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(D) The shunt resistance $s$ required to increase the current capacity of a galvanometer of resistance $G$ by a factor $n$ is given by the formula: $s

Vedclass · Accepted Answer

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

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(D) The shunt resistance $s$ required to increase the current capacity of a galvanometer of resistance $G$ by a factor $n$ is given by the formula: $s = \frac{G}{n-1}$.From this,we can express the galvanometer resistance as: $G = s(n-1) \dots (i)$.When the same galvanometer is shunted by a resistance $s_1$,the new current capacity factor $n_1$ is given by: $s_1 = \frac{G}{n_1-1}$.Rearranging for $n_1$,we get: $n_1 - 1 = \frac{G}{s_1}$,which implies $n_1 = \frac{G}{s_1} + 1 = \frac{G + s_1}{s_1}$.Substituting the value of $G$ from equation $(i)$ into this expression:$n_1 = \frac{s(n-1) + s_1}{s_1}$.

When a galvanometer is shunted by a resistance $s$,its current capacity increases $n$ times. If the same galvanometer is shunted by another resistance $s_1$,its capacity will increase to $n_1$ times the original current. The value of $n_1$ is

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