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A galvanometer of resistance $G$ can be converted into a voltmeter of range $V$ by connecting a resistance $R$ in series with it. The series resistance required to change its range to $\frac{V}{3}$ is
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Verified Answer
The correct answer is:
$\frac{R-2 G}{3}$
Case-1
$\begin{aligned}
& R=\frac{V}{I_G}-G \\
& \Rightarrow I_G=\frac{V}{R+G}
\end{aligned}$
Case-2:
$R^{\prime}=\frac{\left(\frac{V}{3}\right)}{I_G}-G=\frac{\left(\frac{V}{3}\right)}{\left(\frac{V}{R+G}\right)}-G=\left(\frac{R-2 G}{3}\right)$
$\begin{aligned}
& R=\frac{V}{I_G}-G \\
& \Rightarrow I_G=\frac{V}{R+G}
\end{aligned}$
Case-2:
$R^{\prime}=\frac{\left(\frac{V}{3}\right)}{I_G}-G=\frac{\left(\frac{V}{3}\right)}{\left(\frac{V}{R+G}\right)}-G=\left(\frac{R-2 G}{3}\right)$
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