A galvanometer has a resistance of 50 Ω and it allows maximum current of 5 mA. It can be converted into voltmeter to measure up to 100 V by connecting in series a resistor of resistance: (a) 19500 Ω (b) 5975 Ω (c) 20050 Ω (d) 19950 Ω

Answer: (d). Series resistance R = V/Ig - G. R = 1005 × 10⁻³ - 50 = 20000 - 50 = 19950 Ω So the answer is (d) 19950 Ω.

A galvanometer has a resistance of 50 Ω and it allows maximum current… | JEE Main 2024 (1 Feb, Shift 1)

Q47JEE Main 2024 · 1 FEB · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments

A galvanometer has a resistance of $50\ \Omega$ and it allows maximum current of $5\text{ mA}$ . It can be converted into voltmeter to measure up to $100\text{ V}$ by connecting in series a resistor of resistance:

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Answer: (d)

Series resistance $R = \frac{V}{I_g} - G$ .
$R = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19950\ \Omega$
So the answer is (d) $19950\ \Omega$ .

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