---
title: "Zener network with two load values"
subject: "Basic Electrical and Electronics Engineering"
chapter: 3
key: "semester_1/basic_electrical_and_electronics_engineering:zener-regulator-numerical-rl"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/zener-regulator-numerical-rl
site: BCT Engineering
---

# Zener network with two load values

Basic Electrical and Electronics Engineering · Chapter 3: Introduction to Electronics Engineering

- Asked in: 2081 Baishakh (new course), 6 marks
- Video: [Analysis of Circuit Containing Zener Diode Part 2](https://www.youtube.com/watch?v=OyQAIxB8WRI) — EE-Vibes (Electrical Engineering Lessons) 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/zener-regulator-numerical-rl
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-3

## Question

For the Zener diode network shown below, determine $V_L$, $V_R$, $I_Z$ and $P_Z$ for $R_L = 1.2\ k\Omega$ and $R_L = 3\ k\Omega$. [Figure: $V_i = 16$ V source in series with $R = 1\ k\Omega$; Zener diode with $V_Z = 10$ V and $P_{ZM} = 30$ mW in parallel with load $R_L$; $V_R$ across R, $I_Z$ through the Zener.]

## Short answer

For RL=1.2kΩ, VL=8.73V (<10V), Zener off, so VR=7.27V, IZ=0, PZ=0. For RL=3kΩ, VL=10V, Zener on, VR=6V, IZ=2.667mA, PZ=26.67mW.

## Answer

**Given:** $V_i=16$ V, $R=1\ \text{k}\Omega$, $V_Z=10$ V, $P_{ZM}=30$ mW.

**Method.** First assume the Zener is OFF (removed) and find the voltage across the load. If it is at least $V_Z$, the Zener is ON and $V_L=V_Z$.

#### Case 1: $R_L=1.2\ \text{k}\Omega$

$$
V_L=\frac{R_LV_i}{R+R_L}=\frac{1.2\times16}{1+1.2}=8.727\ \text{V}
$$

Since $8.727\ \text{V}<10\ \text{V}$, the Zener is **OFF** (not in breakdown).

$$
V_L=8.73\ \text{V},\quad V_R=16-8.727=7.27\ \text{V},\quad I_Z=0,\quad P_Z=0
$$

#### Case 2: $R_L=3\ \text{k}\Omega$

$$
V_L=\frac{3\times16}{1+3}=12\ \text{V}>10\ \text{V}
$$

The Zener is **ON**, so $V_L=V_Z=10$ V.

$$
\begin{aligned} V_R&=16-10=6\ \text{V}\\ I_R&=\frac{6}{1\ \text{k}\Omega}=6\ \text{mA}\\ I_L&=\frac{10}{3\ \text{k}\Omega}=3.333\ \text{mA}\\ I_Z&=I_R-I_L=6-3.333=2.667\ \text{mA}\\ P_Z&=V_ZI_Z=10\times2.667=26.67\ \text{mW} \end{aligned}
$$

$P_Z=26.67\ \text{mW}<P_{ZM}=30$ mW, so the Zener is safe.

| $R_L$ | $V_L$ | $V_R$ | $I_Z$ | $P_Z$ |
|---|---|---|---|---|
| 1.2 kΩ | 8.73 V | 7.27 V | 0 | 0 |
| 3 kΩ | 10 V | 6 V | 2.667 mA | 26.67 mW |

## Check yourself

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

1. What is the first step in analysing the Zener network?
   A. Assume the Zener is ON and VL = 10 V
   B. Remove the Zener and find the voltage across the load; compare with VZ
   C. Short the load
   D. Compute Iz first
   Answer: B. If the open-circuit load voltage is at least VZ the Zener is on; otherwise it is off. (Key point: Method)

2. For RL = 1.2 kΩ, what happens?
   A. Zener ON, VL = 10 V
   B. Zener ON with Iz = 6 mA
   C. Zener OFF, VL = 8.73 V, Iz = 0
   D. Zener destroyed
   Answer: C. VL = 16 x 1.2/2.2 = 8.727 V < 10 V, so no breakdown. (Key point: Case RL = 1.2 kΩ)

3. For RL = 3 kΩ, what is VL and the logic?
   A. VL = 12 V, Zener off
   B. VL = 16 V
   C. VL = 6 V
   D. Open-circuit VL = 12 V > 10 V so Zener ON and VL = VZ = 10 V
   Answer: D. The Zener conducts and clamps the load at 10 V, so VR = 6 V. (Key point: Case RL = 3 kΩ)

4. For RL = 3 kΩ, what are Iz and Pz, and is it safe?
   A. Iz = 2.667 mA, Pz = 26.67 mW < 30 mW, safe
   B. Iz = 6 mA, Pz = 60 mW, unsafe
   C. Iz = 3.333 mA, Pz = 33.3 mW, unsafe
   D. Iz = 0, Pz = 0
   Answer: A. IR = 6 mA, IL = 3.333 mA, so Iz = IR - IL = 2.667 mA and Pz = 10 x 2.667 mW. (Key point: Zener current and power)

## Related in Basic Electrical and Electronics Engineering

- [Zener circuit output voltage and current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Azener-circuit-120v-5k) — Basic Electrical and Electronics Engineering, chapter 3
- [Zener diode as voltage regulator](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Azener-voltage-regulator) — Basic Electrical and Electronics Engineering, chapter 3
- [Mesh analysis with 3 A source](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Amesh-analysis-3a-source) — Basic Electrical and Electronics Engineering, chapter 1
- [Series AC circuit element parameters](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aseries-ac-circuit-parameters) — Basic Electrical and Electronics Engineering, chapter 1
- [Parallel RL and RC branch currents, power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aparallel-impedance-total-current) — Basic Electrical and Electronics Engineering, chapter 1
- [Two-wattmeter method star load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Atwo-wattmeter-star-load) — Basic Electrical and Electronics Engineering, chapter 1

## Related in other subjects

- [Zener current for two load values](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-current-rl-numeric) — Electronics Devices and Circuit, chapter 6
- [Zener input range, 91 Ω, 8 V, 400 mW](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-range-vi-91ohm) — Electronics Devices and Circuit, chapter 6
- [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: range of load for VL = 5 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-load-range-5v) — Electronics Devices and Circuit, chapter 6