18 V Cw, = 3 pF Cad = 4 pF Cw = 5 pF C= 6 pF C4 = 1 pF 3 ka 4,7 µF ov 1 ka Ipss = 6 mA Vp = -6 V,r= oe 2 0,1 µF 1 MQ 1,2 k2 : 10 uF +

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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For the image circuit, determine:

a) fhiand  fho

18 V
Cw;
= 3 pF Cad = 4 pF
Cw.
= 5 pF C= 6 pF
Cds = 1 pF
3 k2
4,7 µF
1 ka
Ipss = 6 mA
Vp = - 6 V, r, oe 2
0,1 µF
3,9 kM
V,
'1 MQ
1,2 k2
10 µF
Transcribed Image Text:18 V Cw; = 3 pF Cad = 4 pF Cw. = 5 pF C= 6 pF Cds = 1 pF 3 k2 4,7 µF 1 ka Ipss = 6 mA Vp = - 6 V, r, oe 2 0,1 µF 3,9 kM V, '1 MQ 1,2 k2 10 µF
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