In the circuit given below, find the total current drawn from the battery:

\frac{3}{24}

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\frac{14}{25}

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\frac{9}{32}

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\frac{3}{20}

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Solution

The correct option is D $\frac{9}{32}$
$R_{e q} = 45 + \frac{25}{3} = \frac{135 + 25}{3} = \frac{160}{3}$
$i = \frac{V}{R} = \frac{15}{\frac{160}{3}} = \frac{9}{32}$

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