For the power amplifier of Figure 4, calculate:1.Po(ac);2.% n; and3.Power dissipated by both outputtransistors.+Vcc (+40 V)100 2100 µFV,-18 V, rmsVLR3R100 2C282Q2100 pFR2100 2-VEE (-40 V)

Given data RMS value of voltage is VL(rms) = 18 V Line resistance is RL = 8 ohm Calculating P0…

For the power amplifier of Figure 4, calculate: 1. Po(ac); % n; and Power dissipated by both output 2. 3. transistors. +Vcc (+40 V) R1 100 2 100 µF V, - 18 V, rms R3 100 2 82 Q2 100 µF R2 100 2 -VEE (-40 V)

For the power amplifier of Figure 4, calculate: 1. Po(ac); % n; and Power dissipated by both output 2. 3. transistors. +Vcc (+40 V) R1 100 2 100 µF V, - 18 V, rms R3 100 2 82 Q2 100 µF R2 100 2 -VEE (-40 V)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Current Division Method

The generalized current divider formula is given by the following relation below,

Question

For the power amplifier of Figure 4, calculate:
1.
Po(ac);
2.
% n; and
3.
Power dissipated by both output
transistors.
+Vcc (+40 V)
100 2
100 µF
V,
-18 V, rms
VL
R3
R
100 2
C2
82
Q2
100 pF
R2
100 2
-VEE (-40 V)

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