Question

In the Wheat stone's network given, $ P=10\Omega , $ $ Q=20\Omega ,R=15\Omega , $ $ S=30\Omega , $ the current passing through the battery (of negligible internal resistance) is

• A. 0.36 A
• B. zero
• C. 0.ISA
• D. 0.72 A

Answer 1

Answer: (A)

The balanced condition for Wheatstone's bridge is $ \frac{P}{Q}=\frac{R}{S} $ as is obvious from the given values. No, current flows through galvanometer is zero. 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}}=Q+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}{\frac{50}{3}}=0.36A $