Q41 / 60 · KVPY 2012 (SA) · Physics
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
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
- Correct answerA1
- B$\mathrm{n}$
- C$\mathrm{n}^{2}$
- D$1 / \mathrm{n}$
Solution
$\begin{array}{l}
\mathrm{i}_{1}=\frac{\mathrm{nE}}{\mathrm{n} R_{0}+\mathrm{R}}, \mathrm{i}_{2}=\frac{\mathrm{E}}{\left(\mathrm{R}_{0}+\mathrm{n} \mathrm{R}\right)} \\
\mathrm{P}_{1}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{nR}_{0}+\mathrm{R}\right)^{2}}, \mathrm{P}_{2}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{R}_{0}+\mathrm{nR}\right)^{2}} \\
\because \mathrm{P}_{1}=\mathrm{P}_{2} \\
\text { Hence } \mathrm{R}_{0} / \mathrm{R}=1
\end{array}$
Hence, the correct answer is option: 'A'.