JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

A voltmeter with internal resistance of $x Ω$ can be used to measure upto $20$ V. In order to increase its measuring range to $30$ V, the required modification is to ________.

A
connect resistor of $\frac{x}{2} Ω$ , in series with voltmeter.
B
connect resistor of $\frac{x}{2} Ω$ , in parallel to voltmeter.
C
connect resistor of $x Ω$ , in series with voltmeter.
D
connect resistor of $2 x Ω$ , in parallel to voltmeter.

Solution

The voltmeter's full-scale current must be the same in both cases. With internal resistance $x$ giving $20$ V, the full-scale current is $I_{g} = \frac{20}{x}$ .

To extend the range to $30$ V, a series multiplier $R$ is needed so that the same $I_{g}$ now corresponds to $30$ V across the combination:

$\frac{30}{x + R} = \frac{20}{x} \Rightarrow 30 x = 20 (x + R) \Rightarrow R = \frac{x}{2} .$

Hence connect $\frac{x}{2} Ω$ in series.

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