JEE Main 2023 · 24 January · evening · Single correct · +4 / −1
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 :
- (a) 40 V
- (b) 90 V
- (c) 45 V
- (d) 80 V
Answer and solution
Answer (a) 40 V
Solution
Physics sir · 3 steps
A real voltmeter draws current, so treat it as a 400 Ω resistor in parallel with the 100 Ω it measures.
- Voltmeter as resistor. Put the 400 Ω voltmeter across one 100 Ω resistor. The pair acts as a single resistance Rₚ = (100 × 400)/(100 + 400) = 80 Ω.
- Find the current. This 80 Ω is in series with the other 100 Ω, so the total is 180 Ω. The current from the cell is I = 90/180 = 0.5 A.
V = I Rₚ = 0.5 × 80 = 40 V
- So. The voltmeter reads 40 V, option (a). By symmetry it reads the same across either resistor. The reading is less than 45 V because connecting the meter lowers the resistance of that part of the circuit.
The idea to keep. A voltmeter with finite resistance changes the circuit it measures. Combine it in parallel with the element first, then find the voltage across that combination.
Answer from the NTA final answer key. Solution by PrepWiser.