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In the circuit shown, $\mathrm{n}$ identical resistors $\mathrm{R}$ are connected in parallel $(\mathrm{n}>1$ ) and the combination is connected in series to another resistor $\mathrm{R}_{0}$. In the adjoining circuit $\mathrm{n}$ resistors of resistance $\mathrm{R}$ are all connected in series along with $\mathrm{R}_{0}-$
The batteries in both circuits are identical and net power dissipated in the n resistors in both circuits is same. The ratio $\mathrm{R}_{0} / \mathrm{R}$ is
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The batteries in both circuits are identical and net power dissipated in the n resistors in both circuits is same. The ratio $\mathrm{R}_{0} / \mathrm{R}$ is
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1
$\begin{array}{l}
\mathrm{i}_{1}=\frac{\mathrm{nE}}{\mathrm{n} R_{0}+\mathrm{R}}, \mathrm{i}_{2}=\frac{\mathrm{E}}{\left(\mathrm{R}_{0}+\mathrm{n} \mathrm{R}\right)} \\
\mathrm{P}_{1}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{nR}_{0}+\mathrm{R}\right)^{2}}, \mathrm{P}_{2}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{R}_{0}+\mathrm{nR}\right)^{2}} \\
\because \mathrm{P}_{1}=\mathrm{P}_{2} \\
\text { Hence } \mathrm{R}_{0} / \mathrm{R}=1
\end{array}$
\mathrm{i}_{1}=\frac{\mathrm{nE}}{\mathrm{n} R_{0}+\mathrm{R}}, \mathrm{i}_{2}=\frac{\mathrm{E}}{\left(\mathrm{R}_{0}+\mathrm{n} \mathrm{R}\right)} \\
\mathrm{P}_{1}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{nR}_{0}+\mathrm{R}\right)^{2}}, \mathrm{P}_{2}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{R}_{0}+\mathrm{nR}\right)^{2}} \\
\because \mathrm{P}_{1}=\mathrm{P}_{2} \\
\text { Hence } \mathrm{R}_{0} / \mathrm{R}=1
\end{array}$
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