Question

A galvanometer has a resistance $50\Omega$ . A resistance of $5\Omega$ is connected parallel to it. Fraction of the total current flowing through galvanometer is

• A. $\frac{1}{10}$
• B. $\frac{1}{11}$
• C. $\frac{1}{50}$
• D. $\frac{2}{15}$

Answer 1

Answer: (B)

The galvanometer $G$ and shunt $S$ is connected in parallel,
hence potential difference is the same.
$\therefore i_g\times G=\left(i-i_g\right)\times S$
$\Rightarrow \frac{i_g}{i-i_g}=\frac{S}{G}=\frac{5}{50}=\frac{1}{10}$
$\Rightarrow 10i_g=i-i_g$
$\Rightarrow 11i_g=i$
$\Rightarrow \frac{i_g}{i}=\frac{1}{11}$