For the given circuit, the power across Zener diode is .............. mW.

Solution

Physics sir · 4 steps

Fix the Zener at 10 V, then split the current: series resistor feeds both the diode and the load.

1. Is it in breakdown. Without the Zener, the load would get 24 × 5/(1+5) = 20 V. That is above 10 V, so the Zener conducts and holds the load at 10 V.
2. Current in each branch. The 1 kΩ resistor drops 24 - 10 = 14 V, so it carries 14 mA. The load carries (10 V)/(5 kΩ) = 2 mA.

   I_z = Iₛ - I_L = 14 mA - 2 mA = 12 mA

3. Zener power. P_z = V_z I_z = 10 V × 12 mA = 120 mW.
4. So. The power dissipated in the Zener diode is 120 mW.

The common slip: 140. Taking the full series-resistor current of 14 mA as the Zener current, and forgetting the 2 mA that goes to the load, gives 140 mW.

The idea to keep. In a Zener regulator, the Zener and the load share the current through the series resistor. Find that total from the voltage drop, then subtract what the load takes.

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