Physics - Current Electricity Question with Solution | TestHub

Given $V_1=10$ V, $R_v=50\Omega$, $V_2=15$ V.

The current for full scale deflection is $I_g=\frac{V_1}{R_v}=\frac{10}{50}=0.2$ A.

To increase the range of the voltmeter, a resistance $R$ must be connected in series.

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 $25\Omega$ resistance must be connected in series.

Answer: $25\Omega$ resistance in series