Question 538: Current Electricity

Question

When a galvanometer is shunted by a resistance ' $S$ ', its current capacity increases ' $n$ ' times. If the same galvanometer is shunted by another resistance $\mathrm{S}^{\prime}$, its current capacity increases to $\mathrm{n}^{\prime}$. The value of $n^{\prime}$ in terms of $n, S$ and $S^{\prime}$ is

Accepted Answer

Let $I_g$ be the current capacity of the galvanometer without any shunt. When a shunt resistance $S$ is connected, the current capacity increases to $nI_g$.  The fraction of current passing through the galvanometer is given by:
$$ \frac{I_g}{nI_g} = \frac{S}{G+S} $$
where $G$ is the galvanometer resistance. This simplifies to:
$$ \frac{1}{n} = \frac{S}{G+S} $$
Solving for $G$:
$$ G+S = nS $$
$$ G = nS - S = S(n-1) $$
Now, if the galvanometer is shunted by another resistance $S'$, the current capacity increases to $n'I_g$.  Using the same principle:
$$ \frac{I_g}{n'I_g} = \frac{S'}{G+S'} $$
$$ \frac{1}{n'} = \frac{S'}{G+S'} $$
Substitute the expression for $G$ we derived earlier:
$$ \frac{1}{n'} = \frac{S'}{S(n-1) + S'} $$
Solving for $n'$:
$$ n' = \frac{S(n-1) + S'}{S'} $$
Therefore, the value of $n'$ in terms of $n$, $S$, and $S'$ is:
$$ n' = \frac{S(n-1) + S'}{S'} $$
Thus, the correct option is
D
.

Answer

$\frac{\mathrm{n}+\mathrm{S}}{\mathrm{S}^{\prime}}$

Answer

$\frac{\mathrm{S}(\mathrm{n}-1)-\mathrm{S}^{\prime}}{\mathrm{S}^{\prime}}$

Answer

$\frac{(\mathrm{n}+1) \mathrm{S}}{\mathrm{S}^{\prime}}$

Answer

$\frac{\mathrm{S}(\mathrm{n}-1)+\mathrm{S}^{\prime}}{\mathrm{S}^{\prime}}$

Current Electricity

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}^{\prime}, its current capacity increases to \mathrm{n}^{\prime}. The value of n^{\prime} in terms of n, S and S^{\prime} is

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