A cell of emf 90 V is connected across series combination of two resistors each of $100 \Omega$ resistance. A voltmeter of resistance $400 \Omega$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Current Electricity

A cell of emf 90 V is connected across series combination of two resistors each of

100Ω

resistance. A voltmeter of resistance

400Ω

is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

Answer: (A)

Connecting the

400 Ω

voltmeter across one

100 Ω

resistor forms a parallel combination of

\frac{400 \times 100}{500} = 80 Ω

. The total circuit resistance becomes

80 + 100 = 180 Ω

, so the current supplied by the cell is

\frac{90}{180} = 0.5

A. The voltmeter registers the potential across the

80 Ω

section, which is

0.5 \times 80 = 40

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option