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A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is
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Verified Answer
The correct answer is:
$\frac{G^2}{S+G}$
Given, galvanometer resistance, $R_G=G$ and shunt resistance, $R_S=S$,
Let the series resistance is $R$, so the net resistance of circuit after connected $R$,
Since, it is given that the current should not change. So, it should be $G$,
From Eq. (i) and (ii), we get
$$
\begin{aligned}
G & =R+\frac{G \times S}{G+S} \\
\Rightarrow \quad R & =G-\frac{G S}{G+S} \Rightarrow R=\frac{G^2}{G+S}
\end{aligned}
$$
Hence, the correct option is (a).
Let the series resistance is $R$, so the net resistance of circuit after connected $R$,
Since, it is given that the current should not change. So, it should be $G$,
From Eq. (i) and (ii), we get
$$
\begin{aligned}
G & =R+\frac{G \times S}{G+S} \\
\Rightarrow \quad R & =G-\frac{G S}{G+S} \Rightarrow R=\frac{G^2}{G+S}
\end{aligned}
$$
Hence, the correct option is (a).
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