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The maximum current that can be measured by a galvanometer of resistance $40 ~\Omega$, is $10 \mathrm{~mA}$. It is converted into a voltmeter that can read upto $50 \mathrm{~V}$. The resistance to be connected in series with the galvanometer (in $\Omega$ ) is
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Verified Answer
The correct answer is:
$4960$
Given, $G=40 \Omega, I_{g}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$
$$
V=50 \mathrm{~V}
$$
Let $R$ be the resistance connected in series with the galvanometer to convert into voltmeter, then
$$
\begin{aligned}
& & V &=I_{g}(G+R) \\
\Rightarrow & & 50 &=10 \times 10^{-3}(40+R) \\
\Rightarrow \quad & & \frac{50}{10 \times 10^{-3}} &=40+R \\
\Rightarrow & & R &=5000-40=4960 \Omega
\end{aligned}
$$
$$
V=50 \mathrm{~V}
$$
Let $R$ be the resistance connected in series with the galvanometer to convert into voltmeter, then
$$
\begin{aligned}
& & V &=I_{g}(G+R) \\
\Rightarrow & & 50 &=10 \times 10^{-3}(40+R) \\
\Rightarrow \quad & & \frac{50}{10 \times 10^{-3}} &=40+R \\
\Rightarrow & & R &=5000-40=4960 \Omega
\end{aligned}
$$
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