For the network shown in the figure the value of the current $i$ is
AIPMT 2005, Medium
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The given network is a balanced Wheatstone bridge. It's equivalent resistance will be $R = \frac{{18}}{5}\,\Omega $
So current from the battery $i = \frac{V}{R} = \frac{V}{{18/5}} = \frac{{5V}}{{18}}$
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