The circuit contains two diodes each with a forward resistance of 50Ω and with infinite reverse resistance. If the battery voltage is 6V, the current through the 120Ω resistance is ____________ mA.

Solution

Physics sir · 4 steps

First decide which diode conducts. The dead branch drops out and you are left with one series loop.

1. Battery direction. The longer plate of the cell is its positive terminal, and it faces left. Conventional current leaves it, goes up the left wire and reaches both diodes from their left ends.
2. Check each diode. D₁ points to the right, along the current, so it is forward biased and acts as a 50 Ω resistor. D₂ points to the left, against the current, so it is reverse biased. Its infinite resistance means no current flows in the 100 Ω branch.
3. One series loop. Current flows through D₁, the 130 Ω and the 120 Ω resistors, all in series. The total resistance is 50 + 130 + 120 = 300 Ω.

   I = V/R = (6 V)/(300 Ω) = 0.02 A

4. So. The same current passes through the 120 Ω resistor. It is 0.02 A = 20 mA.

The common slip: 24. Treating D₁ as an ideal diode and leaving out its 50 Ω forward resistance gives 6/250 A, which is 24 mA, not 20 mA.

The idea to keep. In diode circuits, find the current direction from the battery first. A reverse-biased diode cuts off its whole branch. A forward-biased diode with a given forward resistance is just another resistor in series.

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