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Alternating Current
Voltage and Power
2011
1.
In
an
AC
circuit,
V
and
I
are
given
by V = 150sin (150t ) volt ,
π

and I = 150sin  150t +  amp . The power dissipated in the circuit is

3
a) 106 W
b) 150 W
c) 5625 W
d) Zero
2010
2.
An AC ammeter is used to measure current in a circuit. When a given direct
current passes through the circuit, the AC ammeter reads 3A. When another
alternating current passes through the circuit, the AC ammeter reads 4A. Then
the reading of this ammeter, if DC and AC flow through the circuit
simultaneously, is
a) 3 A
3.
b) 4 A
c) 7 A
An AC source is 120V – 60Hz. The value of voltage after
a) 20.2 V
b) 42.4 V
d) 5 A
1
s from start will be
720
c) 84.8 V
d) 106.8 V
2009
4.
The oscillating electric and magnetic vectors of an electromagnetic wave are
oriented along
a) The same direction but has a phase difference of 90°
b) The same direction and are in same phase
c) Mutually perpendicular directions and are in same phase
d) Mutually perpendicular directions but has a phase difference of 90°
5.
Alternating current is transmitted to take places
a) At high voltage and low current
b) At high voltage and high current
c) At low voltage and low current
d) At low voltage and high current
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2008
6.
In an AC circuit the emf (e) and the current (i) at any instant are given
respectively by
e = Eo sin ω t
i = I o sin ( ωt − φ )
The average power in the circuit over one cycle of AC is
a)
7.
Eo I o
2
b)
Eo I o
sin φ
2
c)
Eo I o
cos φ
2
d) Eo I o
Alternating current cannot be measured by DC ammeter because
a) AC cannot pass through DC ammeter
b) AC changes direction
c) Average value of current for complete cycle is zero
d) DC ammeter will get damaged
2007
8.
If the power factor changes from
1
1
to then what is the increase in impedance in
2
4
AC?
a) 20%
9.
b) 50%
c) 25%
d) 100%
The instantaneous voltage through a device of impedance 20Ω is e = 80sin100π t .
The effective value of the current is
a) 3 A
e)
10.
b) 2.828 A
c) 1.732 A
d) 4 A
c) Zero resistance
d) None of the
2A
For high frequency, capacitor offers
a) More resistance
b) Less resistance
above
2006
11.
If reading of an ammeter is 10A, the peak value of current is
a)
10
A
2
b)
5
A
2
c) 20 2 A
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d) 10 2 A
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12.
If impedance is
a) Zero
13.
3 times of resistance, find phase difference
b) 30°
c) 60°
d) Data is incomplete
An alternating voltage V = Vo sin ωt is applied across a circuit. As a result the
current I = I o sin ( ωt − π 2) flows in it. The power consumed in the circuit per
cycle is
a) 0.5Vo I o W
14.
b) 0.707 Vo I o W
c) 1.919Vo I o W
d) Zero
An AC is represented by e = 220sin (100π ) t volt and is applied over a resistance
of110Ω . The heat produced in 7 min is
a) 11 × 103 cal
15.
b) 22 × 103 cal
c) 33 × 103 cal
d) 25 × 103 cal
If an AC produces same heat as that produced by a steady current of 4A, then
peak value of current is
a) 4 A
16.
b) 1.56 A
c) 5.6 A
d) 1.41 A
The potential difference across an instrument in an AC circuit of frequencies f is V
and the current through it is I such that V = 5cos 2π ft and I = 2sin 2π ft amp . The
power dissipated in the instrument is
a) Zero
b) 10 W
c) 5 W
d) 2.5 W
2005
17.
An alternating current is given by I = I1 cos ωt + I 2 sin ωt . The root mean square
current is
a)
18.
( I1 + I 2 )
2
2
2
c)
I12 + I 22
2
d)
I12 − I 22
2
The peak value of AC voltage on 220V mains is
a) 240 2 V
19.
b)
( I1 + I 2 )
b) 230 2 V
c) 220 2 V
d) 200 2 V
The time taken by an alternating current of 50Hz in reaching from zero to its
maximum value will be
a) 0.5 s
b) 0.005 s
c) 0.05 s
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d) 5
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Alternating Current
Voltage and Power
Key
1)
c
2)
d
3)
c
4)
c
5)
a
6)
c
7)
c
8)
d
9)
b
b
11) d
12) c
13) d
14) b
15) c
16) a
17) c
Solutions
1.
V = 150 sin (150t) volt
And I = 150sin (150t + π / 3)amp
I 0 = 150amp and
V0 = 150volt
1
P = Vo I 0 cos φ
2
P = 0.5 × 150 × 150 × cos 600
2.
Quantity of heat liberated in the ammeter of resistance R
i) Due to direct current of 3 A,
2
Q = ( 3 ) R / J ) 


2
ii) Due to alternating current of 4 A, Q = ( 4 ) R / J ) 


Total heat produced per second
( 3)
=
2
J
R
( 4)
+
2
J
R
=
I 2 R 25R
But,
=
J
J
I=5A
3.
V = V0 sin ωt
V = Vrms 2 sin ωt
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25 R
J
18) c
19) b
10)
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t=
1
s
720
V = 120 2 sin 2π v t = 120 2 sin 2π × 60 ×
6.
Pav =
Pav =
W
T
( E0 I 0 cos φ ) T / 2
Pav =
8.
T
E0 I 0 cos φ
2
cos φ α
1
Z
Z1 cos φ1 1/ 4 1
=
=
=
Z 2 cos φ2 1/ 2 2
∴
Z 2 = 2 Z1
⇒
Percentage change =
9.
1
= 60 2 = 84.8V
720
2 Z1 − Z1
× 100 =100%
Z1
e = 80 sin 100π t
Here, e0 = 80V
Where e0 is the peak value of voltage
Impedance (Z) = 20 Ω
I0 =
11.
e0
e
I0 = 0
Z
Z
I rms =
I0
4
=
= 2 2 = 2.828 A
2
2
irms =
i0
2
⇒ i0 = 2 × irms = 2 × 10 A = 10 2 A
2011
13.
V = V0 sin ωt and I = I 0 sin (ωt − π / 2 )
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1
∴ P = V0 I 0 cos φ
2
But φ = π / 2 = 900
1
P = V0 I 0 cos 900
2
(
= 0 ∴cos900 = 0
14.
H=
I v2 RT
J
)
( 2)
cal =
2
× 100 × 7 × 60
42
= 22 × 103 Cal
I 0 220 /100


=
= 2
∵ I ms =
2
2


15.
I 0 = 2 I rms
I rms = 4 A
I 0 = 2 × 4 = 5.6 A
16.
P = VI cos φ
V= 5cos P = VI cos 2π ft
π

V = 5sin  2π ft +  and I = 2sin 2π ft
2

Here, φ =
π
2
P = VI cos φ = VI cos
17.
π
2
=0
I = I1 cos ωt + I 2 sin ωt
I 0 = I12 + I 22
I rms
18.
I
= 0 =
2
I12 + I 22
2
=
I12 + I 22
2
Vrms = 220V
But Vrms =
peak voltage VP
=
2
2
VP = 220 2 V
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19.
T=
∴
1
1
=
= 0.0.2 s
f 50
T 0.02
=
= 0.005s
4
4
AC Circuits
2011
1.
In an AC circuit an alternating voltage e = 200 2 sin100t volt is connected to a
capacitor of capacity 1 µ F . The rms value of the current in the circuit is
a) 100 mA
2.
b) 200 mA
c) 20 mA
d) 10 mA
An AC voltage is applied to a resistance R and an inductor L in series. If R and the
inductive reactance are both equal to 3Ω , the phase difference between the applied
voltage and the current in the circuit is
3.
a) π 4
b) π 2
c) Zero
d) π 6
The peak value of an alternating current is 5A and its frequency is 60Hz. Find its
rms value and time taken to reach the peak value of current starting from zero
4.
a) 3.536 A, 4.167 ms
b) 3.536 A, 15 ms
c) 6.07 A, 10 ms
d) 2.536 A, 4.167 m
An L – C – R series current is under resonance. If I m is current amplitude, Vm is
voltage amplitude, R is the resistance, z is impedance, X L is the inductive
reactance and X C is the capacitive reactance then
a) I m =
5.
Vm
Z
Vm
XL
b) I m =
c) I m =
Vm
XC
d) I m =
Vm
R
In the case of an inductor
a) Voltage lags the current by
π
c) Voltage leads the current by
2
π
3
b) Voltage leads the current by
d) Voltage leads the current by
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π
2
π
4
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2010
6.
The power factor of an R – L circuit is
1
. If the frequency of AC is doubled,
2
what will be the power factor?
a)
7.
1
3
b)
1
5
c)
1
7
d)
1
11
An ideal choke draws a current of 8A when connected to an AC supply of 100V,
50Hz. A pure resistor draws a current of 10A when connected to the same source.
The ideal choke and the resistor are connected in series and then connected to the
AC source of 150V, 40Hz. The current in the circuit becomes
a)
8.
15
A
2
b) 8A
c) 18A
d) 10A
In an AC circuit the potential difference V and current i are given respectively by
π

V = 100sin (100t ) volt and i = 100sin  100t +  mA . The power dissipated in the
3

circuit will be
a) 104 W
9.
b) 10 W
c) 2.5 W
d) 5 W
An inductor L, a capacitor of 20 µ F and a resistor of 10Ω are connected in series
with an AC source of frequency 50Hz. If the current in the phase with the voltage,
then the inductance of the inductor is
a) 2.00 H
10.
b) 0.51 H
c) 1.5 H
d) 0.99 H
The impedance of a circuit, when a resistance R and an inductor of inductance L
are connected in series in an AC circuit of frequency f, is
11.
a)
R + 2π 2 f 2 L2
b)
R + 4π 2 f 2 L2
c)
R 2 + 4π 2 f 2 L2
d)
R 2 + 2π 2 f 2 L2
In a series L – C – R circuit, resistance R = 10 Ω and the impedance Z = 10 Ω . The
phase difference between the current and the voltage is
a) 0°
b) 30°
c) 45°
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d) 60°
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12.
The inductive time constant in an electrical circuit is
a) LR
13.
b)
c)
L
R
d)
R
L
In a L – C – R circuit R = 10 Ω . When capacitance C is removed, the current lags
behind the voltage by
voltage by
π
3
a) 50 Ω
14.
L
R
π
3
, when inductance L is removed, the current leads the
. The impedance of the circuit is
b) 100 Ω
c) 200 Ω
d) 400 Ω
In an L – C – R series Ac circuit at resonance
a) The capacitive reactance is more than the inductive reactance
b) The capacitive reactance equals the inductive reactance
c) The capacitive reactance is less than the inductive reactance
d) The power dissipated is minimum
15.
An L – C – R series circuit consists of a resistance of 10 Ω a capacitor of reactance
6.0 Ω and an inductor coil. The circuit is found to resonate when put across a
300V, 100Hz supply. The inductance of coil is (taking π = 3 )
a) 0.1 H
16.
b) 0.01 H
c) 0.2 H
d) 0.02 H
A capacitor or capacitance 1 µ F is charged to a potential of 1V. It is connected in
parallel to an inductor of inductance 10−3 H . The maximum current that will flow
in the circuit has the value
a) 1000 mA
17.
b) 1 A
c) 1 mA
d) 1000 mA
Assertion: For an electric lamp connected in series with a variable capacitor and
AC source, its brightness increases with increase in capacitance
Reason: Capacitive reactance decreases with increase in
1) Both assertion and reason are true and reason is the correct explanation of assertion
2) Both assertion and reason are true but reason is not the correct explanation of
assertion
3) Assertion is true but reason is false
4) Assertion is false but reason is true
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18.
 0.4 
In an L – R circuit, the value of L is 
 H and the value of R is 30Ω . If in the
 π 
circuit, an alternating emf of 200V at 50cycle s is connected, the impedances of
the circuit and current will be
a) 11.4 Ω, 17.5 A
b) 30.7 Ω, 6.5 A
c) 40.4 Ω, 5 A
d) 50 Ω, 4 A
2009
19.
Power dissipated in an L – C – R series circuit connected to an AC source of emf
ε is
a)



20.

2
1  
 
Cω  
ε 2R
d)
R
2
2
b)
2
 2 
1  
 R +  Lω −
 
Cω  


ε 2  R 2 +  Lω −
c)
ε
ε 2R
1 

R +  Lω −

Cω 

R
2
1 

R +  Lω −

Cω 

2
2
An AC voltage is applied to a pure inductor L, drives a current in the inductor.
The current in the inductor would be
21.
a) Ahead of the voltage by π 2
b) Lagging the voltage by π 2
c) Ahead of the voltage by π 4
d) Lagging the voltage by 3π 4
The average power dissipated in a pure capacitance AC circuit is
a) CV
22.
b) Zero
1
CV 2
d)
1
CV 2
4
In a pure inductive circuit, current
a) Lags behind emf by
π
b) Leads the emf by
2
c) Lags behind by π
23.
c)
π
2
d) Leads the emf by π
An alternating current of rms value 10A is passed through a 12 Ω resistor. The
maximum potential difference across the resistor is
a) 20 V
b) 90 V
c) 169.68 V
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d) None of these
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24.
Same current is flowing in to alternating circuits. The first circuit contains only
inductance and the other contains only a capacitor. If the frequency of the emf of
AC is increased, the effect on the value of the current will be
a) Increases in the first circuit and decreases in the other
b) Increases in both the circuits
c) Decreases in both the circuits
d) Decreases in the first circuit and increases
2008
25.
An AC source of angular frequency ω is fed across a resistor R and a capacitor C
in series. The current registered is I. If now the frequency of source is changed to
ω 3 (but maintaining the same voltage), the current in the circuit is found to be
halved. Calculate the ratio of reactance to resistance at the original frequency ω
a)
26.
3
5
b)
2
5
c)
1
5
d)
4
5
What is the value of inductance L for which the current is a maximum in a series
L – C – R circuit with C = 10 µ F and ω = 1000 s −1 ?
27.
a) 100 mH
b) 1 mH
c) Cannot be calculated unless R is known
d) 10 mH
In non – resonant circuit, what will be the nature of the circuit for frequencies
higher than the resonant frequency?
a) Resistive
28.
b) Capacitive
c) Inductive
d) None of the above
A coil has an inductance of 0.7H and is joined in series with a resistance of 20Ω .
When an alternating emf of 220V at 50 cps is applied to it, then the wattles
component of the current in the circuit is
a) 5 A
29.
c) 0.7 A
d) 7 A
In L – C – R circuit what is the phase angle φ ?
a) 90°
30.
b) 0.5 A
b) 180°
c) 0°
d) 60°
In R – C circuit ω = 100 rad s −1 , R = 100 Ω, C = 20 µ F . What is impedance?
a) 510Ω
b) 200Ω
c) 250Ω
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d) 300Ω
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31.
The power factor in a circuit connected to an AC power supply has a value which
is
a) Unity when the circuit contains only inductance
b) Unity when the circuit contains only resistance
c) Zero when the circuit contains an ideal resistance only
d) Unity when the circuit contains an ideal capacitance only
32.
In an AC circuit, a resistance of R ohm is connected in series with an inductor of
self inductance L. If phase angle between voltage and current be 45° , the value of
inductive reactance ( X L ) will be equal to
a) R
33.
b)
R
8
c)
R
4
d)
R
2
A transistor – oscillator using a resonant circuit with an inductor L (or negligible
resistance) and a capacitor C in series produce oscillations of frequency f. If L is
doubled and C is changed to 4C, the frequency will be
a) f 4
34.
d) f 2
The natural frequency ( ω o ) of oscillations in L – C circuit is given by
a)
35.
c) f 2 2
b) 8f
1
2π
1
LC
b)
1
2π
LC
c)
1
LC
d)
LC
In L – C- R circuit the resonance condition in terms of capacitive reactance ( X C )
and inductive reactance ( X L ) is
a) X C + X L = 0
b) X C = 0
c) X L = 0
d) X C − X L = 0
2007
36.
An inductor of inductance L and resistor of resistance R are joined in series and
connected by a source of frequency ω . Power dissipated in the circuit is
 R 2 + ω 2 L2 
a) 
V


b)
(R
V 2R
2
+ ω 2 L2
)
V


c)  2
2
2
 R + ω L 
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d)
R 2 + ω 2 L2
V2
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37.
An inductive coil has a resistance of 100Ω . When an AC signal of frequency 1000
Hz is applied to the coil, the voltage leads the current by 45° . The inductance of
the coil is
a)
38.
1
10 π
2)
1
20 π
c)
1
40 π
d)
1
60 π
(
The reactance of a coil when used in the AC power supply 220V , 50 cycles −1
)
is 50 Ω . The inductance of the coil is nearly
a) 0.16 H
39.
b) 0.22 H
c) 2.2 H
d) 1.6 H
 10−3 
 200 
An inductance of 
mH
,
a
capacitance
of
 π  F and a resistance of 10 Ω
 π 
are connected in series with an AC source 220V, 50Hz. The phase angle of the
circuit is
a)
40.
π
b)
6
π
c)
4
π
2
d)
π
3
In a circuit, the current lags behind the voltage by a phase difference of π 2 , the
circuit will contain which of the following?
a) Only R
41.
b) Only C
c) R and C
d) Only L
A coil of inductive reactance 31Ω has a resistance of 8 Ω . It is placed in series with
a condenser of capacitive reactance 25 Ω . The combination is connected to an AC
source of 110V. The power factor of the circuit is
a) 0.56
42.
b) 0.64
d) 0.33
In a choke coil, the reactance X L and resistance R are such that
a) X L = R
43.
c) 0.80
b) X L >> R
c) X L << R
d) X L = ∞
The power factor of a series L – C – R circuit when at resonance is
a) Zero
b) 0.5
c) 1.0
d) Depends on values of L, C and R
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2006
44.
In a series L – C – R circuit the frequency of a 10V, AC voltage source is adjusted
in such a fashion that the reactance of the inductor measures 15Ω and that of the
capacitor 11Ω . If R = 3Ω , the potential difference across the series combination of
L and C will be
a) 8 V
45.
b) 10 V
c) 22 V
d) 52 V
An L – C circuit is in the state of resonance. If C = 0.1µ F and L = 0.25H,
neglecting ohmic resistance of circuit, what is the frequency of oscillations?
a) 1007 Hz
46.
b) 100 Hz
c) 109 Hz
d) 500 Hz
In an AC circuit, the current lags behind the voltage by π 3 . The components of
the circuit are
1) R and L
47.
3) R and C
4) Only R
What is the ratio of inductive and capacitive reactance in AC circuit?
a) ω 2 LC
48.
2) L and C
b) 1
c) zero
d) ω 2 L
If a circuit made up of a resistance 1Ω and inductance 0.01H, and alternating emf
200V at 50Hz is connected, then the phase difference between the current and the
emf in the circuit is
a) tan −1 ( π )
49.
π
b) tan −1  
 2
π
c) tan −1  
 4
π
d) tan −1  
 3
A circuit draws 330W from a 110V, 60Hz AC line. The power factor is 0.6 and the
current lags the voltage. The capacitance of a series capacitor that will result in a
power factor of unity is equal to
a) 31 µ F
b) 54 µ F
c) 151 µ F
d) 201 µ F
2005
50.
In a circuit, L, C and R are connected in series with an alternating voltage source
of frequency f. The current leads the voltage by 45° . The value of C is
a)
c)
1
2π f ( 2π fL + R )
1
2π f ( 2π fL − R )
b)
1
π f ( 2π fL + R )
d)
1
π f ( 2π fL − R )
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AC Circuits
Key
1)
c
2)
a
3)
a
4)
a
5)
b
6)
b
7)
a
8)
c
9)
b
10) c
11) a
12) b
13) b
14) b
15) a
16) a
17) a
18) d
19) a
20) a
21) b
22) a
23) c
24) d
25) a
26) a
27) c
28) b
29) c
30) a
31) b
32) a
33) c
34) a
35) d
36) b
37) b
38) a
39) b
40) d
41) c
42) b
43) c
44) a
45) a
6)
47) a
48) a
49) b
50) c
a
Solutions
1.
e = 200 2 sin100t
and
C = 1µ F
Erms = 200V
2.
3.
XC =
1
1
=
= 104 Ω
−6
ωC 1× 10 × 100
irms =
Erms
Xc
irms =
200
= 2 × 10−2 A = 20 mA
4
10
tan φ =
X L Lω
=
R
R
tan φ =
3Ω
3Ω
irms =
T=
6.
Or φ =
π
4
rad
i0
5
=
= 2.536 A
2
2
1
= 4.167 ms
4 × 60
cos φ =
R
R 2 + 4ω 2 L2
=
R
R2 + 4R2
=
1
5
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7.
R=
100
= 10Ω
10
X L = 2π fL
100
= 2π × 50 × L
8
⇒ L=
X i = 2π fL = 2π × 40 ×
1
8π
X i = 2π fL = 2π × 40 ×
1
=
8π
i=
8.
1
H
8π
(10 )
2
+ (10 ) = 10 2Ω
2
V
150
15
=
=
A
Z 10 2
2
V0 = 100V , i0 = 100 mA = 100 × 10−3 A and φ =
π
3
= 600
V0i0
100 × 100 × 10−3
P=
× cos 600 = 2.5 W
cos φ =
2
2
9.
tan φ =
ωL −
L=
ωL −
R
1
ωC = 0
1
ωC
1
ω 2C
Here ω = 2π f = 2 × 3.14 × 50s −1 = 314s −1
And C = 20µ F = 20 × 10−6 F
∴L =
11.
(
1
314 s
) (
−1 2
× 20 × 10−6 F
)
= 0.51H
Z = R 2 + X L2
X L = ω L = 2π fL
Z = R 2 + 4π 2 f 2 L2
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(10 )
∴ 10 =
2
+ ( X L − XC )
⇒ 100 = 100 + ( X L − X C )
2
2
⇒ X L − XC = 0
13.
tan
π
3
=
XL
R
When L is removed, tan
π
XL
R
=
3
Z= R = 100Ω
⇒ i0 =
220 × 2
220 × 2
=
100π
100 × 3.14
I 0 = 1A
15.
ω0 = 2π n − 2π × 100
ω0 = 2 × 3 × 100 = 600rad / s
Also, ω0 =
XC =
1
LC
1
= 60ω
Cω0
⇒ C=
1
1
=
ω0 × 60 600 × 60
1
600 −
 1 
L
3
 36 × 10 
36 × 103
36 × 10 =
L
4
36 × 103 1
L=
=
= 0.1H
36 × 104 10
16.
q = q0 sin ωt
Or I 0 = ω q0 = maximum current
ω=
( )
1
1
=
= 109
LC
10−9
1/ 2
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( )
I 0 = 109
18.
1/ 2
(
)
× 1× (10−6 = 1000 mA
X L = ω L = 2π fL = 2π × 50 ×
0.4
π
= −40Ω
R = 30Ω
( 30 )
Z = rR 2 + X L2 =
irms =
25.
2
+ ( 40 ) = 50Ω
2
vrms 200
=
= 4A
X
50
Vrms
I rms =
 1 
R +

 ωC 
2
2
I rms
=
2
Also,
∴ 3R 2 =
26.
Vrms




1
R2 + 
ωc 


 3 
2
Vrms
=
R2 +
9
ω C2
2
5
ω C2
2
⇒
1
3
=
ωC
5
R
i=
V
R2 + ( X L = X C )
2
But, X L = X C
Or ω L =
1
ωC
Or L =
1
ω 2C
ω = 1000 s −1 , C = 10µ F = 10 × 10−6
∴L =
28.
1
(1000 )
2
× 10 × 10−6
=0.1 H
Wattless component of current = I v sin θ =
=
220
R 2 + ω 2 L2
×
ωL
R 2 + ω 2 L2
=
220 × ω L
(R
2
+ ω 2 L2
EV
sin θ
Z
)
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=
34.
220 × ( 2π × 50 × 0.7 )
( 220 )
+ ( 2π × 50 × 0.7 )
2
X L = X C ⇒ ωo L =
1
= 0.5 A
2
1
ω0C
1
LC
ω02 =
1
LC
ω0 =
2π f =
f =
2
=
1
LC
1
2π LC
35.
Resonance condition is X L = X C or X C − X L = 0
37.
tan φ =
XL
R
⇒ tan 450 =
⇒ L=
⇒ L=
38.
R
ω
=
Lω
R
100
2π (1000 )
1
20π
X L = ω L = 2π fL
X L = 50Ω, f = 50cps
L=
39.
XL
ω
tan θ =
=
XL
50
1
=
=
= 0.16 H
2π f 2π × 50 2 × 3.14
X L − XC
R
 200

X L = 2π fL = 2π × 50 × 
× 10−3  = 20Ω
 π

XC =
1
1× π
=
= 10Ω
2π fC 2π × 50 × 10−3
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And R = 10Ω
tan θ =
20 − 10
=1
10
tan θ = tan
π
4
π
θ=
4
1
45.
f =
47.
X L = ωL
2π LC
=
1
2π 0.25 × 0.1 × 10−6
and X C =
≈ 1007 Hz
1
ωC
XL
ωL
=
= ω 2 LC
X C 1/ ωC
48.
XL
and X L = ω L = 2π fL = 2π × 50 × 0.01 = πΩ
R
tan φ =
Also R = 1Ω ∴φ = tan −1 (π )
49.
V2 110 × 110 110
=
=
Ω
P
330
3
R=
Cos R =
V2 110 × 110 110
=
=
Ω = 0.6.
P
330
3
R
R 2 + X L2
R +
2
X L2
X L2
=
XL =
= 0.6
 R 
=

 0.6 
R2
( 0.6 )
2
−R
2
2
⇒
X L2
R 2 × 0.64
=
0.36
0.8 R 4 R
=
0.6
3
And
X L = KC
XC =
4 R 4 100 440
= ×
=
Ω
3
3 3
9
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1
=
2π f
C=
440
Ω
9
9
2 × 3.14 × 60 × 440
−0.000054 F = 54 µ F
50.
ωL −
tan φ =
1
ωC
R
φ = 45°
∴
tan 45° =
⇒ 1=
⇒
ωL −
ωL −
1
ωC
R
1
ωC
R
R = ωL −
⇒
ωC =
⇒
C=
1
ωC
1
(ω L − R)
1
1
=
2 ( ω L − R ) 2π f ( 2π fL − R )
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Growth and Decay of Current
2010
1.
A coil of resistance R and inductance L is connected to a battery of emf E volt. The
final current in the coil is
a)
E
R
b)
E
L
c)
E
R + L2
2
d)
EL
R + L2
2
2008
3.
A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage
2V. The current reaches half of its steady state value in
a) 0.05 s
b) 0.1 s
c) 0.15 s
d) 0.3 s
2007
4.
A coil of 40H inductance is connected in series with a resistance of 8Ω and this
combination is connected to the terminals of 2V battery. The inductive time
constant of the circuit is (in second)
a) 40
5.
b) 20
c) 5
d) 0.2
In an L – R circuit, time constant is that time in which current grows from zero to
the value
a) 0.63 I o
b) 0.50 I o
c) 0.37 I o
d) I o
2004
6.
In a series resonant L – C – R circuit, the voltage across R is 100V and R = 1k Ω
with C = 2µ F . The resonant frequency ω is 200 rad s −1 . At resonance the voltage
across L is
a) 2.5 × 10 −2 V
b) 40 V
c) 250 V
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d) 4 × 10−3 V
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Growth and Decay of Current
Key
1)
a
2)
c
3)
b
4)
c
5)
a
6)
c
Hints
2.
At time t = 0, i.e., when capacitor is charging, current I =
I=
2
= 1mA
2000
(
I = I o 1 − e − Rt
3.
L
(
Io
= I o 1 − e − Rt
2
(
1
= 1 − e − Rt
2
L
)
L
)
)
e − Rt L = 1 2
Rt
= ln 2
L
∴
L
300 × 10 −3
ln 2 =
× 0.693
R
2
= 150 × 0.693 × 10 −3
= 0.10395 s = 0.1 s
L 40
=
= 5s
R 8
4.
t=
5.
I = I o 1 − e −1 τ
(
)
 L
τ  =  Is called time constant.
 R
At t = τ
(
I = I o 1 − e −1
)
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2
= 2mA
1000
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 1
Or I = I o 1 − 
 e
I = 0.63 I o
Or
6.
ωL =
I=
1
ωC
VR 100
=
= 0.1A
R 1000
VL = IX L = I ω L
But
∴ VL =
ωL =
1
ωC
1
0.1
=
= 250V
ωC 200 × 2 × 10−6
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