CBSE 2025, Set 1

Question

A voltmeter of resistance 1000 $\Omega$ can measure up to 25 V. How will you convert it so that it can read up to 250 V ?

Edvaya · Accepted Answer

The correct answer is Subjective. The full-scale deflection current of the voltmeter: $$I_g = \frac{V}{R_v} = \frac{25}{1000} = 0\cdot025 	ext{ A} = 25 	ext{ mA}$$ To extend the range to 250 V, a **series resistance** $R_s$ must be connected in series with the voltmeter so that the full deflection current $I_g$ flows when 250 V is applied: $$I_g = \frac{V'}{R_v + R_s}$$ $$0\cdot025 = \frac{250}{1000 + R_s}$$ $$1000 + R_s = \frac{250}{0\cdot025} = 10000 	ext{ }\Omega$$ $$\boxed{R_s = 9000 	ext{ }\Omega}$$ **Method:** Connect a resistance of **9000 Ω** (9 kΩ) in **series** with the voltmeter. > **Multiplier formula:** $R_s = R_v(n - 1)$ where $n = V'/V = 250/25 = 10$. So $R_s = 1000(10 - 1) = 9000$ Ω. A voltmeter is always extended by adding series resistance (a "multiplier"), never a shunt.

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Physics > Electrostatics And Current Electricity > Current Electricity

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