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A moving coil galvanometer of resistance 100 $\Omega$ shows full scale deflection when a current of 100 micro-amperes passes through it. If it is intended to show full scale deflection when a current of 1 milliampere passes through it, the value of shunt resistance in ohms to be connected to the galvanometer is.
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Verified Answer
The correct answer is:
$\frac{100}{9}$
Resistance of galvanometer
$$
\begin{gathered}
G=100 \Omega \\
I_g=100 \mu \mathrm{A}=100 \times 10^{-6} \mathrm{~A}=0.1 \times 10^{-3} \mathrm{~A} \\
I=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}
\end{gathered}
$$
Shunt resistance
$$
\begin{aligned}
S & =\frac{I_g G}{I-I_g}=\frac{0.1 \times 10^{-3} \times 100}{1 \times 10^{-3}-0.1 \times 10^{-3}} \\
& =\frac{1 \times 10^{-2}}{0.9 \times 10^{-3}}=\frac{10^2}{9}=\frac{100}{9} \Omega
\end{aligned}
$$
$$
\begin{gathered}
G=100 \Omega \\
I_g=100 \mu \mathrm{A}=100 \times 10^{-6} \mathrm{~A}=0.1 \times 10^{-3} \mathrm{~A} \\
I=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}
\end{gathered}
$$
Shunt resistance
$$
\begin{aligned}
S & =\frac{I_g G}{I-I_g}=\frac{0.1 \times 10^{-3} \times 100}{1 \times 10^{-3}-0.1 \times 10^{-3}} \\
& =\frac{1 \times 10^{-2}}{0.9 \times 10^{-3}}=\frac{10^2}{9}=\frac{100}{9} \Omega
\end{aligned}
$$
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