A galvanometer of resistance G is converted into a voltmeter of range 0 – 1 V by connecting a resistance R1 in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0 – 2 V will be :

Current Electricity PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q58 with Solution

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Correct answer: C

Let the galvanometer full-scale current be $I_g$ and its resistance be $G$ .

To convert it into a voltmeter of range $0$ to $1\,\text{V}$ , a series resistance $R_1$ is connected.

At full-scale deflection:

$1 = I_g(G + R_1)$

So,

$G + R_1 = \frac{1}{I_g} \quad \Rightarrow \quad R_1 = \frac{1}{I_g} - G$

Suppose an additional resistance $R$ is connected in series with the existing combination.

Then total series resistance becomes:

$G + R_1 + R$

For full-scale deflection at $2\,\text{V}$ :

$2 = I_g(G + R_1 + R)$

Using

$G + R_1 = \frac{1}{I_g}$

we get:

$2 = I_g\left(\frac{1}{I_g} + R\right) = 1 + I_gR$

Thus,

$I_gR = 1 \quad \Rightarrow \quad R = \frac{1}{I_g}$

But from step 1,

$\frac{1}{I_g} = G + R_1$

Hence,

$R = G + R_1$

The additional resistance required is:

$R_1 + G$

So the correct option is C.