---
title: "Class B power and efficiency, 14 V peak, 16 Ω"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:class-b-numeric-14v-16ohm"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-numeric-14v-16ohm
site: BCT Engineering
---

# Class B power and efficiency, 14 V peak, 16 Ω

Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers

- Asked in: 2074 Chaitra, 4 marks
- Video: [Class B Power Amplifier - High Power Amplifiers - Applied Electronics](https://www.youtube.com/watch?v=T9HvHYsbC3g) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-numeric-14v-16ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

For a class B amplifier providing a 14V peak signal to 16 Ω load and a power supply of Vcc = 24V, determine input power, output power and circuit efficiency.

## Short answer

Output power Po = Vp²/(2RL) = 14²/(2×16) = 6.125 W; input power Pi = 2 Vcc Vp/(π RL) = 2×24×14/(π×16) ≈ 13.37 W; efficiency η = (π/4)(Vp/Vcc) = 45.8%.

## Answer

**Given:** Class B push-pull, peak output voltage Vp = 14 V, load RL = 16 Ω, VCC = 24 V.

#### Formulas
- Peak load current: Im = Vp/RL
- Output power: Po = Vp²/(2RL)
- Input (dc) power: Pi = VCC · Idc = 2 VCC Vp/(π RL), because each supply gives an average current Im/π per half-cycle, and the two together give 2Im/π
- Efficiency: η = Po/Pi × 100 %

#### Calculation
- Im = 14/16 = 0.875 A
- **Output power** Po = 14² / (2 × 16) = **6.125 W**
- **Input power** Pi = 2 × 24 × 14 / (π × 16) = **13.37 W**
- **Efficiency** η = 6.125 / 13.37 = **45.8 %**

Check with η = (π/4)(Vp/VCC) = 0.7854 × 14/24 = 45.8 %.

The power lost as heat in the two transistors is Pi − Po = 7.24 W.

**Answer: Pi = 13.37 W, Po = 6.125 W, η = 45.8 %.**

## Check yourself

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

1. What is the output power for Vp = 14 V, RL = 16 Ω?
   A. 12.25 W
   B. 6.12 W
   C. 3.06 W
   D. 0.875 W
   Answer: B. Po = Vp^2/(2RL) = 196/32 = 6.12 W. (Key point: Output power)

2. What is the input power with VCC = 24 V?
   A. 21.0 W
   B. 6.68 W
   C. 13.37 W
   D. 9.45 W
   Answer: C. Pi = 2 VCC Vp/(π RL) = 2 x 24 x 14/(π x 16) = 13.37 W. (Key point: Input power)

3. What is the efficiency?
   A. 45.8%
   B. 58.3%
   C. 78.5%
   D. 32.6%
   Answer: A. 6.12/13.37 = 45.8% = (π/4)(14/24). (Key point: Efficiency)

4. How much power is dissipated as heat in the two transistors?
   A. 6.12 W
   B. 13.37 W
   C. 19.5 W
   D. 7.24 W
   Answer: D. Pi - Po = 13.37 - 6.12 = 7.24 W. (Key point: Power lost)

## Close variants set in other exams

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