---
title: "Zener current: 16 V, 1 kΩ, RL 3 kΩ"
subject: "Basic Electronics Engineering"
chapter: 2
key: "semester_2/basic_electronics_engineering:zener-current-16v-1k"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-current-16v-1k
site: BCT Engineering
---

# Zener current: 16 V, 1 kΩ, RL 3 kΩ

Basic Electronics Engineering · Chapter 2: Diodes

- Asked in: 2071 Magh, 5 marks
- Video: [Special Diode Circuits | Zener Diode (Fixed Vi and RL) | Solved Problem](https://www.youtube.com/watch?v=n5EbPnk6yXo) — Electronics Insight
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/zener-current-16v-1k
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2

## Question

Find the zener current from the given zener diode network when RL = 3 kΩ and Vo = 10 V. [Figure: Vi = 16 V source, series resistor R = 1 kΩ, Zener diode in parallel with RL; Vo = 10 V.]

## Short answer

The unloaded divider gives 12V > VZ=10V, so the Zener conducts and clamps Vo=10V. IR = (16-10)/1k = 6mA, IL = 10/3k ≈ 3.33mA. IZ = IR - IL = 2.67mA.

## Answer

**Zener regulator** with Vi = 16 V, R = 1 kΩ, RL = 3 kΩ and Vo = 10 V.

Check whether the Zener is ON: with the Zener removed, the voltage across RL would be 16 × 3/(1 + 3) = 12 V, which is above 10 V, so the Zener breaks down and holds Vo = 10 V.

- Current through R: IR = (Vi − Vo)/R = (16 − 10)/1 kΩ = 6 mA
- Load current: IL = Vo/RL = 10/3 kΩ = 3.33 mA
- By KCL at the output node: IZ = IR − IL = 6 − 3.33

```text
 16 V o--[1k]--+-------+--o Vo=10 V
               |       |
               Z      RL 3k
               |       |
 0 V  o--------+-------+
```

**Answer: IZ = 2.67 mA**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is it decided that the Zener is ON for Vi = 16 V, R = 1 kΩ, RL = 3 kΩ and VZ = 10 V?
   A. With the Zener removed, the voltage across RL is 12 V, above 10 V
   B. Because Vi is greater than 10 V
   C. Because RL is greater than R
   D. Because the current is 6 mA
   Answer: A. Divider gives 16 x 3/4 = 12 V > 10 V, so the Zener breaks down. (Key point: Is the Zener on?)

2. What is the current through R?
   A. 10 mA
   B. 6 mA
   C. 3.33 mA
   D. 16 mA
   Answer: B. IR = (16 - 10)/1 kΩ = 6 mA. (Key point: Series current)

3. What is the load current IL?
   A. 3.33 mA
   B. 6 mA
   C. 2.67 mA
   D. 10 mA
   Answer: A. IL = Vo/RL = 10 V/3 kΩ = 3.33 mA. (Key point: Load current)

4. Which relation gives the Zener current and what is its value?
   A. IZ = IR + IL = 9.33 mA
   B. IZ = IL - IR
   C. IZ = IR - IL = 2.67 mA
   D. IZ = Vo/R = 10 mA
   Answer: C. By KCL at the output node, IZ = IR - IL = 6 - 3.33 = 2.67 mA. (Key point: Zener current)

## Close variants set in other exams

- [Zener current: 18 V, 100 Ω, RL 1.2 kΩ](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Azener-current-18v-100) — Basic Electronics Engineering, chapter 2, asked 4×

## Related in Basic Electronics Engineering

- [Zener network: VL, IL, Iz, IR](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Azener-regulator-220-ohm) — Basic Electronics Engineering, chapter 2
- [Zener current: 15 V, 120 Ω, 360 Ω](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Azener-current-15v-120) — Basic Electronics Engineering, chapter 2
- [Thevenin: 4 kΩ resistor, 12 V, 10 mA](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-12v-10ma) — Basic Electronics Engineering, chapter 1, asked 2×
- [Thevenin: current through 10 Ω](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-10v-5a) — Basic Electronics Engineering, chapter 1

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