A Zener diode of breakdown voltage 10 V is used as a voltage regulator as shown in the figure. The current through the Zener diode is :

Solution

Physics sir · 4 steps

The Zener fixes 10 V across the load, so find the series current and the load current, then subtract.

1. Is it in breakdown. Without the Zener, the 500 Ω load would get 20 × 500/700 ≈ 14.3 V. That is more than 10 V, so the Zener breaks down and holds the load at exactly 10 V.
2. Series current. The 200 Ω resistor drops the remaining 20 - 10 = 10 V. So I = 10/200 = 0.05 A = 50 mA flows from the battery.
3. Load current. The load has 10 V across it, so I_L = 10/500 = 0.02 A = 20 mA.

   I_Z = I - I_L = 50 mA - 20 mA = 30 mA

4. So. The Zener carries the leftover current, 30 mA, which is option (b).

The idea to keep. In a Zener regulator, first check that the Zener is in breakdown. Then the load voltage equals V_Z, and the Zener current is the series current minus the load current.

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