The value of power dissipated across the zener diode (V z = 15V) connected in the circuit as shown in the figure is x × 10⁻¹ watt.
The value of x, to the nearest integer, is __________.

Solution

Physics sir · 4 steps

Fix 15 V across the load, find the series current, and the zener takes whatever the load does not.

1. Is it breaking down. Remove the zener and divide the voltage: 22 × 90/(35 + 90) = 15.84 V. This is more than V_z = 15 V, so the zener is in breakdown and holds the load at exactly 15 V.
2. Series current. R_S drops the rest, 22 - 15 = 7 V, so I = 7/35 = 0.2 A flows in from the source.
3. Split the current. The load draws I_L = 15/90 = 1/6 A ≈ 0.167 A. The zener carries the remainder, I_Z = 0.2 - 1/6 = 1/30 A.

   P_Z = V_Z I_Z = 15 × 1/30 = 0.5 W = 5 × 10⁻¹ W

4. So. The zener dissipates 5 × 10⁻¹ W, so x = 5.

The common slip: 30. Multiplying V_Z by the full series current, 15 × 0.2 = 3 W, gives 30. The zener carries only the part of that current the load does not take.

The idea to keep. In a zener regulator, first check that the zener is in breakdown. Then I_S = (Vᵢₙ - V_Z)/R_S, I_L = V_Z/R_L, and the zener carries I_S - I_L.

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