Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the galvanometer be $G$ and the total current in the circuit be $I$.

The shunt resistance is given as $S=\frac{G}{n-1}$.

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

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

Substituting the value of $S$:

$I_{g}G=(I-I_g)\frac{G}{n-1}$

$I_g(n-1)=I-I_g$

$n{I}_g-I_g+I_g=I$

$n{I}_g=I$

$\frac{I_g}{I}=\frac{1}{n}$

The fraction of the total current passing through the galvanometer is \dfrac{1}{n}.

Answer: $\frac{1}{n}$