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, $I=\frac{V}{R}=\frac{6}{\frac{50}{3}}=0.36A$