The current in the given circuit is ............ $A$
AIPMT 1999, Medium
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$6$ $\Omega$ and $6\,\Omega$ are in series,
so effective resistance is $12$ $\Omega$
which is in parallel with $3$ $\Omega$,
so $\frac{1}{R} = \frac{1}{3} + \frac{1}{{12}} = \frac{{15}}{{36}}$
$ \Rightarrow $ $R = \frac{{36}}{{15}}$
$I = \frac{V}{R} = \frac{{4.8 \times 15}}{{36}} = 2\,A$
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