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Q.
In the Wheatstone's network given below $P=10 \, \Omega$
$Q=20 \, \Omega$ $R=15 \, \Omega$ $S=30 \, \Omega$
The current passing through the battery (of negligible internal resistance) is
The balanced condition for Wheatstone's bridge is given by $ \frac{P}{Q} = \frac{R}{S} $
As is obvious from the given values condition, no current flows through galvanometer.
Now, P and R are in series, so
Resistance $ R_1 = P + R = 10 + 15 = 25 \Omega $
Similarly, Q and S are in series, so
Resistance $ R_2 = R + S = 20 + 30 = 50\Omega $
Net resistance of the network as $ R_1 $ and $ R_2 $ are in parallel
$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $
$ \therefore R = \frac{ 25 \times 50}{25 + 50} = \frac{50}{3} \Omega $
Hence, $ I = \frac{V}{R} = \frac{6}{50/3} = 0.36A $
