Download PSPICE Tutorial 2

JEE2300 PSpice Tutorial 2
Problem 1.
Use a Y-to- transformation to find (a) io; (b) i1; (c) i2; and (d) the power delivered by the ideal
current source in the circuit show below. Use PSpice to confirm your work.
320
20 
i1
50 
io
i2
240
1A
100 
600
Figure 1.
(a) Need to mark Y network to perform Y-to- transformation. That is done below:
320
a
1A
R1 = 20  i20
i50
i1
R2 = 50 
i240
i2
240
R3 =100 
c
Figure 2.
1
600
b
io
Now we perform Y-to- transformation between nodes a,b, and c:
R1 R2  R2 R3  R3 R1 1000  5000  2000 8000


 400
R1
20
20
R R  R2 R3  R3 R1 1000  5000  2000 8000
Rb  1 2


 160
R2
50
50
R R  R2 R3  R3 R1 1000  5000  2000 8000
Rc  1 2


 80
R3
100
100
Ra 
Now the circuit looks like this:
320
i240
1A
iab R = 80 
c
a
iac
240
Rb =
160 
c
Figure 3.
Using KVL from Figure 3 we can see that:
i240 240  iac 160 (1)
ibc 400  io 600
(2)
i1 320  iab 80
(3)
We can simplify this circuit doing the following:
Req1 = (240) || Rb = 96 and
i96 = i240 + iac
(4)
2
Ra =
400 
i1
b
ibc
io
600
Req2 = (600) || Ra = 240and
i64 = ibc + io
(5)
Req3 = (320) || Rc = 64and
i64 = iab + i1
(6)
New circuit looks like this:
a
i64 Req3 = 64 
b
i96
1A
Req1 =
96 
c
Figure 4.
From Figure 4 using KCL:
1A  i64  i96
Also using KVL for the Figure 4:
i64 64  240  i96 96
From the last two equations we get
i64  0.24 A
i96  0.76 A
Using (2) and (5) we get:
ibc  144mA
io  96mA
3
Req2 =
240 
(b)
From (3) and (6) we get:
iab  192mA
i1  48mA
(c)
From (1) and (4) we get
iac  456mA
i240  304mA
Using KCL from Figure 2 we have:
1A = i240 + i20 + i1
i20 = 648mA.
Also using KCL from Figure 2 we have:
i50 = io - i1
i50 = 48mA
,and using KCL again from Figure 2 we have:
i20 = i50 + i2
i2 = 600mA.
(d)
From Figure (2) we have that
P1A = - (1A)v240
P1A = - (1A)(240)(i240) = - (1A)(240)(304mA) = - 72.96W – power delivered by the 1A ideal current
source
P240 = (i240)2(240) = (304mA)2 (240) = 22.17984W
P20 = (i20)2(20) = (648mA)2 (20) = 8.39808W
P320= (i1)2 (320) = (48mA)2 (320) = 0.73728W
+
P100= (i2)2 (100) = (600mA)2 (100) = 36W
P50= (i50)2 (50) = (48mA)2 (50) = 0.1152W
P600= (io)2 (600) = (96mA)2 (600) = 5.5296W







72.96W – total power absorbed by the resistive network in
the circuit

4
Problem 2. Use the node-voltage method to find v1 and v2 in the circuit shown in Figure 5.
25 
1
+
2.4A v1
i25
2
i125
i250
250 
125
i375
+
375
v2
-
3.2A
3
Figure 5.
There are 3 essential nodes in the circuit from Figure 5. Hence n = 3 and we need (n-1) = 2 equations to
determine all voltages between essential nodes. In this circuit there are 2 voltages like that and we marked
them as v1 and v2. Essential node 3 is common node and we marked it like that. We also chose current
directions for each branch. Using node voltage method we can write two independent equations:
v
v v
Node 1: 2.4 A  1  2 1  0
125
25
v2
v
v v
 2  2 1 0
375 250
25
We have two equations with two unknown voltages v1 and v2. If we solve these two equations we get:
v1 = 25V
v2 = 90V.
Node 2:  3.2 A 
Problem 3.
(a) Use the mesh-current method to find the branch currents ia,ib, and ic in the circuit Figure 6.
(b) Repeat (a) if the polarity of the 64V source is reversed.
ia
3
ic
4
ib
+ 40V
-
I
45
2
II
1.5
Figure 6.
5
64V
+
(a)
Number of essential nodes is ne = 2.
Number of essential branches with unknown currents is be = 3.
We can write be – (ne -1) = 3 – 1 = 2 independent mesh-current method equations.
Mesh I: (-40V) + (3)iI + (45)(iI – iII) + (2)iI = 0
Mesh II: (-64V) + (4)iII + (45)(iII – iI) + (1.5)iII = 0
____________________________________________________________________________
We have two equations with two unknown mesh currents iI and iII and when we solve them:
iI = 9.8A
iII = 10A.
From here we have that
ia = iI = 9.8A
ib = iI – iII = -0.2A
ic = - iII = -10A.
(b)
If 64V independent voltage source has reversed polarity then we have the following mesh equations:
Mesh I: (-40V) + (3)iI + (45)(iI – iII) + (2)iI = 0
Mesh II: (64V) + (4)iII + (45)(iII – iI) + (1.5)iII = 0
_____________________________________________________________________________
We have two equations with two unknown mesh currents iI and iII and when we solve them:
iI = -1.72A
iII = - 2.8A.
From here we have that
ia = iI = -1.72A
ib = iI – iII = 1.08A
ic = - iII = 2.8A.
Problem 4.
(a) Use the mesh-current method to find how much power the 12A current source delivers to the circuit
shown in Figure 7.
(b) Find the total power delivered to the circuit.
(c) Check your calculations by showing that the total power developed in the circuit equals the total
power dissipated.
6
10
+ 600V
-
8
2
I
40
14
+ 400V
-
II
2
1
4
3
+
v14
-
12A
Figure 7.
(a)
Number of essential nodes is ne = 4.
Number of essential branches with unknown currents is be = 5.
We can write be – (ne -1) = 5 – 3 = 2 independent mesh-current method equations.
Mesh I: (-600V) + (10)iI + (40)(iI – iII) + (14)(iI +12A) = 0
Mesh II: (400V) + (8)iII + (40)(iII – iI) + (2)(iII +12A) = 0
__________________________________________________________________________
We have two equations with two unknown mesh currents iI and iII and when we solve them:
iI = 2.9A
iII = -6.16A.
Therefore the power delivered by the 12A current source is:
P12A = (v14)(12A)
P12A = (v13+v34)(12A)
P12A = (v13+v34)(12A)
P12A = (v13+v34)(12A)
P12A = [-(14)(12A+iI)-(2)(12A+iII)](12A)
P12A = [-(14)(12A+2.9A)-(2)(12A-6.16A)](12A)
P12A = [-(14)(14.9A)-(2)(5.84A)](12A)
P12A = - 2643.36W.
7
(b)
P12A =
- 2643.36W
P600V = - (600V)iI = - 1740W
+
P400V = (400V)iII = - 2464W
________________________________________________________
PTOTAL DEVELOPED = - 6847.36W
(c)
P10 = (iI)2(10) = (2.9A)2(10)
= 84.1W
2
2
P8= (iII) (8) = (-6.16A) (8)
= 303.5648W
2
2
P2= (iII+12A) (2) = (-6.16A +12A) (2) = 68.2112W
+
P40= (iI – iII)2(40) = (2.9A + 6.16A)2(40) = 3283.344W
P14= (iI+12A)2(14) = (2.9A + 12A)2(14) = 3108.14W
________________________________________________________
PTOTAL ABSORBED
= 6847.36W
Problem 5.
(a) Use the principle of superposition to find the voltage v in the circuit shown in Figure 8.
(b) Find the power dissipated in the 20 resistor.
6A
5
1
8
2
3
+
+
-
75V
v
20
4
Figure 8.
8
12
(a)
First we analyze circuit without 6A independent current source. The new circuit will look like this:

5
1
i 1'
2
i 3'
8
3
i 2'
+
-
75V
20
12
4
Figure 9.
Using KCL at node 2 from Figure 9 we have:
i1 ’ = i2 ’ + i3 ’
(1)
Using KVL from Figure 9 we have:
(20)i2’ = (8+12)i3’
i2 ’ = i3 ’
(2)
Also using KVL we have:
75V = (5)i1’ + (20)i2’
(3)
Using (1), (2) and (3) we get:
i1’ = 5A
i2’ = 2.5A
i3’ = 2.5A.
Second we analyze circuit without 75V independent voltage source. The new circuit will look like this:
9
6A
5
1
i 1"
2 i4
i 3"
20
12
Figure 10.
Using KVL from Figure 10 we can see that
(20)i2” = - (5i1”
(4)
Using KCL from Figure 10 we can see that at node 1 we have
6 + i2 ” + i 3 ” = i1 ”
(5)
Combining (4) and (5) we have
6 + 5i2” + i3” = 0
(6)
Using KCL from Figure 10 at node 3
6 - i4 + i3” = 0 (7)
Using KVL from Figure 10 we have that
(20)i2” = (8i4 + (12 i3”
(8)
Combining (7) and (8) we get
20i2” - 20 i3” = 48
3
i 2"
4
i1” = - 4i2”
8
(9)
From (6) and (9) we have:
10
6 + 5i2” + i3” = 0
20i2” - 20 i3” = 48
________________
5i2” + i3” = -6
20i2” - 20 i3” = 48
________________
From here
i2” = -0.6A
If we add both i2’ and i2” we get total current i2 determined by the superposition method:
i2 = i2’ + i2” = 2.5 -0.6 = 1.9A.
And voltage v is:
v = (20)i2 = 38V.
(b)
The power dissipated in the 20 resistor is
v 2 38


 72.2W
20
20
2
P20
Problem 6. The Inverting-Amplifier Circuit Design
The Inverting-Amplifier Circuit Design is shown in Figure 11.
10k
+15V
1k
+
-
vs
+
-15V
Figure 11.
11
+
vo
-
 10k 
vo  
v s  10v s
 1k 
(a) If vs = 1Vdc then vo = -10Vdc (the op amp is operating in linear region).
(b) If vs = 2Vdc then vo = -15Vdc (the op amp is operating in negative saturation region).
For the Inverting-Amplifier Circuit Design shown in Figure 12,
10k
+15V
1k
+
vs
+
-15V
+
vo
-
Figure 12.
10
k



vo  
v s  10v s
 1k 
(c) If vs = 1Vdc then vo = 10Vdc (the op amp is operating in linear region).
(d) If vs = 2Vdc then vo = 15Vdc (the op amp is operating in positive saturation region).
Problem 7. The Summing-Amplifier Circuit Design
The Summing-Amplifier Circuit Design is shown in Figure 13.
Rf
Ra
+ v
a
-
+15V
Rb
Rc
+ v
- b +
v
- c
+
-15V
Figure 13.
12
+
vo
-
Rf
Rf 
 Rf
vo  
va 
vb 
vc 
R
R
R
b
c
 a

(a) If Rf = Ra = Rb = Rc = 1k, va = 1V, vb = 2V, and vc = 3V, then vo = -6V (the op amp is operating in
linear region).
(b) If Rf = Ra = Rb = Rc = 1k, va = 5V, vb = 5V, and vc = 7V, then vo = -15V (the op amp is operating in
negative saturation region).
Problem 8. The Non-Inverting-Amplifier Circuit Design
Rf
Rs
+10V
Rg
+
-10V
+
-
+
vo
-
vg
Figure 14.
For the Non-Inverting-Amplifier Circuit Design shown in Figure 14
 Rs  R f
vo  
 Rs

v g

(a) If vg = 2Vdc, Rs = Rg = 1k, and Rf = 2k, then vo = 6Vdc (the op amp is operating in linear region).
(b) If vg = 5Vdc, Rs = Rg = 1k, and Rf = 2k, then vo = 10Vdc (the op amp is operating in positive
saturation region).
13