Keeping the potential difference constant, the resistance of a circuit is halved. By how much does the current change?
According to ohm's law:
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$
If $\mathrm{V}$ is constant , then $\mathrm{I} \propto \frac{1}{\mathrm{R}}$
So, when $\mathrm{R}$ is halved, $I$ becomes double.
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