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
- A40   V
- B45   V
- C80   V
- 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