Physics - Current Electricity Question with Solution | TestHub

Let $I_g$ be the current for full scale deflection of the galvanometer.

For the initial range $V$ with a series resistance of $100\Omega$:

$V=I_g(R+100)$

When an additional resistance of $1000\Omega$ is connected in series, the new series resistance is $100+1000=1100\Omega$ and the range is doubled to $2V$.

$2V=I_g(R+1100)$

Dividing the second equation by the first equation:

$\frac{2V}{V}=\frac{I_g(R+1100)}{I_g(R+100)}$

$2=\frac{R+1100}{R+100}$

$2(R+100)=R+1100$

$2R+200=R+1100$

$R=1100-200=900\Omega$