JEE Main 2023 · 15 April · morning · Single correct · +4 / −1
In the given circuit, the current (I) through the battery will be
- (a) 1A
- (b) 2.5 A
- (c) 2 A
- (d) 1.5 A
Answer and solution
Answer (d) 1.5 A
Solution
Physics sir · 3 steps
Check each diode's direction first; the reverse-biased one drops right out of the circuit.
- Find the polarity. The longer plate of the cell is on the right, so the bottom-right corner is at +10 V. The whole left wire, including the junction where the slanted branches meet, is at 0 V. Current leaves the bottom-right corner, flows through the network to the left, and returns down the left wire as I.
Branch Diode points Current wants to go Status Lower slant: 10 Ω + D₃ towards left right to left forward, conducts Top: 10 Ω + D₁ towards left right to left forward, conducts Upper slant: 10 Ω + D₂ towards right right to left reverse, open circuit - Two paths remain. Path 1 runs from the bottom-right corner through D₃ and 10 Ω to the left side, giving 10 Ω. Path 2 runs up the right 10 Ω, then through D₁ and the top 10 Ω, giving 20 Ω. These two paths are in parallel across the cell.
R_(eq) = (10 × 20)/(10 + 20) = 20/3 Ω
- So. I = 10/(20/3) = 1.5 A. Path 1 carries 1 A and path 2 carries 0.5 A, so the answer is (d).
The idea to keep. In circuits with ideal diodes, first mark which end is at higher potential and check every diode against it. A forward diode is a plain wire, a reverse one is a break. Then solve the ordinary resistor network.
Answer from the NTA final answer key. Solution by PrepWiser.