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Test: Alternating Current - 1 - CUET Humanities MCQ


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10 Questions MCQ Test - Test: Alternating Current - 1

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Test: Alternating Current - 1 - Question 1

Which of the following devices changes low voltage alternating current to high voltage alternating current and vice-versa?

Detailed Solution for Test: Alternating Current - 1 - Question 1

Transformer changes low voltage alternating current to high voltage alternating current and vice-versa.

Test: Alternating Current - 1 - Question 2

What is the capacitance of a capacitor whose reactance for AC frequency of 400 Hz is 25π?

Detailed Solution for Test: Alternating Current - 1 - Question 2


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Test: Alternating Current - 1 - Question 3

Two resistances of 10Ω and 20Ω and an ideal inductor of inductance 5 H are connected to a battery of 2 V through a key K as shown in the figure. If at t = 0, K is inserted, then the initial potential drop across the inductor is

Detailed Solution for Test: Alternating Current - 1 - Question 3

At t = 0, the inductor acts as an open circuit. So, voltage across the inductor is the same as 10-ohm resistance.
Voltage across 10-ohm resistance = 

Test: Alternating Current - 1 - Question 4

A coil of inductance 300 mH and resistance 2Ω  is connected to a source of voltage 2 V. The current reaches half of its steady state value in

Detailed Solution for Test: Alternating Current - 1 - Question 4

Growth of the current in the inductor,  where 'I' is the steady state value of the current and 

Test: Alternating Current - 1 - Question 5

A coil has an inductance of 4/π henry and is connected in series with a resistance of 30 ohms. What is the current flowing in the circuit when connected to AC mains of 200 V and frequency 50 cycles per second?

Detailed Solution for Test: Alternating Current - 1 - Question 5

L = 4/π
R = 30 ohms
Virtual voltage, Ev = 200
Vn = 50 Hz
I= ?


I= 200/401.1 = 0.4986 A

Test: Alternating Current - 1 - Question 6

If 8Ω  resistance and 6Ω  reactance are present in an AC series circuit, what will be the impedance of the circuit?

Detailed Solution for Test: Alternating Current - 1 - Question 6

Z = (R2 + Xc2)1/2
Z = (64 + 36)1/2 = 10 ohm

Test: Alternating Current - 1 - Question 7

An AC source of variable frequency f is connected to an LCR series circuit. Which of the following graphs represents the variation of current in the circuit with frequency?

Detailed Solution for Test: Alternating Current - 1 - Question 7

The current in an LCR circuit is given by

where ω = 2πf. Thus I increases with increase in ω upto a value of ω = ωc given by

or 
when I becomes maximum. At ω > ωc, I decreases with increase in ω.
Hence, the correct graph is shown in option (4).

Test: Alternating Current - 1 - Question 8

If a resistance of 100 Ω, an inductance of 0.5 H and a capacitance of 10 × 10−6 F are connected in series with a 50 Hz AC supply, then what will be the impedance?

Detailed Solution for Test: Alternating Current - 1 - Question 8

The impedance of an A.C. circuit containing resistance, inductance and capacitance is
Z = 
Z = 
Here : R = 100Ω,
L = 0.5 H
C = 10 × 10-6 F
f = 50 Hz
Z = 



= 189.95 ≈ 190 Ω

Test: Alternating Current - 1 - Question 9

Alternating current is measured by

Detailed Solution for Test: Alternating Current - 1 - Question 9

An induction ammeter works based on electromagnetic induction. Hence, a varying magnetic flux is required for the working of an induction ammeter. In an alternating current, the magnetic flux is continuously varying. Therefore, an induction ammeter is used for measuring the alternating current.
A permanent magnet type ammeter is used for measuring DC only.
An electrostatic voltmeter is used for measuring high voltage.
A moving iron repulsion type voltmeter is used for measuring the voltage.

Test: Alternating Current - 1 - Question 10

An ideal transformer is supplied with a primary input power of 10 kilowatt. When the transformer is on load, the secondary current is 24 A. What is the potential difference applied to the primary if the primary to secondary turns ratio is 4 : 1?

Detailed Solution for Test: Alternating Current - 1 - Question 10

Ep × Ip = 10 kW = 10 × 103 W = 104 W
Is = 24 A
np/ns = 4/1
np/ns = Is/Ip
4/1 = 24/Ip
Ip = 6 A
Ep = 104/6 = 1666.7 V

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