Methods of Analysis

AEE1000 / MEC1151 - Electrical Circuits and Analysis
Dr Hong Fan, University of Glasgow, 2022

Tutorial – Topic 3 : Methods of Analysis
apply KCL @ node 1
,

V, -0 V, -0 ✓' - ✓'
+ + 0
+ 12 =
10 g
[Nodal Analysis] 2


V1 + IV t 5V , 5- V2 + 120 = O
3.1 For the circuit in Figure 3.1, obtain v1 and v2.
-

,

'
R , // R , = SV, - 5V , + 120=0 -
(1)
E. +
÷ 2
apply KCL @ node ,
= 2.55L
124
V1 V2 V2 - °
MM
-


^ ~ 12 + 6 + =




;÷"^
2 4
> > <
"A " ~ ~
IV , -
2 V2 + 72 =
V2
" n R, ,, ,,
" in
.

a" -
•v. + → = ◦ -




☐
6A (2) × 4 ;
2.5h


×
8V , -
12 V2 +
288 = 0 - (3)

✓ :O
! (3) -
( l) :
Figure 3.1
-
7- V2 + 168 = O

V, = Or
V2 = 24 V

Using 0hm 'S law ,
sub V, = 24 into (e) :


V2 =
12124 SV , 120 = 0
-
5 (24 ) +


I 12 ✗ 2
V = O V
,
# #
=
24W

3.2 Use nodal analysis to obtain vo in the circuit of Figure 3.2.




apply KCL , 0


I, t
12 7
13 I 0
Figure 3.2


V0 -
12 V0 - O
V0 - 10 -
O
+ + = 0

4 6 2




3Vo -
36 + IVO + GVO -
60 = 0


It V0 = 96

i.
V0 ≈ 8.727 V 1
#