Q125 / 180 · KCET 2007 · Physics
In the Wheatstone's network given, $P=10 \Omega$, $Q=20 \Omega, R=15 \Omega, S=30 \Omega$ the current have passing through the battery (of negligible internal resistance) is
- Correct answerA$0.36 \mathrm{~A}$
- Bzero
- C$0.18 \mathrm{~A}$
- D$0.72 \mathrm{~A}$
Solution
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
$$
\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}
$$
$$
\begin{array}{ll}
\therefore & R=\frac{25 \times 50}{25+50}=\frac{50}{3} \Omega \\
\text { Hence, } & I=\frac{V}{R}=\frac{6}{50}=0.36 \mathrm{~A}
\end{array}
$$
Hence, the correct answer is option: 'A'.