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Alternating Currents
1.
The power is transmitted from a power house on high voltage ac because
(a) Electric current travels faster at higher volts
(b) It is more economical due to less power wastage
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m
(c) It is difficult to generate power at low voltage
(d) Chances of stealing transmission lines are minimized
The potential difference V and the current i flowing through an instrument in an ac circuit
V = 5 cos ω t
volts and I = 2 sin ωt amperes (where ω = 2πf). The
power dissipated in the instrument is
(b) 10 W
(c) 5 W
(d) 2.5 W
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3.
(a) Zero
n.
of frequency f are given by
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2.
Alternating current(A.C.) can not be measured by D.C. ammeter because
(a) A.C. cannot pass through D.C. ammeter
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i
(b) Average value of complete cycle is zero
(c) A.C. is virtual
4.
In an ac circuit, peak value of voltage is 423 volts. Its effective voltage is
(b) 323 volts
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(a) 400 volts
(c) 300 volts
(d) 340 volts
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In an ac circuit I = 100 sin 200 πt. The time required for the current to achieve its peak
value will be
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5.
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(d)A.C. changes its direction
(a)
1
sec
100
(b)
1
sec
200
(c)
1
sec
300
(d)
1
sec
400
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6.
An alternating current is given by the equation i = i1 cos ω t + i2 sin ω t . The r.m.s. current is given
by
1
(c)
1
2
2
(i1 + i2 )
(b)
(i12 + i22 )1 / 2
(d)
1
2
(ii + i2 )2
1 2 2 1/ 2
(i1 + i2 )
2
The phase angle between e.m.f. and current in LCR series ac circuit is
(a) 0 to π / 2
(b) π / 4
(c) π / 2
(d) π
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7.
(a)
n.
8. Current in the circuit is wattless, if
(b) Resistance in the circuit is zero
(c) Current is alternating
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(d) Resistance and inductance both are zero
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(a) Inductance in the circuit is zero
9. An alternating current of frequency
'f'
is flowing in a circuit containing a resistance R and
a choke L in series. The impedance of this circuit is
R 2 + L2
(d)
R 2 + 4 π 2 f 2 L2
R 2 + 2πfL
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(c)
(b)
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(a)R + 2πfL
10. A resonant ac circuit contains a capacitor of capacitance
10 −6 F
and an inductor of
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.s
frequency of electrical oscillations will be
10 5 Hz
(b) 10 Hz
(c)
10 5
Hz
2π
(d)
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(a)
10
Hz
2π
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11. Power delivered by the source of the circuit becomes maximum when
(a)
ωL = ωC
(c)
ωL = − 
 1 

 ωC 
(b) ωL =
1
ωC
2
(d) ωL =
ωC
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10 −4 H .
the
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12. In a LCR circuit having L = 8.0 henry, C = 0.5 µF and R = 100 ohm in series. The
resonance frequency in per second is
(a) 600 radian
(b) 600 Hz
(c) 500 radian
(d) 500 Hz
13. In LCR circuit, the capacitance is changed from C to 4C. For the same resonant frequency,
(a) 2L
(b) L / 2
(c) L / 4
(d) 4 L
co
m
the inductance should be changed from L to
[R 2 + (L ω − C ω )2 ]1 / 2
(b)
2
 2 
1  
R +  L ω −
 
Cω  



(c)
2
 2 
1  
R +  Lω −
 
Cω  



(d)
2

1  

(R ω )2 +  L ω −
 
Cω  



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io
(a)
n.
14. 4.In a series LCR circuit, operated with an ac of angular frequency ω , the total impedance is
1/ 2
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−1 / 2
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1/2
15. For series LCR circuit, wrong statement is
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(a) Applied e.m.f. and potential difference across resistance are in same phase.
(b) Applied e.m.f. and potential difference at inductor coil have phase difference of π / 2 .
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(c) Potential difference at capacitor and inductor has phase difference of π / 2 .
(d) Potential difference across resistance and capacitor has phase difference of π / 2 .
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16. In a purely resistive ac circuit, the current
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(a) Lags behind the e.m.f. in phase
(b) Is in phase with the e.m.f.
(c) Leads the e.m.f. in phase
(d) Leads the e.m.f. in half the cycle and lags behind it in the other half
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17. The current in series LCR circuit will be maximum when ω is
(a) As large as possible
(b) Equal o natural frequency of LCR system
(c)
LC
(d)
1 / LC
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m
18. An inductor L and a capacitor C are connected in the circuit as shown in the figure. The
frequency of the power supply is equal to the resonant frequency of the circuit. Which
A1
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C
A2
A3
E = E0 sinωt
(a)
A1
(b)
(c)
A3
(d) None of these
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i
A2
[A]: In LCR series circuit, current is maximum at resonance.
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19.
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L
n.
ammeter will read zero ampere
[R]: In LCR series circuit, impedance is minimum at resonance.
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.s
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
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c) Assertion is true but reason is false.
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d) Both assertion and reason are false.
20.
[A]: If Xc < XL, the phase factor is negative.
[R]: If Xc < XL; current lags behind emf.
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.
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21.
[A]: Electric power is transmitted over long distances through conducting wires at high
voltage.
[R]: A Power loss is less when power is transmitted at high voltage.
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
c) Assertion is true but reason is false.
22.
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d) Both assertion and reason are false.
The voltage of an a.c source varies with time according to the relation E = 120 sin 100 π t
n.
cos 100 πt. Then
ii) The peak voltage = 60 V
iii) The peak voltage = 120/
2V
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i) The peak voltage of the source is 120 V
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iv) The frequency of source voltage is 100 Hz
a) i) and ii) are correct
b) ii) and iii) are correct
d) ii) and IV) are correct
c) i) and iii) are correct
b) 1.2 A
c) 4.2 A
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a) 2.2 A
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23. In the circuit shown below what will be the reading of the voltmeter and ammeter?
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d) 0.5 A
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Key
b
2)
a
3)
b
4) c
5)
d
6) c
7)
a
8) b
9) b
10) c
14) b
15) c
16) b
17) d
18) c
19) a
20) a
11) b
12) c
13) c
21) a
22) c
23) a
co
m
1)
2.
π

V = 5 cos ω t = 5 sin ω t + 
2

Power
and
= Vr.m . s. × ir.m . s. × cos φ
i = 2 sin ω t
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Power loss
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io
1
(Voltage) 2
1.
∝
n.
Hints
=0
(Since φ = π , therefore cos φ = cos π
=0)
2
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i
2
3.
In dc ammeter, a coil is free to rotate in the magnetic field of a fixed magnet.
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If an alternating current is passed through such a coil, the torque will reverse its direction
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each time the current changes direction and the average value of the torque will be zero.
Effective voltage
Vr.m . s. =
5.
The current takes
T
4
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4.
6.
7.
irms =
2
=
423
2
= 300 V
sec to reach the peak value.
In the given question
∴ Time
Vo
2π
1
= 200 π ⇒ T =
sec
T
100
to reach the peak value
=
1
sec
400
i12 + i22
1 2 2 1/2
=
(i1 + i2 )
2
2
(a)
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8.
Because power
9.
Z=
10.
R 2 + X L2 , X L = ωL and ω = 2πf
∴Z =
R 2 + 4 π 2 f 2 L2
ν =
1
2π LC
=
1
2π 10
−6
× 10
−4
=
10 5
Hz
2π
co
m
(b)
12.
Resonance frequency in radian/second is
(b)
15.
(c)
16.
(b)
1
L1 C 1
8 × 0 .5 × 10 − 6
=
1
L2 C 2
= 500 rad / sec
⇒ L2 =
L1
4
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14.
LC
1
=
at
io
ω=
13.
1
n.
ω=
17.
if R = 0, then P = 0.
At resonant frequency current in series LCR circuit is maximum.
(c)
19.
a
20.
a
21.
a
22.
c
23.
V = VR2 + (VL − VC )
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18.
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2
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11.
= i 2 R,
220 = VR2 + ( 300 − 300 ) ⇒ VR = 220V
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2
So reading in voltmeter = 220V
w
V = i R ⇒ 220 = i ×100 ⇒ i = 2.2 A.
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D.C Circuits
1.
Why the current does not rise immediately in a circuit containing inductance
(a) Because of induced emf
(b) Because of high voltage drop
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(c) Because of low power consumption
(d) Because of Joule heating
2.
The adjoining figure shows two bulbs B1 and B2 resistor R and an inductor L. When the
B2
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L
S
B1
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R
n.
switch S is turned off
(a) Both B1 and B2 die out promptly.
(b) Both B1 and B2 die out with some delay.
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(c) B1 dies out promptly but B2 with some delay.
(d) B2 dies out promptly but B1 with some delay.
In L-R circuit, for the case of increasing current, the magnitude of current can be
ak
3.
calculated by using the formula
4.
I = I0 e − Rt / L
(b)
I = I0 (1 − e − Rt / L )
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(a)
(c) I = I0 (1 − e Rt / L )
(d)
I = I0 e Rt / L
A capacitor is fully charged with a battery. Then the battery is removed and coil is
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connected with the capacitor in parallel, current varies as
(b) Decreases monotonically
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(a)Increases monotonically
(c)Zero
5.
(d) Oscillates indefinite
The time constant of an LR circuit represents the time in which the current in the circuits
(a) Reaches a value equal to about 37% of its final value
(b) Reaches a value equal to about 63% of its final value
(c) Attains a constant value
(d) Attains 50% of the constant value
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6.
The resistance and inductance of series circuit are
and 20H respectively. At the instant
5Ω
of closing the switch, the current is increasing at the rate 4A-s. The supply voltage is
(a) 20 V
7.
(b) 80 V
(c) 120 V
(d) 100 V
In inductance L and a resistance R are first connected to a battery. After some time the
battery is disconnected but L and R remain connected in a closed circuit. Then the current
(a) RL sec
R
sec
L
(c)
L
sec
R
(d)
1
sec
LR
In an LR-circuit, time constant is that time in which current grows from zero to the
value (where
0 . 63 I0
is the steady state current)
(b) 0 .50 I0
(c)
0 . 37 I0
(d)
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(a)
I0
n.
8.
(b)
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m
reduces to 37% of its initial value in
9.
I0
A solenoid has an inductance of 60 henrys and a resistance of 30 ohms. If it is connected
ed
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to a 100 volt battery, how long will it take for the current to reach
value?
(a)1 second
(c) e seconds
A coil of inductance 300 mH and resistance
2Ω
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10.
(b) 2 seconds
e −1
≈ 63 . 2 %
e
of its final
(d) 2e seconds
is connected to a source of voltage 2 V . The
current reaches half of its steady state value in
(b) 0.3 s
(c)
0.05 s
(d)
0.1 s
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(a) 0.15 s
11. Find the time constant (in µ s ) for the given RC circuits in the given order respectively.
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R1 = 1Ω , R2 = 2Ω , C1 = 4µ F , C2 = 2µ F
V
V
C1
C1
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R1
C2
R2
I)
C2
II)
a) 18, 4,
8
9
b) 18,
V
R1
R1
R2
R2
8
,4
9
C1
C2
III)
c) 4, 18,
8
9
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d) 4,
8
, 18
9
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12. An ideal coil of 10 Henry is joined in series with a resistance of 5 ohm and a battery of 5
volt. 2 second after joining, the current flowing in ampere in the circuit will be
1) e-1
2) (1-e–1)
3) (1-e)
4) e
13. An inductor of L = 400 mH and two resistors of R1 = 2Ω and R2 = 2Ω are connected to a
co
m
12V battery as shown in the figure. The internal resistance of the battery is negligible. The
12 −3t
e V
t
2) 6 (1 − e−t /0.2 )V
3) 12e −5tV
4) 6e−5tV
ed
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1)
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n.
switch is closed at time t = 0. The potential drop across ‘L’ as a function of time is
14. The current in the given circuit is increasing with a rate 4 amp/ s. The charge on the
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capacitor at an instant when the current in the circuit is 2 amp will be
2) 5 µ C
3) 6 µ C
4) 3µC
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1) 4 µ C
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15. The ratio of time constants in charging and discharging in the circuit shown in figure is
1) 1:1
2)1:2
3) 3:2
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4) 2:3
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Key
2) c
3) b
4) d
5) b
6) b
7) c
8) a
9) b
10) d
11) b
12) b
13) c
14) c
15) c
Rt
L

Rt
Rt

−
−
 ⇒ di = −i  − R e L = i0 R .e L
0

dt
L
 L

At t = 0;
(c) When battery disconnected current through the circuit start decreasing exponentially
according to
⇒
8.
i = i0 e − Rt / L
0 .37 i0 = i0 e − Rt / L
⇒
1
= e − Rt / L
e
0 . 37 =
⇒
t =
L
sh
i
7.
di i0 R E
E
=
=
⇒4 =
⇒ E = 80 V
dt
L
L
20
at
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−
ed
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
(b) i = i0  1 − e
R
(a) Current at any instant of time t after closing an L-R circuit is given by
t=
L
R
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Time constant
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6.
n.
Hints
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−R L 

×
1

∴ I = I 0 1 − e L R  = I 0 (1 − e −1 ) = I 0  1 − 
e



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1 

= I0 1 −
 = 0 .63 I 0 = 63 %
2
.
718


9.
co
m
1) a
t =τ =
(b)

10. (d) i = i0 1 − e

⇒
− Rt
L




t = 0 . 693 ×
Of I0
L 60
=
= 2 sec .
R 30
⇒ For
i=
i0
2
,
t = 0 . 693
L
R
300 × 10 −3
= 0 . 1 sec
2
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−R 

t
I = I 0 1 − e L 


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11. t = CR
t1 = ( C1 + C2 )( R1 + R2 ) = 18 µ s
C1C2
RR
8
× 1 2 =
C1 + C2 R1 + R2 9
t3 = ( C1 + C2 )

12. i = i0 1 − e
R1 R2
=4
R1 + R2
R
− t
L




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m
t2 =
n.
5×2
−

5
i = 1 − e 10 
5

R
− t 

12
= 6 , p.d = v − i0 1 − e L  R
2


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13. i0 =
−
t 
12 
−3
p.d = 12 − 1 − e 400×10  2
2 

2
p.d = 12 e −5t volt.
q = 6µ c
q
3
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.s
4=2+4–
sh
i
di q
−
dt c
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14. E = iR + L
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λ1 R2 2 R + R
=
=
λ2 R1
2R
Or
λ1 3
=
λ2 2
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15.
at
io
. i = (1 − e −1 ) A
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