Question

A galvanometer of resistance $G$ is converted into a voltmeter of range $0–1V$ by connecting a resistance $R_1$ in series with it. The additional resistance that should be connected in series with $R_1$ to increase the range of the voltmeter to$0–2V$ will be:

• A. $G$
• B. $R_1-G$
• C. $R_1+G$
• D. $R_1$

Answer 1

The correct option is C

$R_1+G$

Step 1. Given data:

$R_1\&R_2=$Resistance connected in series with the galvanometer

$G=$Resistance of the Galvanometer

$I=$Current at which deflection is maximum

Step 2. Calculating potential difference for $\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathbf{}\mathbf{1}$:

From ohm's law:

$V=I\left(G+R_1\right)$

$\begin{array}{rcl}& \Rightarrow & 1=I\left(G+R_1\right)\\ & \Rightarrow & I=\frac{1}{\left(G+R_1\right)}\end{array}$ [Potential difference in case 1 is $V=1volt$]

Step 3. Calculating the value of $R_2$ for the $\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathbf{}\mathbf{2}$:

From ohm's law,

$\begin{array}{rcl}V& =& I\left(G+R_1+R_2\right)\\ & \Rightarrow & 2=\frac{1}{\left(G+R_1\right)}\left(G+R_1+R_2\right)\\ & \Rightarrow & 2G+2R_1=G+R_1+R_2\\ \therefore R_2& =& R_1+G\end{array}$ [Putting value of $I=\frac{1}{\left(G+R_1\right)}$ from step 1 & Potential difference in case 2 is $V=2volt$]

Thus, the additional resistance that should be connected is $R_1+G$.

Hence, correct option is (C).