Circuit Analysis Exercise 3
2012/03/08
Exercise 3 (Resistive Network Analysis)
Problem 1. (Hambley 2.49)
Problem 2. (Hambley 2.52)
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Circuit Analysis Exercise 3
Problem 3. (Hambley 2.59)
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Problem 4. (Hambley 2.68)
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Circuit Analysis Exercise 3
Problem 5. (Hambley 2.75)
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Circuit Analysis Exercise 3
Problem 6. (Hambley 2.83)
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Circuit Analysis Exercise 3
Problem 7. (Hambley 2.84)
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Problem 8. (Hambley 2.88)
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Circuit Analysis Exercise 3
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Problem 9. (Hambley 2.94)
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Circuit Analysis Exercise 3
Problem 10. (Rizzoni 3.10)
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Using node voltage analysis in the circuit of Figure 1, find the three indicated node voltages. Let
I  0.2 A; R1  200; R2  75; R3  25; R4  50; R5  100; V  10V .
Figure 1
Solution:
Known quantities:
The current source value, the voltage source value and the resistance values for the circuit shown in
Figure P3.10.
Find:
The three node voltages indicated in Figure P3.10 using node voltage analysis.
Analysis:
At node 1:
v1 v1  v2

 0.2 A
200
75
At node 2:

v2 v1 v2 v2 v3
 
i  0
75
25
50
At node 3:

v  v2
v
i  3
 3 0
50
100
For the voltage source we have:
v 3 10  v 2
Solving the system, we obtain:
v1 14.24 V, v2  4.58V , v3  5.42 V and, finally, i 254 mA .


 11. (Rizzoni

Problem
3.20)

For the circuit of Figure P3.20, use mesh current analysis to find the matrices required to solve
the circuit, and solve for the unknown currents. Hint:You may find source transformations useful.
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Circuit Analysis Exercise 3
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Solution:
Circuit shown in Figure P3.20.
Find:
Mesh equation in matrix form and solve for currents.
Analysis:
after source transformation, we can have the equivalent
circuit shown in the right hand side. We can write down the
following matrix
4
 4  I1,2  12  3
6  4  4
 4
4 48
0   I 3    3 

0
2  4  I 4   5 
  4
Solve the equation, we can have
I1,2 1.2661A  I2
I3  0.5040A
I4 1.6774A
I1  2A
Problem 12. (Rizzoni 3.30)
Using mesh current analysis, find the current i in the circuit of FigureP3.30.
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Circuit Analysis Exercise 3
Solution:
2012/03/08
Known quantities:
The values of the resistors in the circuit
of Figure P3.30.
Find:
The current in the circuit of Figure P3.30
using mesh current analysis.
Analysis:
Since I is unknown, the problem will be solved in terms of this current.
For mesh #1, it is obvious that:
i1  I


 
  2 5 

  5 
1 1
1
For mesh #2: i11i21
 

 i3  
 0
For mesh #3:

 1 
 1 
1 1 1 
i1  i 2   i 3    0
 4 
 5 
4 3 5 
i 2  0.645I
 i 3  0.483I
Solving,
Then,
i i3 i2 and i0.483
I0.645
I 0.163
I

Problem
13. (Rizzoni
3.56)


Find the Thevenin equivalet resistance seen by the load resistor RL in the circuit of Figure P3.56.
Problem 3.56
Solution:
Known quantities:
Circuit shown in Figure P3.56.
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Circuit Analysis Exercise 3
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Find:
Thevenin equivalent circuit
Analysis:
To find RT, we need to make the current source an open circuit and
voltage sources short circuits, as follows:
Note that this circuit has only three nodes. Thus, we can re-draw
circuit as shown:
and combine the two parallel resistors to obtain:
Thus,
the
the
RT 50||(5033.3)||10023.81

Problem 14. (Rizzoni 3.58)
Find the Thevenin equivalent of the circuit connected to RL in the Figure 2, where
R1  10; R2  20; Rg  0.1, and R p  1;
Figure 2
Solution:
Known quantities:
Circuit shown in Figure P3.58.
Find:
Thevenin equivalent circuit
Analysis:
To find RT, we short circuit the source
Starting from the left side,
(1  0.1)||10 0.99 ,
(1 0.99 0.1)||201.893
Therefore, we have
RT  1.893 0.1  1 2.993Ω.
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Circuit Analysis Exercise 3
To find voc , we apply mesh analysis:
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Two resistors are omitted because no current flows through them and they, therefore, do not affect
vOC.
1 0.110 i1-10i2  15
1+20+0.1+10 i2 -10 i1=0 Solving for i2,
i2  0.612 A
we obtain,
vT  vOC  20 i2 12.24V
Problem 15. (Rizzoni 3.59)
The Wheatstone bridge circuit shown in Figure 3 is used in a number of practical applications.
One traditional use is in determining the value of an unknown resistor Rx. Find the value of the
voltage Vab=Va-Vb in terms of R, Rx, and Vs. If R  1k, Vs  12V and Vab=12mV, what is the value
of Rx?
Figure 3
Solution:
Known quantities:
Circuit shown in Figure P3.59.
Find:
Value of resistance Rx
Analysis:
a)
We have Va b  Va-Vb 
Rx
R
VS VS
R+ R
R+ Rx
Rx
1
Vab = VS VS
2
R + Rx
b)
For R  1 kW , Vs  12 V , Vab  12 mV,
0.012  6-
Rx
12
1000+Rx
R x  996 Ω
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Circuit Analysis Exercise 3
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Problem 16. (Rizzoni 3.43)
With reference to Figure 4, using superposition, determine the component of the current through
R3 that is due to Vs2. VS1  VS 2  450V ; R1  7; R2  5; R3  10
R4  R5  1
Figure 4
Solution:
Known quantities:
The values of the voltage sources and of the resistors in the circuit of Figure P3.43:
VS1  VS2  4 5 0V
R1  7 
R2  5
R3  1 0
R4  R5  1
Find:

The component of the current through R3 that is due to VS2, using superposition.
Analysis:
Suppress VS1 by replacing it with a short circuit. Redraw the circuit. A solution using equivalent
resistances looks reasonable. R1 and R4 are in parallel:
R1 4 
R1R4
71  0.875

R1  R4 71
R1 4 is in series with R3 :

R1 4 3 R1 4  R3  0.87 5 10  10 .87 5
Req  R5  R2 R1 4 3  R5 
OL:



IS 
R2 R1 4 3
510 .87 5  4.4 25
 1
R2  R1 4 3
5  10 .87 5
VS 2
450

 101.695 A
Req
4.425
CD: I R32 
I S R2
101.6955  32.03 A

R2  R1 4 3 510.875
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Circuit Analysis Exercise 3
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Problem 17. (Rizzoni 3.21)
In the circuit the Figure 5, assume the source voltage and source current and all resistances are
known.
a. Write the node equations required to determine the node voltages.
b. Write the matrix solution for each node voltage in terms of the known parameters.
Figure 5
Solution:
Known quantities:
Circuit of Figure P3.21 with voltage source,
VS , current source, IS, and all resistances.
Find:
a. The node equations required to determine the node voltages.
b. The matrix solution for each node voltage in terms of the known parameters.
Analysis:
a) Specify the nodes (e.g., A on the upper left corner of the circuit in Figure P3.10, and B on the
right corner). Choose one node as the reference or ground node. If possible, ground one of the
sources in the circuit. Note that this is possible here. When using KCL, assume all unknown current
flow out of the node. The direction of the current supplied by the current source is specified and
must flow into node A.
KCL:
b)

V  VS Va  Vb
I S  a

0
R2
R1
 1
 1 
1 
V
Va    Vb   I S  S
R2
R2 R1 
 R1 
KCL:
Vb  Va Vb  VS Vb  0


0
R1
R3
R4
 1 
 1
1
1  VS
Va   Vb  


 R1 
R1 R3 R4  R3
Matrix solution:

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Circuit Analysis Exercise 3
IS 
Va 

VS
R2
VS
R3
1
1

R1 R2
1

R1
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1
R1

1
1
1
V  1
1
1  VS  1 



I S  S  
   
R1 R3 R4
R2 R1 R3 R4  R3  R1 


 1
1
1  1
1
1   1  1 


 
 
   
R1
R1 R2 R1 R3 R4   R1  R1 
1
1
1


R1 R3 R4

1
1
V

IS  S
R1 R2
R2
 1
1
VS
1 VS   1 
V 

 
   I S  S 
R1
R3
R2 
R1 R2 R3   R1 
Vb 

 1




1
1
1
1
1
1
1
1  1 



 
 
   
R1 R2
R1
R1 R2 R1 R3 R4   R1  R1 
1
1
1
1



R1
R1 R3 R4
Notes:

1. The denominators are the same for both solutions.
2. The main diagonal of a matrix is the one that goes to the right and down.
3. The denominator matrix is the "conductance" matrix and has certain properties:
a) The elements on the main diagonal [i(row) = j(column)] include all the conductance connected
to node i=j.
b) The off-diagonal elements are all negative.
c) The off-diagonal elements are all symmetric, i.e., the i j-th element = j i-th element. This is true
only because there are no controlled (dependent) sources in this circuit.
d) The off-diagonal elements include all the conductance connected between node i [row] and
node j [column].
Problem 18. (Rizzoni 3.24)
Using KCL, perform node analysis on the circuit shown in Figure 6, and determine the voltage
across R4. Note that one source is a controlled voltage source! Let VS  5V ; AV  70;
R1  2.2k; R2  1.8k; R3  6.8k; R4  220
Figure 6
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Circuit Analysis Exercise 3
Solution:
2012/03/08
Known quantities:
Circuit shown in Figure P3.24
VS  5 V
R2 1.8 k
AV  70
R1  2.2 k
R3  6.8 k
R4  220
Find:
The voltage across R4 usingKCL and node voltage analysis.
Analysis:
Node analysis is not a method of choice because the dependent source is [1] a voltage source and [2]
a floating source. Both factors cause difficulties in a node analysis. A ground is specified. There are
three unknown node voltages, one of which is the voltage across R4. The dependent source will
introduce two additional unknowns, the current through the source and the controlling voltage
(across R1) that is not a node voltage. Therefore 5 equations are required:
1KCLV1  VS
R1

V1  V3 V1  V2
V V

 0 2KCL 2 1  I CS  0
R3
R2
R2
3KCLV3  V1  I CS  V3
R3
R4
 0 4KVL  VS  VR1  V1  0VR1  VS  V1
5KVL  V3  AV VR1  V2  0V2  V3  AV VR1  V3  AV VS  V1 
Substitute using Equation [5] into Equations [1], [2] and [3] and eliminate V2 (because it only
appears twice in these equations). Collect terms:
 1
 1
1
1 AV 
1 
V
V A
V1 



 V3 
 I C S0  S  S V
R
R
R
R
R
R
R
R2
 1
 3
3
2
2 
2 
1
 1 A 
 1 
V A
V1
 V  V3   I C S1   S V
R
R
R
R2
 2
 2 
2 
 1 
 1
1 
V1  V3  
 I C S1  0
 R3 
R3 R4 

1
1
1
1

 555.6 10 -6 -1

 147.1 10 -6 -1
3
R2 1.8 10
R3 6.8 10 3
1
1
1
1



 702.6 10 -6 -1
R3 R2 6.8 10 3 1.8 10 3
1
1
1
1
1 AV
1 70



 4.69 10 -3 -1


 39.44 10 -3 -1
R3 R4 6.8 10 3 0.22 10 3
R2 R2 1.8 10 3
1
1
1 AV
1
1
1 70






 40.05 10 -3 -1
3
3
R1 R3 R2 R2 2.2 10
6.8 10 1.8 10 3

VS AV (5)70

194.4mA
R2
1.8103
Solving, we have:

(5)70
VS VS AV
5



196.7mA
R1
R2
2.2103 1.8103
VR 4  V3  5.1 mV
Notes:
1. This solution was not difficult in terms of theory, but was terribly long and arithmetically
cumbersome. This was because the wrong method was used. There are only 2 mesh currents in the
circuit; the sources were voltage sources; therefore, a mesh analysis is the method of choice.
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Circuit Analysis Exercise 3
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2. In general, a node analysis will have fewer unknowns (because one node is the ground or
reference node) and will, in such cases, be preferable.
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