If each diode has a forward bias resistance of 25 Ω in the below circuit,
Which of the following options is correct :

Solution

Physics sir · 3 steps

Check which way each diode points first, because the reverse-biased branch carries no current at all.

1. Diode directions. The long plate of the cell is on the left, so current I₁ goes up the left wire and enters each branch from the left. D₁ and D₃ point to the right and are forward biased. D₂ points left, so it is reverse biased and I₃ = 0.

   Branch	Diode state	Branch resistance	Current
   D₁ (I₄)	forward	25 + 125 = 150 Ω	I₄ = I₂
   D₂ (I₃)	reverse	open circuit	I₃ = 0
   D₃ (I₂)	forward	25 + 125 = 150 Ω	I₂

2. Total current. Two equal 150 Ω branches in parallel give 75 Ω. Adding the 25 Ω in series gives 100 Ω, so I₁ = 5/100 = 0.05 A.
3. So. I₁ splits equally between the two conducting branches: I₂ = I₄ = 0.025 A. Hence I₁/I₂ = 2, which is option (b).

The idea to keep. In a diode circuit, first mark each diode as forward or reverse. A reverse-biased diode is an open switch, so delete that branch and solve the rest as an ordinary resistor network.

Answer from the NTA final answer key. Solution by PrepWiser.