The current in the adjoining circuit will be
IIT 1983, Medium
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(c) ${R_{{\rm{equivalent}}}} = \frac{{(30 + 30)30}}{{(30 + 30) + 30}} = \frac{{60 \times 30}}{{90}} = 20\,\Omega $
$\therefore i = \frac{V}{R} = \frac{2}{{20}} = \frac{1}{{10}}\,ampere$
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