Physics - Current Electricity Question with Solution | TestHub

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

When a resistance of $980\Omega$ is connected in series, the range of the voltmeter is $V$.

$V=I_g(G+980)$

When a resistance of $470\Omega$ is connected in series, the range is halved to $\frac{V}{2}$.

$\frac{V}{2}=I_g(G+470)$

Dividing the first equation by the second equation:

$2=\frac{G+980}{G+470}$

$2G+940=G+980$

$G=40\Omega$

Answer: $40\Omega$