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
B
100Ω and 0.86 H
C
200Ω and 1.0 H
D
200Ω and 0.93 H

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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