A potential divider circuit is connected with a dc source of 20 V, a light emitting diode of glow in voltage 1.8 V and a zener diode of breakdown voltage of 3.2 V. The length (PR) of the resistive wire is 20 cm. The minimum length of PQ to just glow the LED is _________ cm.

Solution

Physics sir · 4 steps

The LED and the reverse-biased zener sit across PQ, so PQ must supply both their voltages together.

1. Read the branch. The battery's positive plate is on top, so P is at the higher potential. Current in the side branch runs from P through the LED, which is forward biased, and then down through the zener, which is reverse biased. The branch then returns to the wire at Q.
2. Voltage needed. The LED glows only above 1.8 V, and the zener conducts in reverse only once it breaks down at 3.2 V. Both are in series, so the voltage across PQ must be at least 1.8 + 3.2 = 5 V.
3. Potential gradient. The full 20 V is spread uniformly over the 20 cm wire PR, so the gradient is 1 V cm⁻¹.

   l_(PQ) = V_(PQ)/gradient = (5 V)/(1 V cm⁻¹) = 5 cm

4. So. The minimum length of PQ that just makes the LED glow is 5 cm.

The common slip: 1.8. Counting only the LED's glow voltage and leaving out the zener's breakdown voltage gives 1.8 cm, not 5.

The idea to keep. In a divider, the tapped length sets the voltage. Add up every threshold in the branch, including a zener's breakdown voltage when it is reverse biased, then divide by the gradient.

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