14 V Cw = 5 pF Che = 12 pF Cw, = 8 pF Cbe = 40 pF Cee %3D %3D Cce = 8 pF %3D 5.6 k2 68 k2 0.47 µF 0.82 k2 B = 120 0.47 uF 3.3 k2 V, 10 ΚΩ 1.2 k2 20 µF Figure 1

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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For the network shown of Figure 1 below:
a. Determine re.
b. Find Avmid = VO/Vi
c. Calculate Zi.
d. Find AvSmid = VO/VS.
e. Determine fLS, fLC, and fLE.

14 V
Cw = 5 pF Chc = 12 pF
Cw = 8 pF Cbe = 40 pF
Cce = 8 pF
5.6 k2
68 k2
0.47 µF
0.82 k2
B = 120
0.47 µF
3.3 k2
10 k2
1.2 k2
20 µF
Figure 1
O
Transcribed Image Text:14 V Cw = 5 pF Chc = 12 pF Cw = 8 pF Cbe = 40 pF Cce = 8 pF 5.6 k2 68 k2 0.47 µF 0.82 k2 B = 120 0.47 µF 3.3 k2 10 k2 1.2 k2 20 µF Figure 1 O
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