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In the Wheatstone's network given, $\mathrm{P}=10 \Omega$, $Q=20 \Omega, R=15 \Omega, S=30 \Omega$, the current passing through the battery (of negligible internal resistance) is
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The correct answer is:
$0.36 \mathrm{~A}$
Balanced wheatstone bridge condition
$$
\frac{P}{Q}=\frac{R}{S}
$$
No, current flows through galvanometer Now, $\mathrm{P}$ and $\mathrm{R}$ are in series, so Resistance $\mathrm{R}_{1}=\mathrm{P}+\mathrm{R}$
$$
=10+15=25 \Omega
$$
Similarly, Q and S are in series, so Resistance $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}$ $=20+30=50 \Omega$
Net resistance of the network as $R_{1}$ and $R_{2}$ are in parallel
$$
\begin{array}{l}
\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \\
\therefore \mathrm{R}=\frac{25 \times 50}{25+50}=\frac{50}{3} \Omega
\end{array}
$$
Hence, current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{50 / 3}=0.36 \mathrm{~A}$
$$
\frac{P}{Q}=\frac{R}{S}
$$
No, current flows through galvanometer Now, $\mathrm{P}$ and $\mathrm{R}$ are in series, so Resistance $\mathrm{R}_{1}=\mathrm{P}+\mathrm{R}$
$$
=10+15=25 \Omega
$$
Similarly, Q and S are in series, so Resistance $\mathrm{R}_{2}=\mathrm{R}+\mathrm{S}$ $=20+30=50 \Omega$
Net resistance of the network as $R_{1}$ and $R_{2}$ are in parallel
$$
\begin{array}{l}
\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \\
\therefore \mathrm{R}=\frac{25 \times 50}{25+50}=\frac{50}{3} \Omega
\end{array}
$$
Hence, current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{50 / 3}=0.36 \mathrm{~A}$
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