When a galvanometer is shunted by a resistance. 's', its current capacity increases ' $n$ ' times. If the…

When a galvanometer is shunted by a resistance. 's', its current capacity increases ' $n$ ' times. If the same galvanometer is shunted by another resistance ' $\mathrm{s}_1$ ', its capacity will increase to ' $\mathrm{n}_1$ ' times original current. The value of ' $n_1$ ' is
  1. $\frac{(\mathrm{n}+\mathrm{s})}{\mathrm{s}_1}$
  2. $\frac{\mathrm{s}_1(\mathrm{n}-\mathrm{s})-\mathrm{s}_1}{\mathrm{~s}_1}$
  3. $\frac{(\mathrm{n}+1) \mathrm{s}}{\mathrm{s}_1}$
  4. $\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1}$

Solution

For galvanometer shunt resistance is given by, $s=G\left(\frac{1}{1-I / I_g}\right)$
Current capacity is given by, $\mathrm{n}=\mathrm{I} / \mathrm{Ig}$ $\begin{aligned} & \mathrm{s}=\mathrm{G}\left(\frac{1}{1-\mathrm{n}}\right) \\ & \therefore \quad \mathrm{G}=\mathrm{s}(\mathrm{n}-1)...(i) \end{aligned}$
When same galvanometer is shunted with resistance ' $\mathrm{s}_1$ ' $\begin{aligned} & \mathrm{s}_1=\mathrm{G}\left(\frac{1}{1-\mathrm{n}_1}\right) \\ & \mathrm{n}_1=\frac{\mathrm{G}+\mathrm{s}_1}{\mathrm{~s}_1}=\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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