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A galvanometer of $50 \mathrm{ohm}$ resistance has 25 divisions. A current $4 \times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of:
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Verified Answer
The correct answer is:
$2450 \Omega$ in series
Total current in the galvanometer $=$
$$
\begin{aligned}
& 25 \times 4 \times 10^{-4} \mathrm{~A} \\
& \mathrm{I}_g=10^{-2} \mathrm{~A}
\end{aligned}
$$
The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is
$$
\begin{aligned}
R & =\frac{V}{I_g}-G \\
& =\frac{25}{10^{-2}}-50 \\
& =2450 \Omega
\end{aligned}
$$
$$
\begin{aligned}
& 25 \times 4 \times 10^{-4} \mathrm{~A} \\
& \mathrm{I}_g=10^{-2} \mathrm{~A}
\end{aligned}
$$
The value of resistance connected in series to convert galvanometer into voltmeter of $25 \mathrm{~V}$ is
$$
\begin{aligned}
R & =\frac{V}{I_g}-G \\
& =\frac{25}{10^{-2}}-50 \\
& =2450 \Omega
\end{aligned}
$$
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