JEE2300 PSpice Tutorial 2
Problem 1.
Use a Y-to-
transformation to find (a) i
o
; (b) i
1
; (c) i
2
; and (d) the power delivered by the ideal
current source in the circuit show below. Use PSpice to confirm your work.
1A
i
o
20
100
320
50
600
240
i
2
i
1
Figure 1.
(a) Need to mark Y network to perform Y-to-
transformation. That is done below:
1A
i
o
R
1
= 20
R
3
=100
320
R
2
= 50
600
240
i
2
i
1
a
b
c
i
20
i
50
i
240
Figure 2.
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Now we perform Y-to-
transformation between nodes a,b, and c:
400
20
8000
20
2000
5000
1000
1
1
3
3
2
2
1
R
R
R
R
R
R
R
R
a
160
50
8000
50
2000
5000
1000
2
1
3
3
2
2
1
R
R
R
R
R
R
R
R
b
80
100
8000
100
2000
5000
1000
3
1
3
3
2
2
1
R
R
R
R
R
R
R
R
c
Now the circuit looks like this:
1A
i
o
R
b
=
160
320
R
c
= 80
600
240
i
ac
i
1
a
b
c
R
a
=
400
i
bc
i
ab
i
240
Figure 3.
Using KVL from Figure 3 we can see that:
160
240
240
ac
i
i
(1)
600
400
o
bc
i
i
(2)
80
320
1
ab
i
i
(3)
We can simplify this circuit doing the following:
R
eq1
= (240
) || R
b
= 96
and
i
96
= i
240
+ i
ac
(4)
R
eq2
= (600
) || R
a
= 240
and
2

i
64
= i
bc
+ i
o
(5)
R
eq3
= (320
) || R
c
= 64
and
i
64
= i
ab
+ i
1
(6)
New circuit looks like this:
1A
R
eq1
=
96
R
eq3
= 64
a
b
c
R
eq2
=
240
i
96
i
64
Figure 4.
From Figure 4 using KCL:
96
64
1
i
i
A
Also using KVL for the Figure 4:
96
240
64
96
64
i
i
From the last two equations we get
A
i
24
.
0
64
A
i
76
.
0
96
Using (2) and (5) we get:
mA
i
bc
144
mA
i
o
96
(b)
From (3) and (6) we get:
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mA
i
ab
192
mA
i
48
1
(c)
From (1) and (4) we get
mA
i
ac
456
mA
i
304
240
Using KCL from Figure 2 we have:
1A = i
240
+ i
20
+ i
1
i
20
= 648mA.
Also using KCL from Figure 2 we have:
i
50
= i
o
- i
1
i
50
= 48mA
,and using KCL again from Figure 2 we have:
i
20
= i
50
+ i
2
i
2
= 600mA.
(d)
From Figure (2) we have that
P
1A
= - (1A)v
240
P
1A
= - (1A)(240
)(i
240
) = - (1A)(240
)(304mA) =
- 72.96W – power delivered by the 1A ideal current
source
P
240
= (i
240
)
2
(240
)
= (304mA)
2
(240
)
=
22.17984W
P
20
= (i
20
)
2
(20
)
= (648mA)
2
(20
)
=
8.39808W
P
320
= (i
1
)
2
(320
)
= (48mA)
2
(320
)
=
0.73728W
+
P
100
= (i
2
)
2
(100
)
= (600mA)
2
(100
)
=
36W
P
50
= (i
50
)
2
(50
)
= (48mA)
2
(50
)
=
0.1152W
P
600

- Fall '09