MHTCET 22 Apr 2025 Shift 1 Paper

Concept:A voltmeter measures up to its full-scale current. To increase its range, a multiplier resistance must be connected in series with it.Explanation:The given voltmeter has a range of $10\,V$ and internal resistance $R = 50\,\Omega$.Its full-scale deflection current is:$$I = \frac{V}{R} = \frac{10}{50} = 0.2\,A$$To extend the range to $15\,V$, the same full-scale current $0.2\,A$ must flow through the meter when $15\,V$ is applied across the combination.Therefore, the extra voltage to be dropped across the series resistance is:$$V_{\text{extra}} = 15\,V - 10\,V = 5\,V$$Now, the required series resistance is:$$R_{\text{series}} = \frac{V_{\text{extra}}}{I} = \frac{5}{0.2} = 25\,\Omega$$Thus, a $25\,\Omega$ resistance must be connected in series with the voltmeter.Answer:$25\,\Omega$ resistance in series, i.e., Option D.