Q120 / 150 · MHT CET 2025 (22 Apr Shift 1) · Physics

The voltmeter has range 10 V and its internal resistance is $50 \Omega$. To increase the range of voltmeter to 15 V , the resistance which is to be connected is

  • A
    $125 \Omega$ resistance in parallel
  • B
    $125 \Omega$ resistance in series
  • C
    $25 \Omega$ resistance in parallel
  • Correct answer
    D
    $25 \Omega$ resistance in series

Solution

To extend the voltmeter range to 15 V, a series resistance is required. The full-scale deflection current remains $I_g = V / R_v = 10 \text{ V} / 50 \Omega = 0.2 \text{ A}$.

For the new range $V' = I_g (R_v + R_s)$, so $15 = 0.2(50 + R_s)$.
Solving gives $R_s = 15 / 0.2 - 50 = 75 - 50 = 25 \Omega$.

A $25 \Omega$ series resistance corresponds to option D.

Hence, the correct answer is option: 'D'.

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