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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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When the speed of a dc motor increases the armature current

Increases

Decreases

Does not change

Increases and decreases continuously

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Explanation

With the increasing speed, ω increases. Thus current reduces due to increase in the back e.m.f.

Moreover $i = \frac{V - K ω}{R}$ . More ω will lead to the lesser current.

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An electric motor operating on a 60 V dc supply draws a current of 10 A. If the efficiency of the motor is 50%, the resistance of its winding is

3Ω

6Ω

15Ω

30Ω

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Explanation

Efficiency $η = 50 %$

So e = E/2

and $i = \frac{E - e}{R} \Rightarrow i = \frac{E - E / 2}{R} = \frac{E}{2 R}$

$\Rightarrow R = \frac{E}{2 i} = \frac{60}{2 \times 10} = 3 Ω$

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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 copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is

$\frac{1}{2} B ω l^{2}$

$\frac{3}{4} B ω l^{2}$

$B ω l^{2}$

$2 B ω l^{2}$

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