The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 Ω is __________ mA.

Solution

Physics sir · 4 steps

Find which diode the battery forward-biases. Only that branch carries current, so it is one loop.

1. Current direction. The longer plate of the cell is the positive terminal, and it faces left. So conventional current leaves the cell, goes up the left wire, crosses the parallel branches from left to right, and returns through the 40 Ω resistor.
2. Check each diode. D₁ points left to right, along the current, so it is forward biased and conducts with a 0.6 V drop. D₂ points right to left, against the current, so it is reverse biased and its branch carries no current.
3. Single loop. The only path for current is the cell, the upper 60 Ω, D₁ and the 40 Ω, all in series. Kirchhoff's loop rule gives:

   I = (1 - 0.6)/(60 + 40) = 0.4/100 A = 4 × 10⁻³ A

4. So. The current through the 40 Ω resistor is 4 mA.

The common slip: 10. Forgetting the 0.6 V drop across D₁ gives 1/100 A, which is 10 mA, not 4 mA.

The idea to keep. With diodes in a circuit, first mark the current direction from the cell and decide which diodes conduct. Drop the reverse-biased branches, then subtract each cut-off voltage from the emf.

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