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Zener current: 15 V, 120 Ω, 360 Ω

Short answer

Without the Zener, VL=11.25V > 9V, so the Zener breaks down at 9V. IR = (15-9)/120 = 50mA, IL = 9/360 = 25mA. IZ = IR - IL = 25mA.

Find Iz, assuming Vz = 9 V. [Figure: Vin = 15 V source, series resistor R = 120 Ω, Zener diode Dz in parallel with RL = 360 Ω; output Vout.]

Given: Vin = 15 V, R = 120 Ω, RL = 360 Ω, VZ = 9 V.

Check: without the Zener, VL = 15 × 360/(120 + 360) = 11.25 V, which is greater than 9 V, so the Zener is in breakdown and Vout = 9 V.

  • IR = (15 − 9)/120 = 0.05 A = 50 mA
  • IL = 9/360 = 25 mA
  • IZ = IR − IL = 50 − 25

Answer: IZ = 25 mA

✓ Answer checked by Claude Opus 5.5 (high effort)

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