Question

In the Wheatstone'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. 0.18 A
• D. 0.72 A

Answer 1

Answer: (A)

0.36 A

Answer 2

The balanced condition for Wheatstones 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 ${{R}_{2}}=R+S$ $=20+30=50\,\Omega$ Net resistance of the network as ${{R}_{1}}$ and ${{R}_{2}}$ are in parallel $i=\frac{V}{R}=\frac{0.6}{50}=0.36\,A$