JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Current Electricity

Two resistors

400Ω

and

800Ω

are connected in series across a 6 V battery. The potential difference measured by a voltmeter of

10 k Ω

across

400Ω

resistor is close to:

Answer: (C)

Solution

A real voltmeter of

10 k Ω

placed across the

400 Ω

resistor sits in parallel with it, giving

R_{eq} = \frac{400 \times 10000}{400 + 10000} \approx 384.6 Ω

. This combination is in series with the

800 Ω

resistor across

6 V

, so the voltmeter reads

\frac{384.6}{384.6 + 800} \times 6 \approx 1.95 V

, a little below the ideal

2 V

because the meter loads the circuit.

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option