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A voltmeter of $250 \mathrm{mV}$ range having a resistance of $10 \Omega$ is converted into an ammeter of $250 \mathrm{~mA}$ range. The value of necessary shunt is (nearly)
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The correct answer is:
$1 \Omega$
A voltmeter of $250 \mathrm{mV}$ range having resistance of $10 \Omega$
$$
\begin{aligned}
& \therefore & V_g & =G I_g \\
\Rightarrow & & 250 \times 10^{-3} & =10 \times I_g \\
\Rightarrow & & I_g & =25 \times 10^{-3} \mathrm{~A}
\end{aligned}
$$
Shunt required to convert this voltmeter into ammeter.
$$
\begin{aligned}
S & =\frac{G I_g}{I-I_g} \\
& =\frac{10 \times\left(25 \times 10^{-3}\right)}{250 \times 10^{-3}-25 \times 10^{-3}}=\frac{10}{9}=1 \Omega
\end{aligned}
$$
$$
\begin{aligned}
& \therefore & V_g & =G I_g \\
\Rightarrow & & 250 \times 10^{-3} & =10 \times I_g \\
\Rightarrow & & I_g & =25 \times 10^{-3} \mathrm{~A}
\end{aligned}
$$
Shunt required to convert this voltmeter into ammeter.
$$
\begin{aligned}
S & =\frac{G I_g}{I-I_g} \\
& =\frac{10 \times\left(25 \times 10^{-3}\right)}{250 \times 10^{-3}-25 \times 10^{-3}}=\frac{10}{9}=1 \Omega
\end{aligned}
$$
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