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 10 i_{g}=i-i_{g} $
$\Rightarrow 11 i_{g}=i$
$\Rightarrow \frac{i_{g}}{i}=\frac{1}{11}$