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JEE Main202324 Jan 2023Evening ShiftPhysicsCurrent ElectricityActual

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

Options

  1. A40   V
  2. B45   V
  3. C80   V
  4. D90   V

Correct answer

A. 40   V

Step-by-step solution

As we know, voltmeter is connected in parallel to the element being measured. The equivalent resistance of the above circuit is R e q = 400 × 100 400 + 100 + 100 = 80 + 100 = 180   Ω Current through the circuit is i = 90 180 = 1 2   A Reading of the voltmeter is V R = i 400 × 100 400 + 100 = 1 2 × 80 = 40   V

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