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Alternating Current MCQs for NEET — Physics Questions with Answers

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In a series resonant LCR circuit, the voltage across R is 100 V and $R= 1k \Omega $ with $C =2 \mu F $ . The resonant frequency $ \Omega $ is 200 rad/s.At resonance the voltage across L is.

A. 40 V B. 250 V C. $ 4 \times 10^{-3} V $ D. $ 2.5 \times 10^{-2} V $

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Explanation

$ \upsilon _L = \upsilon_C = I_{X_C} ={\upsilon \over R\omega C} $

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The core of a transformer is laminated to reduce energy losses due to

Eddy currents

Hysteresis

Resistance in winding

None of these

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Explanation

Circulation of eddy currents is prevented by use of laminated core.

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A loss free transformer has 500 turns on its primary winding and 2500 in secondary. The meters of the secondary indicate 200 volts at 8 amperes under these conditions. The voltage and current in the primary is

100 V, 16 A

40 V, 40 A

160 V, 10 A

80 V, 20 A

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Explanation

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}} = \frac{500}{2500} = \frac{1}{5} \Rightarrow V_{p} = \frac{200}{5} = 40 V$

Also $i_{p} V_{p} = i_{s} V_{s} \Rightarrow i_{p} = i_{s} \frac{V_{s}}{V_{p}} = 8 \times 5 = 40 A$

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A transformer connected to 220 volt line shows an output of 2 A at 11000 volt. The efficiency is 100%. The current drawn from the line is

100 A

200 A

22 A

11 A

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Explanation

$\frac{V_{s}}{V_{p}} = \frac{i_{p}}{i_{s}} \Rightarrow i_{p} = \frac{11000 \times 2}{220} = 100 A$

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A power transformer is used to step up an alternating e.m.f. of 220 V to 11 kV to transmit 4.4 kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary ? Assume 100% efficiency for the transformer

4 A

0.4 A

0.04 A

0.2 A

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Explanation

$i_{s} = \frac{P_{s}}{V_{s}} = \frac{4 .4 \times 10^{3}}{11 \times 10^{3}} = 0 .4 A$

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The primary winding of transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an ac supply of 20 V, 50 Hz. The secondary will have an output of

200 V, 50 Hz

2 V, 50 Hz

200 V, 500 Hz

2 V, 5 Hz

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Explanation

$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \Rightarrow \frac{V_{s}}{20} = \frac{5000}{500}$

$\Rightarrow V_{s} = 200 V$

Frequency remains unchanged.

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A step-down transformer is connected to main supply 200V to operate a 6V, 30W bulb. The current in primary is

3 A

1.5 A

0.3 A

0.15 A

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Explanation

$V_{p} = 200 V, V_{s} = 6 V$

$P_{o u t} = V_{s} i_{s} \Rightarrow 30 = 6 \times i_{s} \Rightarrow i_{s} = 5 A$

From $\frac{V_{s}}{V_{p}} = \frac{i_{p}}{i_{s}} \Rightarrow \frac{6}{200} = \frac{i_{p}}{5} \Rightarrow i_{p} = 0 .15 A$

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A transformer has 100 turns in the primary coil and carries 8 A current. If input power is one kilowatt, the number of turns required in the secondary coil to have 500V output will be

100

200

400

300

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Explanation

$P_{P} = V_{P} i_{P} \Rightarrow 1000 = V_{P} \times 8 \Rightarrow V_{P} = \frac{1000}{8}$

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}} \Rightarrow \frac{(1000 / 8)}{500} = \frac{100}{N_{s}} \Rightarrow N_{s} = 400$

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A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively are

300V,15A

450V,15A

450V,13.5A

600V,15A

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Explanation

Initial power=3000W

As efficiency is 90% then final power

=3000x90/100=2700W

=>V₁I₁=3000W]

V₁I₁=2700W] ...(i)

So, V₂=2700/6=900/2=450V and I₁=3000/200=15A

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In an ideal transformer, the voltage and the current in the primary are 200 volt and 2 amp. respectively. If the voltage in the secondary is 2000 volt. Then value of current in the secondary will be –

0.2 amp.

2 amp.

10 amp.

20 amp.

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Explanation

1. Given : Voltage in primary $V_{p}$ = 200 volt

Current in primary $i_{p}$ = 2 amp

Voltage in secondary $V_{s}$ = 2000 volt

The relation for the current in the secondary is

$\frac{V_{s}}{V_{p}} = \frac{i_{p}}{i_{s}} \Rightarrow \frac{2000}{200} = \frac{2}{i_{s}} or, i_{s} = \frac{2 \times 200}{2000} = 0.2 amp .$

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