JEE MAIN - Physics (2006 - No. 38)
The current $${\rm I}$$ drawn from the $$5$$ volt source will be
$$0.33$$ $$A$$
$$0.5$$ $$A$$
$$0.67$$ $$A$$
$$0.17$$ $$A$$
Explanation
The network of resistors is a balanced wheatstone bridge. The equivalent circuit is
$${{\mathop{\rm R}\nolimits} _{eq}} = {{15 \times 30} \over {15 + 30}} = 10\Omega $$
$$ \Rightarrow I = {V \over R} = {5 \over {10}} = 0.5\,A$$
$${{\mathop{\rm R}\nolimits} _{eq}} = {{15 \times 30} \over {15 + 30}} = 10\Omega $$
$$ \Rightarrow I = {V \over R} = {5 \over {10}} = 0.5\,A$$
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