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A galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly)
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$6 \Omega$
Given:
Resistance of galvanometer is $G=200 \Omega$.
Current through the galvanometer is $i_g=3 \%$ of $i$.
To convert into an ammeter, the galvanometer with resistance $G$ has to be shunted with $S$. The potential drop across the galvanometer and shunt is the same:
$\begin{aligned} & i_g G=\left(i-i_g\right) S \\ & \Rightarrow \frac{i}{i_g}=1+\frac{G}{S}\end{aligned}$
Given, $\frac{i_g}{i}=\frac{3}{100}$
$\begin{aligned} & \therefore \frac{100}{3}=1+\frac{200}{S} \\ & \Rightarrow S=\frac{200 \times 3}{97} \approx 6 \Omega\end{aligned}$
Resistance of galvanometer is $G=200 \Omega$.
Current through the galvanometer is $i_g=3 \%$ of $i$.
To convert into an ammeter, the galvanometer with resistance $G$ has to be shunted with $S$. The potential drop across the galvanometer and shunt is the same:
$\begin{aligned} & i_g G=\left(i-i_g\right) S \\ & \Rightarrow \frac{i}{i_g}=1+\frac{G}{S}\end{aligned}$
Given, $\frac{i_g}{i}=\frac{3}{100}$
$\begin{aligned} & \therefore \frac{100}{3}=1+\frac{200}{S} \\ & \Rightarrow S=\frac{200 \times 3}{97} \approx 6 \Omega\end{aligned}$
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