs
(10-3)(25)=(2-10-3)Rs
Rs=1.25×10-2R
Solution 52
V=Ig(Rg+R)
12=Ig(50+1150)
Ig=0.01A
Now, IgRg=(I-Ig)Rs
(0.01)(50)=(2-0.01)Rs
Rs=0.25R
Solution 53
At null deflection, it will be in wheat-stone symmetry.
Let l be the balancing length from end A
Electric Current in Conductors Exercise 202
Solution 54
Solution 55
(a)
VA=6V
VC=2V
(b) Potential across AD=Potential across AC=4V
⇒ Potential across DB=2V
(c) No current will flow as VC=VD
(d) Potential difference across AC=potential at A-potential at C
7.5=6-Vc
VC=-1.5V
No points exist on wire AB as potential difference. Wire isalways greater than or equal to zero.
Solution 56
(a) Reqfor primary circuit=r+15r=16r
Current=
VAB=IR
VAB=
Potential gradient 
Let balancing length be l from end A.
Emf=xl
(b)
Resistance of 560cm wire = 
Resistance of 40cm wire=15r-14r=r
Nodal analysis at X
Potential difference across resistance r
Solution 57
In steady state, no current flows through capacitor
Req=10+20=30Ω
V=IR
2=I(30)
I=
Voltage across capacitor=voltage across 10Ω
=IR
=
=
Q=CV
Q=6
Solution 58
Nodal analysis at X
For 20Ω resistor
Solution 59
1Ω and 2Ω resistors are in series combination
So, potential across 1
=potential for 1Ω
=1×2=2volt
So, charge=CV=(1)(2)=2μC
Potential across 2μF=potential diff for 2Ω
=2×2=4V
So, charge=CV=(2)(4)=8μC
Now, for 3Ω-3Ω resistors,
Potential drop across each resistance is 3V
Charge on 3μF capacitor=CV
=(3μF)(3)=9μC
Charge on 4μF capacitor=CV
=(4μF)(3)=12μC
Solution 60
No current flows in steady state
Nodal analysis at X
(X-100)3+(X-100)3+(X-0)1+(X-0)1=0
8X=600
X=75
VA-VB=100-75
=25V
VB-VC=75-0
=75V
Solution 61
(a) Maximum potential difference across resistor=ε
When charge on capacitor is 0
(b) At t=0, capacitor acts as short circuit
VR=ε=IR
(c) At t=∞, capacitor acts as open circuit
Vmax-cap.=ε
(d) 
(at t=∞,Vcap=ε)
(e) Power=εI
Pmax=εImax
Pmax=
(f) Power dissipated=
Solution 62
Solution 63
Solution 64
Q = Q0(1-
)
= (20
= 12
= 76μC
Solution 65
Q = Q0
Q = 60 
Q = 6
(a) t = 0
Q = 6
(b) t = 30μs
Q = 6 
Q = 44μC
(c) t = 120μs
Q = (6 
= 18μC
(d) t = 1ms
Q = (6 
= 0.003μC
Solution 66
(a) 
(b)
Electric Current in Conductors Exercise 203
Solution 67
Capacitance
Time constant
Electric field in capacitor =
)
Solution 68
Capacitance
charge
time constant =RC
Energy stored=
Solution 69
(a)
at t=0,
At t=10×60=600sec.
(b)
at 
at 
Solution 70
For discharging, so, answer will be same.
Solution 71
Solution 72
Solution 73
Power=
VI=
I=
Solution 74
Energy stored,
Rate of energy stored=
For maximum rate of energy stored
Put in (i)
Solution 75
(a) 
(b) P=VI
=(6)(2.21)
P=13.25W
(c) Power dissipated=I2R
=(2.21)2(1)
=4.87W
(d) Energy stored 
Solution 76
Heat dissipated 
By energy method:-
Heat dissipated= energy stored at t=0-energy stored at time t
=
Solution 77
Current in circuit at time t,
Heat dissipated
Solution 78
Capacitance
resistance of dielectric material,
time constant
It is independent of geometric parameters
Solution 79
So, charge on each capacitor=18.4/2=9.2μC
Solution 80
Initially, the capacitor was charged completely
Potential across capacitor=Potential across 10Ω
charge
=
Now, capacitor is discharged
Solution 81
During charging, charge developed in 4sec
Now discharging
Solution 82
Equivalent capacitance
Solution 83
Let charge on capacitor B at time t be q
Applying KVL,
Solution 84
Initial charge given to the capacitor=Q
When the capacitor is connected by a battery, it will charge through the battery. The initial charge will also delay.
Growth of charge by battery
Delay of charge through capacitor
Net charge on capacitor at time 
