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, $S=2.5\, \Omega,\, G=25\, \Omega$
$\therefore \frac{I_{g}}{I_{0}}=\frac{2.5}{25+2.5}$
$=\frac{2.5}{27.5} =\frac{1}{11}$
$I_{g}=I$
$\therefore \frac{I_{g}}{I}=\frac{1}{11}$