A galvanometer of resistance $50\Omega$ gives a full scale deflection for a current of $0.05\text{ A}$. To convert it into a voltmeter of range $0$ to $5\text{ V}$, the required series resistance is

Solution

1. The total resistance required is $R_{total}=V/I_g$.
2. Given $V=5\text{ V}$ and $I_g=0.05\text{ A}$, $$\begin{array}{rl}{R}_{total}& =5/0.05\\ & =100\Omega \end{array}$$.
3. The series resistance $R_s$ is $R_{total}-G$, where $G$ is the galvanometer resistance.
4. Thus $$\begin{array}{rl}{R}_s& =100-50\\ & =50\Omega \end{array}$$.
5. Hence the answer is (A).