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An inductor coil wound uniformly has self-inductance ' $L$ ' and resistance ' $\mathrm{R}$ '. The coil is broken into two identical parts. The two parts are then then connected in parallel across a battery of ' $E$ ' volt of negligible internal resistance. The current through battery at steady state is
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The correct answer is:
$\frac{4 \mathrm{E}}{\mathrm{R}}$
In DC, there is no use of inductance.
Since the coil is broken into two identical points, each part will have resistance $\frac{\mathrm{R}}{2}$.
When these are connected in parallel, their equivalent resistance will be $\frac{\mathrm{R}}{4}$.
Hence the current $I$ is given by $=\frac{E}{R / 4}=\frac{4 E}{R}$
Since the coil is broken into two identical points, each part will have resistance $\frac{\mathrm{R}}{2}$.
When these are connected in parallel, their equivalent resistance will be $\frac{\mathrm{R}}{4}$.
Hence the current $I$ is given by $=\frac{E}{R / 4}=\frac{4 E}{R}$
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