JEE Main 2022 · 25 July · evening · Single correct · +4 / −1
A current of 15 mA flows in the circuit as shown in figure. The value of potential difference between the points A and B will be:
- (a) 50 V
- (b) 75 V
- (c) 150 V
- (d) 275 V
Answer and solution
Answer (d) 275 V
Solution
Physics sir · 4 steps
Follow the 15 mA from A to B: it passes the 5 kΩ, then the parallel pair, then the 10 kΩ.
- Read the path. From A the current goes through the bottom 5 kΩ to the left wire. There it splits between the 10 kΩ and the 5 kΩ, which share both end nodes, so they are in parallel. The current rejoins on the right and goes through the last 10 kΩ to B.
- Parallel pair. Rₚ = (10 × 5)/(10 + 5) = 10/3 kΩ. All of the 15 mA passes through this combination, so it acts like one resistor in series with the other two.
V_(AB) = I(5 + 10/3 + 10) kΩ = 15 mA × 55/3 kΩ
- Add the drops. Across the 5 kΩ the drop is 15 mA × 5 kΩ = 75 V. Across the parallel pair it is 15 × 10/3 = 50 V, and across the 10 kΩ it is 150 V.
- So. V_(AB) = 75 + 50 + 150 = 275 V, which is option (d).
The idea to keep. Mark which resistors share both end nodes. Those are in parallel. Collapse them to one resistor, then everything left on the path carries the same current and their drops just add.
Answer from the NTA final answer key. Solution by PrepWiser.