Question

We have a galvanometer of resistance $25\Omega$. It is shunted by a $2.5\Omega$ wire, the part of total current that flows through the galvanometer is given as

• A. $\frac{I_g}{I}=\frac{4}{11}$
• B. $\frac{I_g}{I}=\frac{3}{11}$
• C. $\frac{I_g}{I}=\frac{2}{11}$
• D. $\frac{I_g}{I}=\frac{1}{11}$

Answer 1

Answer: (D)

In the circuit $G$ and $S$ are in parallel, the potential difference across them is same, hence

$I_g\times G=\left(I_0-I_g\right)\times S$
$\Rightarrow \frac{I_g}{I_0}=\frac{S}{S+G}$
Given, $S=2.5\Omega ,G=25\Omega$
$\therefore \frac{I_g}{I_0}=\frac{2.5}{25+2.5}$
$=\frac{2.5}{27.5}=\frac{1}{11}$
$I_g=I$
$\therefore \frac{I_g}{I}=\frac{1}{11}$