In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of I_z ?

Solution

Physics sir · 4 steps

Fix the Zener at 3 V, find the series current, then take away what the load draws.

1. Check breakdown. Without the Zener, the 2 kΩ load would get 10 × 2/3 = 6.67 V. That is more than 3 V, so the Zener does break down and holds V_(AB) = 3 V.
2. Series current. The 1 kΩ resistor has 10 - 3 = 7 V across it, so I = 7/1000 = 7 mA.
3. Load current. The load is in parallel with the Zener, so it also has 3 V across it: I_L = 3/2000 = 1.5 mA.

   I_z = I - I_L = 7 mA - 1.5 mA = 5.5 mA

4. So. The Zener current is 5.5 mA, option (c).

The idea to keep. In a Zener regulator, the Zener sets the voltage across itself and the load. Find the series current from the leftover voltage, then split it: I_z = I - I_L.

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