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A uniform metallic wire of lenght $L$ is mounted in two configurations. In configuration I (triangle), it is an equilateral triangle and a voltage $V$ is applied to corners A and B. In configuration 2 (circle), it is bent in the form of a circle, and the potential $v$ is applied at diameterically opposite points $P$ and $Q$. The ratio of the power dissipated in configuration 1 to configuration 2 is.
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The correct answer is:
$9 / 8$
$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}$
$\left(\mathrm{R}_{1}=\frac{\rho \frac{\ell}{3}}{\mathrm{~A}}\right)$
$\frac{1}{\mathrm{R}_{\mathrm{ep}}} \Rightarrow \frac{1}{2 \mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{1}} \Rightarrow 2$
$\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}_{1}}{3}=\frac{2 \rho \frac{\ell}{3}}{3 \mathrm{~A}}$
$\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}}{2 \mathrm{e} \ell} 9 \mathrm{~A}$
$\mathrm{R}_{\mathrm{eq}} \Rightarrow \frac{\mathrm{R}}{2} \Rightarrow \frac{1}{2}\left(\frac{\rho . \ell / 2}{\mathrm{~A}}\right)$
$\mathrm{i}_{2}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}} \Rightarrow \frac{\mathrm{V}}{\rho \ell} 4 \mathrm{~A}$
$\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{9}{8}$
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