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This mock test of Alternating Current (CBSE Level With Solutions) for Class 12 helps you for every Class 12 entrance exam.
This contains 30 Multiple Choice Questions for Class 12 Alternating Current (CBSE Level With Solutions) (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Domestic power supply in India is

Solution:

Explanation:The standard voltage and frequency of alternating current supply in India is set to 220 V and 50Hz respectively by the government of India becausewhen power has to be transmitted from a power plant, the biggest challenge is to cut the transmission losses. For this purpose, the current value should be small and potential difference(Voltage) should be more. Also losses are minimal at 50 Hz/60 Hz frequency.

QUESTION: 2

For a series LCR circuit the input impedance at resonance

Solution:

Resonance occurs when X_{L} = X_{C} and the imaginary part of the transfer function is zero. At resonance the impedance of the circuit is equal to the resistance value as Z = R.

QUESTION: 3

A resistor of 100 Ω and a capacitor of 10μF are connected in series to a 220 V 50 Hz ac source. Voltage across the capacitance and resistor are

Solution:

QUESTION: 4

The current amplitude in a pure inductor in a radio receiver is to be 250 μA when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). If the voltage amplitude is kept constant, what will be the current amplitude through this inductor at 16.0 MHz?

Solution:

I_{0}=250μA, v_{0}=3.6v . v=1.6x10^{6} Hz

Here,

(Reactance of inductance) X_{L}=ωL

X_{L}=2πv X L

v_{0}/I_{0}=2πv x L

3.6/2.5x10^{-4}=2πx1.6x10^{-6} x L

0.14x10^{4-6}=L

L=0.14x10^{-2}H

Now for v=16.0x10^{6}Hz

XL=2πv X L

=2πx16x10^{6}x14x10^{-4}

X_{L}=1407x10^{2}Ω

Now,

v0=I0 x XL

3.6/1407x10^{2}=I_{0} [∵v0=kept constant.]

I_{0}=0.00256x10^{-2}

I_{0}=25.6μA

QUESTION: 5

A 0.180-H inductor is connected in series with a 90.0 Ω resistor and an ac source. The voltage across the inductor is (-12.0 V)sin(480 rad/s)t What is VR at t = 2ms ?

Solution:

QUESTION: 6

The main reason for preferring usage of AC voltage over DC voltage

Solution:

Explanation:By using the phenomenon of mutual induction, transformers allow us to easily change voltage of AC. This is necessary to cut down poer losses while supplying electricity to our homes

QUESTION: 7

At resonance the current in an LCR circuit

Solution:

Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R

QUESTION: 8

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 μF. Power dissipated in the circuit; and the power factor are

Solution:

Angular frequency of the ac signal w=2πν

∴ w=2π(50)=100π

Capacitive reactance X_{c}=1/wC

∴ X_{c}=1/(100π×786×10^{−6})=4Ω

Inductive reactance X_{L}=w_{L}

∴ X_{L}=100π×(25.48×10−3)=8Ω

Impedance of the circuit Z=√[R^{2}+(X_{L}−X_{c})^{2}]

∴ Z=√[3^{2}+(8−4)^{2}]=5Ω

Phase difference ϕ=tan^{−1}[(XL−Xc)R]

Or ϕ=tan^{−1}((8−4)/3)=tan^{−1}(4/3)

⟹ ϕ=53.13o

Power factor cosϕ=cos53.13o=0.6

Power dissipated in the circuit

P=I_{v}^{2}R

Now, Iv=I_{0}/√2=E_{0}/√2Z=283/(1.414×5)=40A

∴P=I_{v}^{2}R=(40)^{2}×3=4800 watt

QUESTION: 9

A resistor is connected in series with a capacitor. The voltage across the resistor is vR = (1.20 V) cos(2500 rad/s)t . Capacitive reactance is

Solution:

QUESTION: 10

Domestic power supply in India is

Solution:

The voltage in India is 220 volts, alternating at 50 cycles (Hertz) per second. This is the same as, or similar to, most countries in the world including Australia, Europe and the UK. However, it's different to the 110-120 volt electricity with 60 cycles per second that's used in the United States for small appliances.

QUESTION: 11

Alternating current is so called because

Solution:

Explanation:Alternating Current is the current in which the polarity of source continuously changes on a fixed frequency. So the positive and the negative terminals ‘alternate’

QUESTION: 12

Transformer uses the principle of

Solution:

The Voltage Transformer can be thought of as an electrical component rather than an electronic component. A transformer basically is very simple static (or stationary) electro-magnetic passive electrical device that works on the principle of Faraday’s law of induction by converting electrical energy from one value to another.

The transformer does this by linking together two or more electrical circuits using a common oscillating magnetic circuit which is produced by the transformer itself. A transformer operates on the principles of “electromagnetic induction”, in the form of Mutual Induction.

Mutual induction is the process by which a coil of wire magnetically induces a voltage into another coil located in close proximity to it. Then we can say that transformers work in the “magnetic domain”, and transformers get their name from the fact that they “transform” one voltage or current level into another.

QUESTION: 13

In the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor.

Solution:

At any time, t

energy stored in capacitor, EC=q^{2}/2C

Total energy stored in C and L at any time t

E=E_{C}+E_{L}

E=(q^{2}/2C)+(LI^{2}/2)

If q=q_{0}sinωt

I=dq/dt=q_{0}cosωtω=ωq0cosωt

Put in (i), E=[(q_{0}sinωt)^{2}/2C] + L(ωq_{0}cosωt)^{2}/2

E=(q_{0}^{2}/2C) sin^{2}ωt+(q_{0}^{2}/2)cos2ωt.ω^{2}L

But ω^{2}=12/LC

∴E=(q_{0}^{2}/2C) sin^{2}ωt+(q_{0}^{2}/2) (L/LC) cos^{2}ωt

=q_{0}^{2}/2C(sin^{2}ωt+cos^{2}ωt)

=q_{0}^{2}/2C= constant in time.

QUESTION: 14

In a series RLC circuit R = 300 Ω , L = 60 mH, C = 0.50 μF applied voltage V= 50 V and ω = 10,000 rad/s. voltages VR,VL and VC are

Solution:

QUESTION: 15

In an L-R-C series circuit, the rms voltage across the resistor is 30.0 V, across the capacitor it is 90.0 V, and across the inductor it is 50.0 V. Rms voltage of the source is

Solution:

Changes(if required): The answer provided in the forum is wrong, The correct answer would be option B.

Solution:

Given,

V_{R-rms} =30.0V

V_{C-rms}=90.0V

V_{L-rms}=50.0V

To find the root mean square voltage across the source, we will use the formula,

V_{rms}=√[V_{r}^{2}+(V_{L}-V_{C})^{2}]

=√[(30.0V)^{2} +(50.0V-90.0V)^{2}]

=50.0 V

QUESTION: 16

Phasor diagrams show

Solution:

Explanation:By convention phasor diagrams are made to show the rotation of phasor in counterclockwise direction with constant speed

QUESTION: 17

For a transformer the ratio of output to input equals

Solution:

QUESTION: 18

A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

Solution:

** So the range is 88 - 198 pF**

QUESTION: 19

Average power supplied to an inductor over one complete cycle

Solution:

Explanation:In the first half of the cycle, power is supplied to the inductive circuit. In the next half the inductor dissipates the energy supplied by circuit.Thus overall no power is supplied to the circuit

QUESTION: 20

A light bulb is rated at 50W for a 220 V supply. The resistance of the bulb, the peak voltage of the source and the rms current through the bulb are

Solution:

Given P=50W, V_{rms}=220V,

by P=V_{rms}^{2}/R

or R=V_{rms}^{2}/P= 220^{2}/50=968Ω

The peak voltage of the source is,

E_{rms}=E_{0}/√2

=>E_{0}=Erms x √2=220√2=311.13V

Now,

I_{rms}=V_{rms}/R

=220/968

=0.227A

Hence option B is correct.

QUESTION: 21

An electric hair dryer is rated at 1500 W (the average power) at 120 V (the rms voltage). Calculate (a) the resistance, (b) the rms current, and (c) the maximum instantaneous power. Assume that the dryer is a pure resistor.

Solution:

Given,

P_{avg}=1200W

V_{rms}=120 Volts

A,

The resistance is,

R=V_{rms}^{2}/P_{avg}

So,R=120^{2}/1500=9.6ohm

B,

The rms current is,

I_{rms}=V_{rms/}R

So, I_{rms}=120/9.6

=>I_{rms}=12.5A

C,

The maximum instantaneous power is,

P_{m}=2P_{avg}

So,P_{m}=2(1500)

=>P_{m}=3000W.

Hence option D is the correct answer.

QUESTION: 22

You have a 200.0 ΩΩ resistor, a 0.400-H inductor, a 5.0 μF capacitor, and a variable frequency ac source with an amplitude of 3.00 V. You connect all four elements together to form a series circuit. Frequency at which current in the circuit is greatest and its amplitude are

Solution:

The resistance is R=200ohm the inductor is L=0.4 H, the capacitance is C=5 µF and the amplitude voltage is V=CV

The frequency depends on the inductance and the capacitance and it is given by,

f_{0}=1/ (2π√(LC))

So, plug the values of L and C into equation 1 to get f_{0}

f_{0}=1/ (2π√(LC))=2/(2π√[0.4H)(5x10^{-6})]=113Hz

for the current, we use ohm’s law to get the current,

I=V/R

Now, plug the values for V and R to get I

I=V/R=3V/200Ω=0.015A=15mA

QUESTION: 23

An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Natural frequency of the circuit is

Solution:

Natural frequency of the circuit, f = 1/2π√(LC)

here, L = 20mH = 20 × 10^{-3} H, C = 50μF = 50 × 10^{-6} C

so, f = 1/2π√(20 × 10^{-3} × 50 × 10^{-6})

= 1/2π√(1000 × 10^{-9})

= 1/2π√(10^{-6})

= 1000/2π = 500/π Hz ≈ 159 Hz

QUESTION: 24

A hair dryer meant for 110V 60Hz is to be used in India . If 220 V is the supply voltage in India , the turns ratio for a transformer would be

Solution:

Here V_{p}=220V V_{s}=110V

As we know the relation between V and n,

As,

V_{e}/V_{s}=n_{p}/ns ->220/110

Np/ns=2/1=2:1

Therefore, no. of turns in primary is greater than no. of turns in secondary,

Hence, it is a step-down transformer.

QUESTION: 25

You have a 200.0 ΩΩ resistor, a 0.400-H inductor, 5.0 μF a capacitor, and a variable frequency ac source with an amplitude of 3.00 V. You connect all four elements together to form a series circuit. Current amplitude at an angular frequency of 400 rad/s is

Solution:

Given,

The resistance is R=200ohm the inductor is L=0.4 H, the capacitance is C=5 µF and the amplitude voltage is V=CV

The frequency depends on the inductance and the capacitance and it is given by,

f_{0}=1/ (2π√(LC)) (1)

So, plug the values of L and C into equation 1 to get f_{0}

f_{0}=1/ (2π√(LC))=2/(2π√[0.4H)(5x10-6)]=113Hz

The inductance reactance of the coil and could be calculated by,

X_{L}=ωL

Plug the values for L and ω to get XL

X_{L}= ωL=2π(113Hz)(0.4)=160ohm

To get the capacitive reactance we use the formula,

X_{C}=1/ωC

Now, we plus the values for ω and C to get XC

X_{C}=1/ ωC=1/2π(113Hz)(5x10^{-6})=500ohm

We use the inductor, the resistor and the capacitor, so the impedance of the circuit is given by the equation,

Z=√(R^{2}+(X_{L}-XC)^{2} ) (2)

Now, we use the values for XR,XL and R into equation (2) to get Z

Z= √ (R2+(XL-XC)2 )

=√ ((200Ω)2+(160Ω-500Ω)2 )

=394.5Ω

The impedance represents the total resistance in the circuit, so we can use the value in Ohm’s law to get the current amplitude I in the circuit,

I=V/Z

So, plus the values of V and Z to get I,

I=V/Z=3V/394.5Ω=0.0076A=7.61mA

QUESTION: 26

In an inductance the current

Solution:

In an inductor, current lags behind the input voltage by a phase difference of π/2.

Current and voltage are in the same phase in the resistor whereas current leads the voltage by π/2 in a capacitor.

So, the circuit must contain an inductor only.

QUESTION: 27

A lamp is connected in series with a capacitor and connected to an AC source. As the capacitance value is decreased.

Solution:

When frequency decreases, capacitive reactance XC=1/2πνC increases and hence impedance in the circuit Z=√(R2+XC2) increases and so current I=V/Z decreases. As a result, the brightness of the bulb is reduced.

QUESTION: 28

A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum are

Solution:

Given,

Capacitance, C=100μF=100x10^{-6}F

10^{-4}F

Resistance, R=40Ω

Rms voltage, E_{v}=110 volt

Peak voltage, E_{0}=√2 .E_{v}=√2 x 110V

Frequency of Ac supply, v=60Hz.,

ω=2πv=120π rad/s

peak current, I_{0}=?

a. In RC,as

Z=√(R^{2}+X_{C}^{2})= √[R^{2}+(1/ ω^{2}C^{2})]

Therefore,

I_{0}=E_{0}/√[R^{2}+(1/ ω^{2}C^{2})]

=√2x100/√[1600+(1/120πx10^{-4})^{2}]

I_{0}=3.23amp

b. In RC circuit voltage lags behind the current by phase angle Φ, where Φ is given by,

tanΦ=(1/ω)/R

=1/ωCR

=1/(120πx10-4x40)

=0.6628

⇒ Φ=tan^{-1}(0.6628) =33.5o

=(33.5π/180)rad

Time lag= Φ/ω

=33.5π/180x120π

=1.55x10^{-3} sec.

=1.55ms

QUESTION: 29

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 A, even for an instant. What is the largest root-mean-square current you can run through this bulb?

Solution:

Given,

We are given the current I=1.5Am where the wire will break.

We are asked to determine the root mean square of the current Irms. The maximum current here represents the current that just after it the wire will break. The maximum value of the current is the amplitude of the current wave and it should be larger than the root mean square of the current. Using equation, we can get the I_{rms} in the form,

I_{rms} = I_{max}/√2

The term, 1/√2, times any factor represents the root mean square of this factor, Now, plug the value for Imax into equation 1 and get I_{rms}

I_{rms}=I_{max}/√2

=1.5A/√2

=1.06A.

QUESTION: 30

Average power supplied to a capacitor over one complete cycle

Solution:

Explanation:We know that in capacitor Current leads Voltage by 90degree. Over one complete cycle, in first quarter cycle Capacitor charges and next quarter cycle it's discharge. This will continue in next negative half cycle. So the NET POWER ABSORB IS ZERO.

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