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A coil has an inductance of 2.5 H and a resistance of 0.5 r. If the coil is suddenly connected across a 6.0 volt battery, then the time required for the current to rise 0.63 of its final value is

3.5 sec

4.0 sec

4.5 sec

5.0 sec

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Explanation

$t = τ = \frac{L}{R} = \frac{2 .5}{0 .5} = 5 \sec$

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If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is

$\frac{2}{5} H$

$\frac{4}{5} H$

$\frac{5}{4} H$

1 H

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Explanation

$| e | = L \frac{d i}{d t} \Rightarrow 10 = L \times \frac{10}{1} \Rightarrow L = 1 H$

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An inductance L and a resistance R are first connected to a battery. After some time the battery is disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in

RL sec

$\frac{R}{L} s e c$

$\frac{L}{R} s e c$

$\frac{1}{L R} s e c$

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Explanation

When battery disconnected current through the circuit start decreasing exponentially according to $i = i_{0} e^{- R t / L}$

⇒ $0 .37 i_{0} = i_{0} e^{- R t / L}$

⇒ $0 .37 = \frac{1}{e} = e^{- R t / L} \Rightarrow e = e^{\frac{L}{R t}} \Rightarrow \frac{L}{R t} = 1 \Rightarrow \frac{L}{R} = t$

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In an LR-circuit, the time constant is that time in which current grows from zero to the value (where I₀ is the steady-state current)

0.63 I₀

0.50 I₀

0.37 I₀

I₀

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Explanation

Current at any instant of time t after closing an L-R circuit is given by $I = I_{0} [1 - e^{\frac{- R}{L} t}]$

Time constant $t = \frac{L}{R}$

$∴ I = I_{0} [1 - e^{\frac{- R}{L} \times \frac{L}{R}}] = I_{0} (1 - e^{- 1}) = I_{0} (1 - \frac{1}{e})$

$= I_{0} (1 - \frac{1}{2 .718}) = 0 .63 I_{0} = 63 %$ of I₀

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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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