JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Current Electricity

A galvanometer has a resistance of

50Ω

and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance

Answer: (C)

Solution

To extend a galvanometer of resistance

G = 50 Ω

and full-scale current

I_{g} = 5

mA to read up to

V = 100

V, a series resistance

R

is added such that

V = I_{g} (G + R) ⟹ R = \frac{V}{I _{g}} - G = \frac{100}{5 \times 1 0 ^{- 3}} - 50 = 20000 - 50 = 19950 Ω

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option