Physics - Current Electricity Question with Solution | TestHub

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

For the first voltmeter of range $V_1=15\text{ V}$:
$V_1=I_g(G+X)$
$15=I_g(100+X)$

For the second voltmeter, the range is doubled, so $V_2=30\text{ V}$. A resistance of $1500\Omega$ is connected in series with $X$:
$V_2=I_g(G+X+1500)$
$30=I_g(100+X+1500)$
$30=I_g(1600+X)$

Dividing the second equation by the first equation:
$\frac{30}{15}=\frac{1600+X}{100+X}$

$2=\frac{1600+X}{100+X}$

$2(100+X)=1600+X$

$200+2X=1600+X$

$X=1400\Omega$

Answer: 1400