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Alternating Current Applied to an Inductor

In the circui...

Question

In the circuit show below,the key K is closed at t=0. the current through the battery is:

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Solution

From the figure we have,

$i = \frac{V}{R_{e q}}$

$= \frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$

At $t = 0$ inductor behaves like an infinite resistance.

$i = \frac{V}{R_{2}}$

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