Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the galvanometer be $G$ and its original current capacity be $I_g$.

When shunted by a resistance $s$, the new current capacity is $I=n{I}_g$.

Since the galvanometer and shunt are in parallel, the potential difference across them is equal:

$I_{g}G=(I-I_g)s$

$I_{g}G=(n{I}_g-I_g)s$

$G=(n-1)s$

When shunted by another resistance $s_1$, the new current capacity is $I_1=n_1{I}_g$.

Similarly, equating the potential difference:

$I_{g}G=(I_1-I_g)s_1$

$I_{g}G=(n_1{I}_g-I_g)s_1$

$G=(n_1-1)s_1$

Equating the two expressions for $G$:

$(n-1)s=(n_1-1)s_1$

$n_1-1=\frac{s(n-1)}{s_1}$

$n_1=\frac{s(n-1)}{s_1}+1$

$n_1=\frac{s(n-1)+s_1}{s_1}$

Answer: $\frac{\mathrm{s}(\mathrm{n}-1)+{\mathrm{s}}_1}{{\mathrm{s}}_1}$