EXAMPLE 12.4 Calculate the ac power delivered to the 8-N speaker for the circuit of Fig. 12.10. The circuit component values result in a de base current of 6 mA, and the input signal (V;) results in a peak base current swing of 4 mA. 4lc (mA) Vcc = 10 V 14 mA 400- 350 - 12 mA 300- R82 10 mA R1 250- 8 mA N1 = 3:1 N2 200- V, HE 150H 6 mA 100 4 mA Iz =6 mA 50 I-2 mA Due to V;: = 4 mA RE CE 10 15 20 25 V) "peak

EXAMPLE 12.4 Calculate the ac power delivered to the 8-N speaker for the circuit of Fig. 12.10. The circuit component values result in a de base current of 6 mA, and the input signal (V;) results in a peak base current swing of 4 mA. 4lc (mA) Vcc = 10 V 14 mA 400- 350 - 12 mA 300- R82 10 mA R1 250- 8 mA N1 = 3:1 N2 200- V, HE 150H 6 mA 100 4 mA Iz =6 mA 50 I-2 mA Due to V;: = 4 mA RE CE 10 15 20 25 V) "peak

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

Maximum power transfer theorem

A resistive load will consume maximum power from a network when the load resistance is equal to the resistance of the network as viewed from the open-circuit terminals, with all energy sources removed leaving only the internal resistances. Physicist Moritz von Jacobi introduced the maximum power transfer theorem in 1940. Therefore, this theorem is also known as Jacobi's law.

Question

EXAMPLE 12.4 Calculate the ac power delivered to the 8-N speaker for the circuit of Fig.
12.10. The circuit component values result in a de base current of 6 mA, and the input
signal (V;) results in a peak base current swing of 4 mA.
Alc (mA)
Vcc = 10 V
14 mA
400
350
12 mA
300
R¸ =8Q
10 mA
R1
250
71
8 mA
200
V; A
= 3:1
N2
150
6 mA
C
100
4 mA
Ig = 6 mA
R2
50
Ig =2 mA
II
Due to V;:
RE
CE
이
10
15
20
25
pcak = 4 mA

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