Physics - Current Electricity Question with Solution | TestHub

Current for full scale deflection of the voltmeter is given by $I_g=\frac{V_1}{R_v}$.

Substituting the given values, $I_g=\frac{10}{50}=0.2\text{A}$.

To increase the range of the voltmeter to $V_2=15\text{V}$, a resistance $R$ must be connected in series with it.

The new range is given by $V_2=I_g(R_v+R)$.

Substituting the values, $15=0.2(50+R)$.

$50+R=\frac{15}{0.2}=75$

$R=75-50=25\Omega$.

Thus, a $25\Omega$ resistor must be connected in series.

Answer: $25\Omega$ resistor in series.