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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Physics question on Current Electricity

A cell of emf 90 V is connected across series combination of two resistors each of 100Ω100\Omega resistance. A voltmeter of resistance 400Ω400\Omega is used to measure the potential difference across each resistor. The reading of the voltmeter will be :
  1. A.
    45 V
  2. B.
    80 V
  3. C.
    90 V
  4. D.
    40 V
Show answer and solution

Correct option: D

Solution

The voltmeter (400Ω) in parallel with one 100Ω resistor gives an effective resistance of (400×100)/(400+100) = 80Ω. The circuit now has 80Ω and 100Ω in series across 90V. Current from cell = 90/(80+100) = 0.5A. Voltmeter reading = 0.5 × 80 = 40V.

[JEE Main 2023 · 24 Jan · Shift 2]

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