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A galvanometer has a coil of resistance $100 \Omega$ showing a full scale deflection at $50 \mu \mathrm{A}$. The resistance that should be added to use it as an ammeter of range $10 \mathrm{~mA}$ is
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Verified Answer
The correct answer is:
$0.5 \Omega$
Given, Resistance of galvanometer
$$
\begin{aligned}
R_g & =100 \Omega \\
I_g & =50 \mu \mathrm{A}=5 \times 10^{-5} \mathrm{~A} \\
I & =10 \mathrm{~mA}=10^{-2} \mathrm{~A} \\
R_s & =?
\end{aligned}
$$
We know that,
$$
R_s=\frac{I_g}{I-I_g} \times R_g .
$$
$\begin{aligned} & =\frac{5 \times 10^{-5}}{10^{-2}-5 \times 10^{-5}} \times 100 \\ & =\frac{5 \times 10^{-3}}{(1000-5) 10^{-5}}=\frac{5 \times 10^2}{995} \\ & =\frac{500}{995}=0.503 \\ & \simeq 0.5 \Omega\end{aligned}$
$$
\begin{aligned}
R_g & =100 \Omega \\
I_g & =50 \mu \mathrm{A}=5 \times 10^{-5} \mathrm{~A} \\
I & =10 \mathrm{~mA}=10^{-2} \mathrm{~A} \\
R_s & =?
\end{aligned}
$$
We know that,
$$
R_s=\frac{I_g}{I-I_g} \times R_g .
$$
$\begin{aligned} & =\frac{5 \times 10^{-5}}{10^{-2}-5 \times 10^{-5}} \times 100 \\ & =\frac{5 \times 10^{-3}}{(1000-5) 10^{-5}}=\frac{5 \times 10^2}{995} \\ & =\frac{500}{995}=0.503 \\ & \simeq 0.5 \Omega\end{aligned}$
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