In the given circuit, the value of current I L will be ____________ mA. (When R L = 1kΩ)

Solution

Physics sir · 4 steps

First check whether the Zener actually breaks down. Once it does, the load sees exactly 5 V.

1. Remove the Zener. Imagine the Zener is not there. The 800 Ω and 1 kΩ form a divider, so the load would get 10 × 1000/1800 = 5.56 V.
2. Breakdown check. The Zener is reverse biased: its cathode is on the positive node. The load would get 5.56 V, which is more than the 5 V Zener voltage. So the Zener breaks down and holds the load voltage at V_L = 5 V.

   I_L = V_Z/R_L = (5 V)/(1000 Ω) = 5 × 10⁻³ A

3. Consistency check. The series resistor carries (10-5)/800 = 6.25 mA. Then the Zener carries 6.25 - 5 = 1.25 mA. That is positive, so the Zener really is in breakdown.
4. So. I_L = 5 mA.

The common slip: 5.56. Ignoring the Zener and taking I_L = 10/1800 A gives about 5.56 mA, not 5. Once the Zener conducts, the load voltage is fixed at 5 V.

The idea to keep. For a Zener regulator, find the load voltage with the Zener removed. If it exceeds V_Z, the load voltage is pinned at V_Z and I_L = V_Z/R_L. If it does not, the Zener is off.

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