Node 3
2
4 + Vz V+
Exercise 1 (Example E5.8 from the book1)
-
Find 𝑉 in Fig. E5.8 using Thévenin’s theorem.
-
M
6-363
12
+
3Vz
-
=
V +zVz
↓
3 6+2
.
=
5 6kr
.
&
Exercise 2 (Learning Assignment E5.9 from the book)
Find 𝐼 in Fig. E5.9 using Norton’s theorem.
=
10 Fr
t
R S 8th
Er
I
↓
=
Ri + R
=
=
11 875
=
·
/
.
-
V
V 3(19 2136 =
=
.
Vin 21 6 + 4
=
=
.
V
256V
(R)
=
Vin
=
25 6
.
(g
Mon)
+
At
↓
.
9VB
=
92
↓
A
XB
=
10 ZZV
.
Mos
I
↓
Vio
1
24
Vi 19 2V
V
Sa
.
Kn
=
1
Vi + 2(3V 36) 24
=
d
12
=
↓
Ra =3.
3V+ h = 0
=
1
1
A
Basic Engineering Circuit Analysis 12th Edition by J. David Irwin, R. Mark Nelms, Wiley 2021.
In
+ 4
↓
-
Rim
+
-
SVz-V-3un
Vz
↓
-
UV
=
=
34) 12
-
-
,
G
-
Y + Sk
-
NodezK
V y + Vz
V+h
V
↓
Nodel
y
-
-
SH-2MA
=
↓
I_
0.45S-1 703=1 258m
.
.
Fr
=
1
.
258)
2