Q. In an ammeter, $4\%$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $ 5\,\Omega $ resistance, the resistance of the galvanometer is:
Jharkhand CECEJharkhand CECE 2004
Solution:
Potential difference across galvanometer resistance and shunt is same.
Let $i_{g}$ be the current across galvanometer
and $i-i_{g}$ across shunt,
then Potential difference across $G=$ potential difference across $S$
i.e., $i_{g} \times G=\left(i-i_{g}\right) \times S$
Given,
$\frac{i_{g}}{i}=\frac{4}{100}=0.04$,
$ S=5 \,\Omega$
$ \therefore \frac{G}{5}=\frac{1}{0.04}-1=24$
$ \Rightarrow G=24 \times 5=120 \,\Omega$
