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Question

When $$100\space V$$ dc is applied across a solenoid, a current of $$1.0\space A$$ flows in it. When $$100\space V$$ ac is applied across the same coil, the current drops to $$0.5\space A$$. If the frequency of the ac source is $$50\space Hz$$, the impedance and inductance of the solenoid are


A
200Ω and 0.55 H
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B
100Ω and 0.86 H
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C
200Ω and 1.0 H
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D
200Ω and 0.93 H
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Solution

The correct option is A $$200\Omega\space and\space 0.55\space H$$
impedance= $$\frac{V_{ac}}{I_{ac}}= \frac {100 V}{0.5 A} = 200 \Omega $$
R=$$ \frac{V_{dc}}{I_{dc}}=\frac{100}{1}=100 \Omega $$
$$L \omega $$ = $$ \sqrt{ {impedance}^{2} - R^2 } = \sqrt { 200^2-100^2 } =173.2 $$
$$L=\frac{173.2}{\omega}=\frac{173.2}{2\pi 50} =0.55 H $$

Physics

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