Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the galvanometer be $G$ and the full-scale deflection current be $I_g$.

When a resistance of $990\Omega$ is connected in series, the range of the voltmeter is $V$.

$V=I_g(G+990)$

When a resistance of $480\Omega$ is connected in series, the range is halved, i.e., $\frac{V}{2}$.

$\frac{V}{2}=I_g(G+480)$

Dividing the first equation by the second equation:

$\frac{V}{\frac{V}{2}}=\frac{I_g(G+990)}{I_g(G+480)}$

$2=\frac{G+990}{G+480}$

$2(G+480)=G+990$

$2G+960=G+990$

$G=30\Omega$

Answer: $30\Omega$