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Q.

A coil has L = 0.04 H and $R = 12 Ω .$ When it is connected to 220V, 50Hz supply, the current flowing through the coil, in amperes, is

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a

10.7

b

11.7

c

14.7

d

12.7

answer is D.

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Detailed Solution

Impedance, $Z = \sqrt{R^{2} + 4 π^{2} v^{2} L^{2}}$

$= \sqrt{(12)^{2} + 4 \times (3 .14)^{2} \times (50)^{2} \times (0 .04)} = 17 .37 A$

Now, current, $i = \frac{V}{Z} = \frac{220}{17 .37} = 12 .7 Ω$

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A coil has L = 0.04 H and R=12 Ω. When it is connected to 220V, 50Hz supply, the current flowing through the coil, in amperes, is