JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
In the given figure, each diode has a forward bias resistance of 30Ω and infinite resistance in reverse bias. The current I₁ will be :
- (a) 3.75 A
- (b) 2.35 A
- (c) 2 A
- (d) 2.73 A
Answer and solution
Answer (c) 2 A
Solution
Physics sir · 4 steps
Check which way each diode points first. Only the branches that conduct go into the circuit.
- Find the current direction. The long plate of the 200 V cell is on the left, so current leaves it and flows up the left wire, as I₁ shows. It then tries to cross each branch from left to right.
- Sort the diodes. The top two diodes point left to right, so they are forward biased and each acts as a 30 Ω resistor. The third diode points right to left, so it is reverse biased. Its branch carries no current and drops out.
- Total resistance. Each conducting branch has 30 + 130 = 160 Ω. Two of them in parallel give 80 Ω. Adding the 20 Ω in series gives 100 Ω.
I₁ = (200 V)/(80 Ω + 20 Ω) = 2 A
- So. I₁ = 2 A, option (c).
The idea to keep. In diode circuits, first find the direction current wants to flow. Then treat a forward diode as its forward resistance and remove a reverse-biased branch completely.
Answer from the NTA final answer key. Solution by PrepWiser.