Chapter‐07Chapter 07

A C CIRCUITSA 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 voltageEffective voltage is rms voltage

∴ Answer is (1)

il h i i i d i2. A coil having zero resistance is connected in series with a 90Ω resistance and the

bi i i d 120 60 licombination is connected to 120V, 60 Hz line. A voltmeter reads 36 V across the resistance

d 114 V h il Th lf i dand 114 V across the coil. The self inductance of the coil is

1) 0.076 H 2) 0.76 H

3) 7.6 H 4) 76 H

Given VR= 36 V

R = 90ΩR = 90 Ω

∴ I = V/R =36/90

= 0.4 A

Also given VL= 114V

∴ X = V /I∴ XL = VL/I

2πυL = 114/0.4

L = 114/2x3.14x60x0.4 = 0.76 H

3 A di t f h 660 t f i i3. A radio transformer has 660 turns of wire in the primary coil which is connected to 220 V AC Th d il li 6 3VAC source. The secondary coil supplies 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/VpNs/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 connectedto 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 isflows. 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 A h i k l 28 A i d t5. An a.c. having a peak value 28 A is used to heat a metal wire. To produce the same h ti ff t t t t I th t bheating effect a constant current I that can be used is

1) 28A 2) about 20A

3) 7A 4) about 10A) )

Answer : Idc = Irms = Io /√2 = 28/1 41 =20 AAnswer : Idc = Irms = Io /√2 = 28/1.41 =20 A

6 A circuit has a resistance of 30Ω in series6. A circuit has a resistance of 30 Ω in series with an inductive reactance of 40Ω . They are connected in series with an ac source Ifare 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 thevoltage is 220 V, the power consumed by the circuit is

• 1) 66 W 2) 6.6. W

• 3) 0.66 W 4) 33 W

R=30 Ω XL = 40Ω . Z= √(R2 + XL 2)= 50Ω

P = ½ (VoIo cos φ)

( / )= ½ VoIo (R/Z)

= ½ x1x 220 x0.6 = 66 W ½ x1x 220 x0.6 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 andare 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 10 0 5• = ½ x10x0.5 • = 2.5 W

9 I i LCR i it R 100 Ω d li d9. In a series LCR circuit, R = 100 Ω and applied ac voltage is 200 V. When the capacitance l i d th lt l d thalone is removed, the voltage leads the current by 300. When the inductance alone is

d th lt l th t b 300removed, the voltage lags the current by 300. Then the current in the circuit is

1) 0.5 A 2) 1 A

3) 2A 4) 4 A

• Net phase difference between V & INet phase difference between V & I

φ= φ1 + φ2 = 30 – 30 = 0

S k i i• So ckt is in resonance

• 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 2) 2 mA• 1) 20 mA 2) 2 mA

• 3) 1 mA 4) 1 A

Given, C = 4 µF V= 50 V 3L = 10 x 10‐3 H.

• Energy stored in the capacitor is convertedEnergy stored in the capacitor is converted into energy in the inductor

• ½ LI2 = ½ CV2• ½ LI2 = ½ CV2

• ½ x 10 x 10‐3 I2 = ½ 4x 10‐6 x 2500• ½ x 10 x 10 3 I2 = ½ 4x 10 6 x 2500

• I2 = 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% 2) 84%

3) 83.3 % 4) 48%3) 83.3 % 4) 48%

Given ; Ps =140 watt, Vs=24 volt Vp = 240 V Ip= 0.7 A.

• The efficiency of the transformerThe efficiency of the transformer

• η = Ps/Pp / 140/(24 0 ) 20/24• η = Ps/Vs Is =140/(24 x0.7) = 20/24

• = 0.833

• Answer is (3) = 83 3 %Answer is (3) 83.3 %

f12.In a LRC circuit, capacitance is changed from C to 2C. For the resonant frequency to

h d h d h ld bremain unchanged, the inductance should be changed from L to

• 1) 4L 2) 2L

• 3) L/2 4) L /4

• For the resonant frequency to remainFor the resonant frequency to remain unchanged

LC = constLC = const

Lα1/C If C ---------2C, then L--------- L/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 100tcomparing with E = Eo sin ωt

Eo = 200 √ 2 & ω = 100Eo = 200 √ 2 & ω = 100• Reading of the ammeter I = E/Xc

/ å I = Eo/ à 2 Xc

= 200 x ω C = 200x100 X10‐6

= 2X10‐2

= 20X10‐3 A= 20X10 A

Answer is (2) = 20mA

14.The impedance of a coil of resistance R to AC f f i √ h lfof frequency 50Hz is √3R. The self 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 R υ= 50Hz & R = 6.28ΩZ √3 R υ 50Hz & R 6.28Ω

• L =√( Z2 – R2) / 2πυL =√( Z R ) / 2πυ• = √( 3R2 – R2) / 2πυ

√2 R /• = √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 =RGiven XL R

• The phase difference between the voltageThe phase difference between the voltage across the coil to the current through the coil tan φ = (XL –Xc)/Rtan φ (XL Xc)/R

• = XL /R = 1

Φ /• Φ = π/4• Answer is 4) π/4

16.The choke coil has a resistance 8 Ω and the

i d ti t 6Ω Th f t finductive reactance 6 Ω. The power factor of

the coil is

• 1) 0 6 ` 2) 0 8• 1) 0.6 2) 0.8

• 3) 0.4 4) 0.3

Given, R =8Ω XL= 6Ω.Given, R 8 Ω XL 6 Ω.

• The power factor of the coilThe power factor of the coil ,

Cos φ = R/Z8/10 0 8=8/10 = 0.8

Answer is (2) = 0.8

17.An alternating current is given byI= I1 cos wt + I2 sin wtI= I1 cos wt + I2 sin wt .the rms current is given by

n

φφ

I √ (I 2+ I 2 +2 I I ) b t) 900Io = √ (I12+ I2

2 +2 I1 I2cosφ) but) φ=900

I √ (I 2+ I 2 )Io = √ (I12+ I2

2 )

I I /√2 √ (I 2+ I 2 )/2Irms = Io/√2 = √ (I12+ I2

2 )/2

Answer is

18.In the circuit shown, the inductor used is an ideal one. The current in the

circuit when theswitch 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 beSince the inductor is ideal, the circuit can be written as

∴ I = E/(R +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 4) resistivity• 3) conductivity 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 isthe power factor is ½ . Then the actual power is

• 1) 50 W 2) 100 W

• 3) 25 W 4) 75 W

P = VI cosφPav VI cosφ

• P = P i cosφPav = Pvir cosφ

0 ½• Pav = 50 x ½

• Answer is (3) = 25 W

21 Th i f f h 100021.The primary of a transformer has 1000 turns and secondary has 2000 turns. For any input l h i h d ivoltage, the power in the secondary is

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 primary4) 1.5 times that in primary

Answer is (2)

22 In an AC circuit containing a pure inductor22.In an AC circuit containing a pure inductor and a capacitor, the current and voltage

areare

1) in phase1) in phase

2) out of phase by 900

) f h b 03) out of phase by 1800

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 therespect 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= 3Given, B.W 48 kHz. Q 3

Q = υ / BWQ = υo/ BW

kυo = Q xBW = 3x 48 kHz.

= 144 kHz

Answer is (3) = 144 kHz

24.An inductor of inductance 2 H and a resistance of 10Ω are connected to a batteryresistance of 10 Ω are connected to a battery of 5 V in series. The initial rate of change of current iscurrent 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 Theconnected 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 C25.A bulb B and a capacitor C are connected to a batteryas shown in figureas 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 all4) the bulb glows intermittently due to to the4) 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 flowsthe circuit. In the given circuit, current flows during charging and discharging of the capacitor When the switch S is closed thecapacitor. When the switch S is closed, the process is charging

• Hence the answer is (2) = the bulb glows d i h i f th itduring charging of the capacitor

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 nature4) impedance is purely resistive in nature.

A i (3)Answer is (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 = 0, q ybut XL =2πυL = 0• Answer is (1)Answer is (1)

28 Power in an ac circuit depends on28.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 currentcurrent 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 Whi h f h f ll i h h di i f29.Which of the fallowing has the dimension of resistance R (υis the frequency of AC)

1)υC 2) =2πυL

3) C/υ 4) L/υ

[ R] = [ XL ] but XL =2πυL

∴ [ R] = [υL ]∴ [ R] [υL ]

30. For long distance transmission, the AC is stepped up because transmission at highstepped up because transmission at high voltage is

1) faster 2) economical1) faster 2) economical

3)not damped 4)not dangerous

Transmission power loss P = I2R

As V increases I decreases∴P decreases SoAs 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 output31.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 transformer4)decrease the weight of the transformer

Answer is (1)

33 I d h l f d33.Iron core decreases the loss of energy due to

1)heating 2) eddy currents

3)flux leakage 4)hysteresis

Answer is (4)Answer is (4)

34. The AC cannot be used for34. The AC cannot be used for

1) heating

2) li h i2) lighting

3) electrolysis

4) 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 4) 2Io

Answer is (1)

36 The power factor varies between36. 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 & 1Ans ; PF = cos φ which varies from 0 & 1

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 linearly3)It decreases linearly

4)It decreases exponentially

Answer is (4)

38 Th f f ‘ ’ i 50H H38.The frequency of ‘ac’ is 50Hz. How many

times in one second does the voltage in thetimes in one second does the voltage in the

current does becomes zero?

1)25 2)50

3)100 4)150

Answer is (3)Answer is (3)

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 forinductive reactance and frequency of ac for

series RCL circuit?

1)Straight line 2)Parabola

3)Hyperbola 4)Bell shaped

Answer is (1)

X =2πυL It is of the form y = mx which is aXL =2πυL It is of the form y = mx which is a

Straight line passing through the origion with

slope m= 2πL

Answer is (1)

h d f f d41. The impedance of a 10microfarad capacitor for 50 rad/s ac is

1)2Ω 2) 20Ω1)2 Ω 2) 20Ω

3) 200Ω 4) 2000Ω3) 200Ω 4) 2000Ω

The impedance Xc = 1/ ωC

Xc = 1/ (50x10x10‐6 ) =2000ΩXc = 1/ (50x10x10 6 ) =2000Ω

Answer is (4)Answer is (4)

42 A 1 0mH inductance a 10µF capacitance and42. 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

hThe circuit is in resonance.

But resonant frequency ω = 1/√LCBut 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 sinusoidalinductance 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 4)160V3)140V 4)160V

V2 = VR2+(VC‐ VL )2V = VR +(VC VL )

= 802 + (10‐70)2 = 802 + 602 = 1002= 80 + (10 70) = 80 + 60 = 100

V = 100VV = 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) ) )

X ω L 0xL 0 ∴Answer is (1)XL = ω 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) RX 2)R=X

3)R≈X 4)RX

Answer is (4)Answer is (4)