JEE Main 2020 · 9 January · evening · Single correct · +4 / −1
The current i in the network is :
- (a) 0.6 A
- (b) 0.3 A
- (c) 0 A
- (d) 0.2 A
Answer and solution
Answer (b) 0.3 A
Solution
Physics sir · 4 steps
Check each diode first: both are reverse biased, so only one series path carries the current.
- Battery polarity. The long plate of the 9 V cell is on the left, so the left junction of the bridge is at the higher potential. Current goes up through the external 5 Ω into the left junction and returns from the right junction. This matches the direction of the arrow i.
- Check both diodes. The upper-left diode can conduct only from the top junction toward the left junction, but the left junction is at the higher potential, so it is reverse biased. The lower-right diode can conduct only from the right junction toward the bottom junction, but the bottom junction is at the higher potential, so it is also reverse biased. Neither diode branch carries current.
- Remaining path. The only route left is: left junction, 10 Ω down to the bottom junction, the middle 5 Ω up to the top junction, then 10 Ω to the right junction. These three resistors are in series with the external 5 Ω.
i = 9/(5 + 10 + 5 + 10) = 9/30 = 0.3 A
- So. The current in the network is i = 0.3 A, which is option (b).
The idea to keep. With diodes in a circuit, find the higher-potential side first and decide forward or reverse for each diode. A reverse-biased diode removes its whole branch. Then solve what is left.
Answer from the NTA final answer key. Solution by PrepWiser.