Chapter 07
Chapter‐07
A C CIRCUITS
A C CIRCUITS
1. Effective voltage is given by expression 1) Ve = Vo/√2 2) Ve = √2 Vo
3) Vo/π
4) π Vo
Effective voltage is rms voltage Effective
voltage is rms voltage
∴ Answer is (1)
2. A coil having zero resistance is connected in il h i
i
i
di
series with a 90Ω resistance and the combination is connected to 120V, 60 Hz line. bi i i
d 120 60 li
A voltmeter reads 36 V across the resistance and 114 V across the coil. The self inductance d 114 V
h
il Th
lf i d
of the coil is
1) 0.076 H 2) 0.76 H
3) 7.6 H
4) 76 H
Given VR= 36 V
R = 90 Ω
R = 90 Ω
∴ I = V/R =36/90 = 0.4 A
Also given VL= 114V
∴ XL = V
= VL/I /I
2πυL = 114/0.4
L = 114/2x3.14x60x0.4 = 0.76 H
3. A radio transformer has 660 turns of wire in 3
A di t
f
h 660 t
f i i
the primary coil which is connected to 220 V AC
AC source. The secondary coil supplies 6.3V Th
d
il
li 6 3V
for the filament of a bulb. What is the n mber of t rns in the secondar ?
number of turns in the secondary ? • 1) 2
2) 4
• 3) 19
)
4) 57
)
• Ns/Np = Vs/Vp
Ns/Np = Vs/Vp
• Ns = Vs Np /Vp
/
•
= 6.3 x660 / 220
•
= 6.3x 3 =18.9 =19
4. A current of 4A flows in a coil when connected to a 12 V dc source If the same coil is connected
to a 12 V dc source. If the same coil is connected to a 12V, 50 rad s‐1 ac source, acurrent of 2.4 A flows The inductance of the coil is
flows. The inductance of the coil is
• 1) 80 H
2) 8 H
• 3) 80 mH
4) 8 mH.
Idc = 4A Vdc = 12 V ∴R = Vdc / Idc = 12 /4 =3Ω
Vac = 12V, Iac = 2.4 A ∴Z = Vac / Iac = 12 /2.4 =5Ω
ω= 50 rad s‐1
L =√( Z2 – R2) / ω = 4/50 =0.08 = 80 mH
5. An a.c. having a peak value 28 A is used to 5
A
h i
k l 28 A i
dt
heat a metal wire. To produce the same h ti
heating effect a constant current I that can be ff t
t t
t I th t
b
used is 1) 28A
2) about 20A
3) 7A
)
4) about 10A
)
Answer : Idc = Irms = Io /√2 = 28/1.41
Answer : Idc = Irms = Io /√2
28/1 41 =20 A
6. A circuit has a resistance of 30 Ω
6
A circuit has a resistance of 30 Ω in series in series
with an inductive reactance of 40 Ω . They are connected in series with an ac source If
are connected in series with an ac source. If the peak value of current is 1 A and the peak voltage is 220 V the power consumed by the
voltage is 220 V, the power consumed by the circuit is • 1) 66 W
• 3) 0.66 W
2) 6.6. W
4) 33 W
R=30 Ω XL = 40 Ω . Z= √(R2 + XL 2)= 50Ω
P = ½ (VoIo cos φ)
= ½ VoIo (R/Z) ( / )
= ½ x1x 220 x0.6 ½ x1x 220 x0.6 = 66 W
66 W
Answer is ( 1)
7. In the circuit shown VAB = VAC , then P is 1) an inductor 2) a capacitor 3) a suitable combination of inductor and capacitor
4) resistor
VBC = 0 It is possible iff X is a combination L & C at resonance
& C at resonance
8. In an ac circuit, the voltage V and the current I are given by V 100 sin ( 100 t ) volt and
are given by V = 100 sin ( 100 πt ) volt and I = 100 sin ( 100t + π/3 ) mA. The power (
)
p
dissipated in the circuit is
1) zero
2) 10 3) 5 W
4) 2.5 W
•
•
•
•
P = ½(VoIo cos φ)
P
= ½(VoIo cos φ)
= ½(100x100x10-3 cos 60)
= ½ x10x0.5
½ 10 0 5
= 2.5 W
9. In a series LCR circuit, R = 100 Ω
9
I
i LCR i it R 100 Ω and applied d
li d
ac voltage is 200 V. When the capacitance alone is removed, the voltage leads the l
i
d th
lt
l d th
current by 300. When the inductance alone is removed, the voltage lags the current by 30
d th
lt
l
th
t b 300. Then the current in the circuit is
1) 0.5 A
2) 1 A
3) 2A
4) 4 A
• Net
Net phase difference between V & I phase difference between V & I
φ= φ1 + φ2 = 30 – 30 = 0
• So ckt is in resonance
S k i i
• I = V/R = 200/100 = 1 A
10.A 4 µF capacitor charged to 50 V is connected across an ideal inductor of inductance 10 mH.
across an ideal inductor of inductance 10 mH. The maximum value of current in the circuit is
• 1) 20 mA
1) 20 mA
2) 2 mA
2) 2 mA
• 3) 1 mA
4) 1 A
Given, C = 4 µF V= 50 V L = 10 x 10‐33 H. • Energy
Energy stored in the capacitor is converted stored in the capacitor is converted
into energy in the inductor
• ½ LI
½ LI2 = ½ CV
= ½ CV2
• ½
½ x 10 x 10
x 10 x 10‐33 I2 = ½ 4x 10
= ½ 4x 10‐66 x 2500
x 2500
2 = 1
• I2 =1
• Answer is (4) = 1 A
11.A transformer is used to light 140 watt, 24 volt lamp from 240 V ac mains. The current in the main cable is 0.7 A. The efficiency of the transformer is
•
1) 63.8%
3) 83.3 %
3) 83.3 %
2) 84%
4) 48%
4) 48%
Given ; Ps =140 watt, Vs=24 volt Vp = 240 V Ip= 0.7 A. • The efficiency of the transformer
The efficiency of the transformer
• η = Ps/Pp
• η = Ps/Vs
/ Is =140/(24
140/(24 x0.7)
0 ) = 20/24
•
= 0.833
• Answer is (3) = 83.3 %
83 3 %
12.In a LRC circuit, capacitance is changed from f
C to 2C. For the resonant frequency to remain unchanged, the inductance should be h
d h
d
h ld b
changed from L to • 1) 4L 2) 2L
• 3) L/2
4) L /4
• For
For the resonant frequency to remain the resonant frequency to remain
unchanged
LC = const
LC = const
Lα1/C
If C --------- 2C, then L---------ÆL/2
Answer is ( 3) = L/2
13.An ac source E = 200√
√ 2 sin 100t is connected across a circuit containing a AC ammeter and a capacitor of capacity 1µ F. Then the reading of the ammeter is • 1) 10mA 2) 20mA
• 3) 40mA
4) 80mA
E = 200 √ 2 sin 100t
comparing with E = Eo sin ωt
Eo = 200
Eo
= 200 √ 2 & 2 & ω = 100
= 100
• Reading of the ammeter I = E/Xc
• I = Eo// √ 2 Xc = 200 x ω C = 200x100 X10‐6 = 2X10‐2 = 20X10‐3 A
= 20X10
Answer is (2) = 20mA
14.The impedance of a coil of resistance R to AC of frequency 50Hz is √3R. The self ff
i √
h
lf
inductance of the coil is (if R = 6.28Ω) • 1) 0.25 H
2) 2.5 H
• 3) 0.028 H
4) 0.025 H
Z = √3
Z √3 R υ
υ= 50Hz 50Hz & R = 6.28Ω
•
•
•
•
•
•
•
LL =√(
=√( Z2 – R2) / 2πυ
= √( 3R2 – R2) / 2πυ
= √2 x R / 2πυ
= √2 x6.28 / 2x 3.14 x50
=1.41/50 = 0.0282 H
Answer is (3)
15.The reactance offered by a coil to AC of certain frequency is equal to its resistance. The phase difference between the voltage across the coil to the current through the coil in radians is • 1) π/3
2) π/6
• 3) π/2 4) π/4
Given XL =R
Given X
R
• The
The phase difference between the voltage phase difference between the voltage
across the coil to the current through the coil tan φ = (XL –Xc)/R
tan φ
Xc)/R •
= XL /R = 1
• Φ = π/4
/
• Answer is 4) π/4
16.The choke coil has a resistance 8 Ω and the i d ti
inductive reactance 6 Ω. The power factor of t
6 Ω Th
f t
f
the coil is • 1) 0.6
06
• 3) 0.4
`
2) 0 8
2) 0.8
4) 0.3
Given, R =8
Given, R 8 Ω
Ω
The power factor of the coil ,
power factor of the coil
• The
Cos φ = R/Z
=8/10
8/10 = 0.8
08
Answer is (2) = 0.8
XL= 6 Ω. 6 Ω.
17.An alternating current is given by
I= I1 cos wt + I
I= I
cos wt + I2 sin wt .
sin wt
the rms current is given by
n
φ
I √ (I12+ I22 +2 I1 I2cosφ)) b
Io = √
but)
t) φ=90
900
I √ (I12+ I22 )
Io = √
I
Irms = Io/√2 = √
I /√2 √ (I12+ I22 )/2
Answer is
18.In the circuit shown, the inductor used is an ideal one. The current in the circuit when the
switch is closed is • 1) 0.2 A
2) 0.3 A
3) 1 A
4) 0.14 A
Since the inductor is ideal, the circuit can be Since
the inductor is ideal the circuit can be
written as ∴ I = E/(R
I = E/(Re+r)
I =10/10 = 1A
Answer is (3) I = 1A
19.Reciprocal of impedance is called • 1) conductance
2) admittance
• 3) conductivity
3) conductivity
4) resistivity
4) resistivity
• Answer is (2) 20.In an ac circuit the virtual power is 50 W and the power factor is ½ Then the actual power is
the power factor is ½ . Then the actual power is
• 1) 50 W
2) 100 W
• 3) 25 W
4) 75 W Pav = VI cosφ
VI cosφ
• Pav = P
= Pviri cosφ
• Pav = 50 x ½
0 ½
• Answer is (3) = 25 W
21.The primary of a transformer has 1000 turns 21
Th
i
f
f
h 1000
and secondary has 2000 turns. For any input voltage, the power in the secondary is l
h
i h
d
i
1) double that in primary 2) less or equal to that in primary 3) half of that in primary 4) 1 5 times that in primary
4) 1.5 times that in primary Answer is (2) 22.In an AC circuit containing a pure inductor 22
In an AC circuit containing a pure inductor
and a capacitor, the current and voltage are are
1) in phase
in phase
2) out of phase by 900
0
3) out of phase by 180
)
f h
b
4) out of phase by 450
Answer is (1) 23.The band width of a resonance curve with respect to a series RLC circuit is 48 kHz If the
respect to a series RLC circuit is 48 kHz. If the quality factor is 3, the resonant frequency is • 1) 51 kHz 2) 45 kHz
• 3) 144 kHz
4) 16 kHz
Given, B.W = 48 kHz. Q
Given, B.W 48 kHz. Q= 3
3
Q = υo/ BW
Q = υ
/ BW
υo = Q xBW = 3x 48 kHz.
k
= 144 kHz
Answer is (3) = 144 kHz
24.An inductor of inductance 2 H and a resistance of 10 Ω are connected to a battery resistance of 10 Ω
are connected to a battery
of 5 V in series. The initial rate of change of current is
current is • 1) 0.5 A /s
2) 1.25 A/s
• 3) 2.5 A/s
4) 2 A/s
Given : L = 2 H and a resistance of 10 Ω are connected to a battery of 5 V in series The
connected to a battery of 5 V in series. The initial rate of change of current is
Applying kirchof’s loop rule to the ckt
E ‐ L.(dI/dt) = IR
( / )
Initially, I = 0
∴dI/dt = E/L = 5/10= 0 5A/s
∴dI/dt = E/L = 5/10= 0.5A/s
∴Answer (1) = 0.5 A /s
25.A bulb B and a capacitor C 25
A bulb B and a capacitor C
are connected to a battery
as shown in figure
as shown in figure.
When the switch S is closed. 1) the bulb will glow when the capacitor fully gets charged.
gets charged.
2) the bulb glows during charging of the capacitor. capacitor.
3) the bulb will not glow at all
4) the bulb glows intermittently due to to the the bulb glows intermittently due to to the
charging and discharging of the capacitor. Bulb glows only when current flows through the circuit In the given circuit current flows
the circuit. In the given circuit, current flows during charging and discharging of the capacitor When the switch S is closed the
capacitor. When the switch S is closed, the process is charging • Hence the answer is (2) = the bulb glows d i
during charging of the capacitor
h i
f th
it
27.Which of the following statements is not true at the resonant frequency of a series LCR circuit.
1) inductive reactance = capacitive reactance
2) resonant frequency is independent of resistance 3) current in the circuit is minimum 4) impedance is purely resistive in nature.
impedance is purely resistive in nature
A
Answer is (3)
i (3)
26. The inductance of a coil is 5H.What is its effective reactance in dc circuit?
1)0 2) infinity 3) 5Ω
4)0.2Ω
For DC, frequency υ
, q
y = 0
but XL =2πυL = 0
• Answer is (1)
28.Power in an ac circuit depends on 28
Power in an ac circuit depends on
(A) rms value of voltage
(B) rms value of current
(B) rms value of current (C) phase difference between voltage and current current
1) Only A is correct 2) Only B is correct
3) Both A & B are correct
4) All A,B & C are correct
Power in an ac circuit P = (Vrms Irms cos φ)
• Answer is (4)
29.Which of the fallowing has the dimension of 29
Whi h f h f ll i h h di
i
f
resistance R (υis the frequency of AC)
1)υC 2) =2πυL 3) C/υ
[ R] = [ XL ] but XL =2πυL ∴ [ R] [ R] = [υL
[υL ] ]
4) L/υ
30. For long distance transmission, the AC is stepped up because transmission at high
stepped up because transmission at high voltage is 1) faster
1) faster 2) economical 2) economical
3)not damped 4)not dangerous
Transmission power loss P = I2R As V increases I decreases ∴P decreases So As V increases I decreases ∴P decreases . So
transmission at high voltage is economical. Answer is (2)
Answer is (2)
Transmission power loss P = I2R As V increases I decreases ∴P decreases
∴ transmission at high voltage is economical. Answer is (2)
31.The transformer varies the output
31.The transformer varies the output
1) Energy 2) Power 3) Frequency 4) Current
Answer is (4) = current
32.The core of the transformer is laminated to 1) reduce eddy current 2)reduce self induction 3)increase the efficiency 4)decrease the weight of the transformer
4)decrease the weight of the transformer
Answer is (1)
33 I
33.Iron core decreases the loss of energy due to
d
h l
f
d
1)heating 2) eddy currents 3)flux leakage 4)hysteresis
Answer is (4)
Answer is (4)
34. The AC cannot be used for
34. The AC cannot be used for
1)
2)
3)
4)
heating li h i
lighting electrolysis generate mechanical energy
Answer is (3)
35.What is the average value of ‘ac’ over a complete cycle?
complete cycle?
1)zero 2)Io/√2 3)2Io/π
Answer is (1)
4) 2Io
36 The power factor varies between
36. The power factor varies between 1)zero to 0.5 2)0.5 to 1 3)zero to 1 4)1 and 2
Ans ; PF = cos φ which varies from 0 & 1
Ans ; PF = cos φ
37.How does the current in an RC circuit vary when the charge on the capacitor builds up?
1)It increases linearly 2)It increases exponentially 3)It decreases linearly
3)It decreases linearly 4)It decreases exponentially
Answer is (4)
38 Th f
38.The frequency of ‘ac’ is 50Hz. How many f ‘ ’ i 50H H
times in one second does the voltage in the
times in one second does the voltage in the current does becomes zero? 1)25 3)100 Answer is (3)
Answer is (3)
2)50 4)150
39.Why 220V ‘ac’ is more dangerous than 220V dc?
1)) The dc attracts 2)Peak voltage for ac is much larger 3)The body offers less resistance to ac
4) Due to some other reason
Answer is (2)
40.What is the nature of graph between inductive reactance and frequency of ac for
inductive reactance and frequency of ac for series RCL circuit?
1)Straight line 2)Parabola 3)Hyperbola 4)Bell shaped
Answer is (1) XL =2πυL It is of the form y = mx
=2πυL It is of the form y = mx which is a which is a
Straight line passing through the origion with slope m= 2πL Answer is (1) 41. The impedance of a 10microfarad capacitor h
d
f
f d
for 50 rad/s ac is 1)2 Ω
1)2 Ω
2) 20Ω
2) 20Ω
3) 200Ω
3) 200Ω
4) 2000Ω
4) 2000Ω
The impedance Xc = 1/ ωC
Xc = 1/ (50x10x10
= 1/ (50x10x10‐66 ) =2000Ω
) =2000Ω
Answer is (4)
Answer is (4)
42 A 1 0mH inductance a 10µF capacitance and
42. A 1.0mH inductance a 10µF capacitance and a ac source. It is found that the inductor and the capacitor show equal resistances. The reactance should be nearest to
1)100Ω
2)32Ω
3)10Ω
4)3.2Ω
The circuit is in resonance.
h
But resonant frequency ω = 1/√LC
But resonant frequency ω
1/√LC
ω = 1/√1x10-3X10x10-6 = 104
XL = ω L = 104x10-3= 10
Answer is (3) =10Ω
43. In a circuit resistance R, capacitance C, and inductance L are in series with a sinusoidal
inductance L are in series with a sinusoidal power source. If the voltage across R, C and L are 80V, 10V and 70V respectively, then the voltage across the source terminals will be
1)100V 2)110V 3)140V
3)140V 4)160V
4)160V
V2 = V
= VR2+(VC‐ VL L )2
== 80
802 + (10
+ (10‐70)
70)2 = 80
= 802 + 60
+ 602 == 100
1002
V = 100V
V = 100V
Answer is (1) = 100V
44. The inductance of a coil is 5H.What is its effective reactance in dc circuit?
1) 0 2) infinity )
2)) 3) 5Ω
4) 0.2Ω
)
XL = ω
ω L = 0xL = 0 ∴Answer is (1)
L 0xL 0 ∴Answer is (1)
45. For a choke coil the resistance is R and reactance is X. Which of the following relations is valid?
relations is valid?
1) R X 2)R=X 3)R≈X 4)R
Answer is (4)
Answer is (4)
X