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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:

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:

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:

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:

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:

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:

QUESTION: 13

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

Solution:

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:

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:

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:

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:

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:

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:

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:

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:

QUESTION: 26

In an inductance the current

Solution:

QUESTION: 27

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

Solution:

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:

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:

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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