CHAPTER 17 Fourier Method of Waveform Analysis 420
4.10 MAXIMUM POWER TRANSFER THEOREM
At times it is desired to obtain the maximum power transfer from an active network to an external load resistorRL. Assuming that the network is linear, it can be reduced to an equivalent circuit as in Fig. 4-16. Then
I¼ V0 R0þRL and so the power absorbed by the load is
PL¼ V02RL
ðR0þRLÞ2¼V02
4R0 1 R0RL R0þRL
2
" #
It is seen thatPL attains its maximum value,V02=4R0, whenRL¼R0, in which case the power inR0is alsoV02=4R0. Consequently, when the power transferred is a maximum, the efficiency is 50 percent.
It is noted that the condition for maximum power transfer to the load is not the same as the condition for maximum power delivered by the source. The latter happens when RL¼0, in which case power delivered to the load is zero (i.e., at a minimum).
Solved Problems
4.1 Use branch currents in the network shown in Fig. 4-17 to find the current supplied by the 60-V source.
Fig. 4-15
Fig. 4-16
KVL and KCL give:
I2ð12Þ ¼I3ð6Þ ð10Þ I2ð12Þ ¼I4ð12Þ ð11Þ 60¼I1ð7Þ þI2ð12Þ ð12Þ
I1¼I2þI3þI4 ð13Þ
Substituting (10) and (11) in (13),
I1¼I2þ2I2þI2¼4I2 ð14Þ Now (14) is substituted in (12):
60¼I1ð7Þ þ14I1ð12Þ ¼10I1 or I1¼6 A
4.2 Solve Problem 4.1 by the mesh current method.
Applying KVL to each mesh (see Fig. 4-18) results in 60¼7I1þ12ðI1I2Þ
0¼12ðI2I1Þ þ6ðI2I3Þ 0¼6ðI3I2Þ þ12I3 Rearranging terms and putting the equations in matrix form,
19I112I2 ¼60 12I1þ18I2 6I3¼ 0 6I2 þ18I3¼ 0
or
19 12 0
12 18 6
0 6 18
2 4
3 5 I1
I2 I3 2 4
3 5¼
60 0 0 2 4
3 5
Using Cramer’s rule to findI1,
I1¼
60 12 0
0 18 6
0 6 18
19 12 0
12 18 6
0 6 18
¼17 2802880¼6 A Fig. 4-17
Fig. 4-18
4.3 Solve the network of Problems 4.1 and 4.2 by the node voltage method. See Fig. 4-19.
With two principal nodes, only one equation is necessary.
V160 7 þV1
12þV1 6 þV1
12¼0 from whichV1¼18 V. Then,
I1¼60V1 7 ¼6 A
4.4 In Problem 4.2, obtainRinput;1 and use it to calculateI1.
Rinput;1¼ R
11¼ 2880
18 6
6 18
¼2880 288 ¼10
I1¼ 60 Rinput;1¼60
10¼6 A Then
4.5 ObtainRtransfer;12 andRtransfer;13 for the network of Problem 4.2 and use them to calculateI2and I3.
The cofactor of the 1,2-element inRmust include a negative sign:
12¼ ð1Þ1þ2 12 6 0 18
¼216 Rtransfer;12¼R 12¼2880
216 ¼13:33 Then,I2¼60=13:33¼4:50 A:
13¼ ð1Þ1þ3 12 18
0 6
¼72 Rtransfer;13¼R 13¼2880
72 ¼40 Then,I3¼60=40¼1:50 A:
4.6 Solve Problem 4.1 by use of the loop currents indicated in Fig. 4-20.
The elements in the matrix form of the equations are obtained by inspection, following the rules of Section 4.2.
Fig. 4-19
19 7 7
7 13 7
7 7 19
2 4
3 5
I1 I2 I3 2 4
3 5¼
60 60 60 2 4
3 5
R¼
19 7 7
7 13 7
7 7 19 2
64
3 75¼2880 Thus,
Notice that in Problem 4.2, too,R¼2880, although the elements in the determinant were different. All valid sets of meshes or loops yield the same numerical value forR. The three numerator determinants are
N1¼
60 7 7
60 13 7
60 7 19
¼4320 N2¼8642 N3¼4320 Consequently,
I1¼N1 R¼4320
2880¼1:5 A I2¼ N2
R¼3 A I3¼N3
R¼1:5 A
The current supplied by the 60-V source is the sum of the three loop currents,I1þI2þI3¼6 A.
4.7 Write the mesh current matrix equation for the network of Fig. 4-21 by inspection, and solve for the currents.
7 5 0
5 19 4
0 4 6
2 4
3 5
I1 I2 I3 2 4
3 5¼
25 25 50 2 4
3 5
Solving,
I1¼
25 5 0
25 19 4
50 4 6
7 5 0
5 19 4
0 4 6
¼ ð700Þ 536¼ 1:31 A Fig. 4-20
Fig. 4-21
Similarly,
I2¼N2 R¼1700
536 ¼3:17 A I3¼ N3 R¼5600
536 ¼10:45 A
4.8 Solve Problem 4.7 by the node voltage method.
The circuit has been redrawn in Fig. 4-22, with two principal nodes numbered1and2and the third chosen as the reference node. By KCL, the net current out of node1must equal zero.
V1
2 þV125
5 þV1V2 10 ¼0 Similarly, at node2,
V2V1 10 þV2
4 þV2þ50
2 ¼0
Putting the two equations in matrix form, 1 2þ1
5þ 1
10 1
10 1
10 1 10þ1
4þ1 2 2
66 4
3 77 5
V1 V2 2 66 4
3 77 5¼
5 25
The determinant of coefficients and the numerator determinants are
¼ 0:80 0:10
0:10 0:85
¼0:670 N1¼ 5 0:10
25 0:85
¼1:75 N2¼ 0:80 5 0:10 25
¼ 19:5
From these,
V1¼ 1:75
0:670¼2:61 V V2¼19:5
0:670 ¼ 29:1 V In terms of these voltages, the currents in Fig. 4-21 are determined as follows:
I1¼V1
2 ¼ 1:31 A I2¼V1V2
10 ¼3:17 A I3¼V2þ50
2 ¼10:45 A
4.9 For the network shown in Fig. 4-23, findVs which makes I0¼7:5 mA.
The node voltage method will be used and the matrix form of the equations written by inspection.
Fig. 4-22
1 20þ1
7þ1
4 1
4 1
4 1 4þ1
6þ1 6 2
66 4
3 77 5
V1 V2 2 66 4
3 77 5¼
Vs=20 0 2 66 4
3 77 5
Solving forV2,
V2¼
0:443 Vs=20 0:250 0
0:443 0:250 0:250 0:583
¼0:0638Vs
7:5103¼I0¼V2
6 ¼0:0638Vs Then 6
from whichVs¼0:705 V.
4.10 In the network shown in Fig. 4-24, find the current in the 10-resistor.
The nodal equations in matrix form are written by inspection.
1 5þ 1
10 1
5 1
5 1 5þ1
2 2
66 4
3 77 5
V1 V2 2 66 4
3 77 5¼
2 6 2 66 4
3 77 5
V1¼
2 0:20 6 0:70
0:30 0:20 0:20 0:70
¼1:18 V Fig. 4-23
Fig. 4-24
Then,I ¼V1=10¼0:118 A.
4.11 Find the voltageVab in the network shown in Fig. 4-25.
The two closed loops are independent, and no current can pass through the connecting branch.
I1¼2 A I2¼30 10¼3 A
Vab¼VaxþVxyþVyb¼ I1ð5Þ 5þI2ð4Þ ¼ 3 V
4.12 For the ladder network of Fig. 4-26, obtain the transfer resistance as expressed by the ratio ofVin toI4.
By inspection, the network equation is
15 5 0 0
5 20 5 0
0 5 20 5
0 0 5 5þRL 2
66 4
3 77 5
I1 I2 I3 I4 2 66 4
3 77 5¼
Vin 0 0 0 2 66 4
3 77 5
R¼5125RLþ18 750 N4¼125Vin I4¼N4
R¼ Vin
41RLþ150ðAÞ Rtransfer;14¼Vin
I4 ¼41RLþ150ðÞ and
4.13 Obtain a The´venin equivalent for the circuit of Fig. 4-26 to the left of terminalsab.
The short-circuit currentIs:c:is obtained from the three-mesh circuit shown in Fig. 4-27.
Fig. 4-25
Fig. 4-26
15 5 0 5 20 5
0 5 15
2 64
3 75
I1 I2 Is:c:
2 64
3 75¼
Vin 0 0 2 64
3 75
Is:c:¼
Vin 5 20
0 5
R ¼Vin
150
The open-circuit voltageVo:c:is the voltage across the 5-resistor indicated in Fig. 4-28.
15 5 0
5 20 5
0 5 20
2 64
3 75
I1 I2 I3 2 64
3 75¼
Vin 0 0 2 64
3 75
I3¼25Vin 5125 ¼Vin
205 ðAÞ Then, the The´venin sourceV0¼Vo:c:¼I3ð5Þ ¼Vin=41, and
RTh¼Vo:c:
Is:c: ¼150
41
The The´venin equivalent circuit is shown in Fig. 4-29. WithRLconnected to terminalsab, the output current is
I4¼ Vin=41
ð150=41Þ þRL¼ Vin 41RLþ150 ðAÞ agreeing with Problem 4.12.
4.14 Use superposition to find the currentIfrom each voltage source in the circuit shown in Fig. 4-30.
Loop currents are chosen such that each source contains only one current.
Fig. 4-27
Fig. 4-28
54 27 27 74
I1 I2
¼
¼ 460
200
From the 460-V source,
I10¼I0¼ð460Þð74Þ
3267 ¼ 10:42 A and for the 200-V source
I100¼I00¼ð200Þð27Þ
3267 ¼1:65 A I¼I0þI00¼ 10:42þ1:65¼ 8:77 A Then,
4.15 Obtain the current in each resistor in Fig. 4-31(a), using network reduction methods.
As a first step, two-resistor parallel combinations are converted to their equivalents. For the 6and 3,Req¼ ð6Þð3Þ=ð6þ3Þ ¼2. For the two 4-resistors,Req¼2. The circuit is redrawn with series resistors added [Fig. 4-31(b)]. Now the two 6-resistors in parallel have the equivalentReq¼3, and this is in series with the 2. Hence,RT ¼5, as shown in Fig. 4-31(c). The resulting total current is
Fig. 4-29
Fig. 4-30
Fig. 4-31(a)
IT ¼25 5 ¼5 A
Now the branch currents can be obtained by working back through the circuits of Fig. 4-31(b) and 4-31(a)
IC¼IF¼12IT¼2:5 A ID¼IE¼12IC¼1:25 A
IA¼ 3
6þ3IT ¼5 3A IB¼ 6
6þ3IT ¼10 3 A
4.16 Find the value of the adjustable resistanceRwhich results in maximum power transfer across the terminalsabof the circuit shown in Fig. 4-32.
First a The´venin equivalent is obtained, withV0¼60 V andR0¼11. By Section 4.10, maximum power transfer occurs forR¼R0¼11, with
Pmax¼V02
4R0¼81:82 W
Supplementary Problems
4.17 Apply the mesh current method to the network of Fig. 4-33 and write the matrix equations by inspection.
Obtain currentI1by expanding the numerator determinant about the column containing the voltage sources to show that each source supplies a current of 2.13 A.
Fig. 4-31 (cont.)
Fig. 4-32
4.18 Loop currents are shown in the network of Fig. 4-34. Write the matrix equation and solve for the three currents. Ans. 3.55 A,1:98 A,2:98 A
4.19 The network of Problem 4.18 has been redrawn in Fig. 4-35 for solution by the node voltage method. Ob- tain node voltagesV1andV2and verify the currents obtained in Problem 4.18.
Ans. 7.11 V,3:96 V
4.20 In the network shown in Fig. 4-36 currentI0¼7:5 mA. Use mesh currents to find the required source voltageVs. Ans. 0.705 V
4.21 Use appropriate determinants of Problem 4.20 to obtain the input resistance as seen by the source voltage Vs. Check the result by network reduction. Ans: 23:5
Fig. 4-33
Fig. 4-34
Fig. 4-35
4.22 For the network shown in Fig. 4-36, obtain the transfer resistance which relates the currentI0to the source voltageVs. Ans: 94:0
4.23 For the network shown in Fig. 4-37, obtain the mesh currents. Ans. 5.0 A, 1.0 A, 0.5 A
4.24 Using the matrices from Problem 4.23 calculateRinput;1,Rtransfer;12, andRtransfer;13. Ans: 10;50;100
4.25 In the network shown in Fig. 4-38, obtain the four mesh currents.
Ans. 2.11 A,0:263 A,2:34 A, 0.426 A
4.26 For the circuit shown in Fig. 4-39, obtainVo:c:,Is:c:, andR0at the terminalsabusing mesh current or node voltage methods. Consider terminalapositive with respect tob. Ans: 6:29 V;0:667 A;9:44
Fig. 4-36
Fig. 4-37
Fig. 4-38
4.27 Use the node voltage method to obtainVo:c:andIs:c:at the terminalsabof the network shown in Fig. 4- 40. Considerapositive with respect tob. Ans: 11:2 V;7:37 A
4.28 Use network reduction to obtain the current in each of the resistors in the circuit shown in Fig. 4-41.
Ans. In the 2.45- resistor, 3.10 A; 6.7, 0.855 A; 10.0, 0.466 A; 12.0, 0.389 A; 17.47, 0.595 A;
6.30, 1.65 A
4.29 Both ammeters in the circuit shown in Fig. 4-42 indicate 1.70 A. If the source supplies 300 W to the circuit, findR1andR2. Ans: 23:9;443:0
Fig. 4-39
Fig. 4-40
Fig. 4-41
Fig. 4-42
4.30 In the network shown in Fig. 4-43 the two current sources provide I0 and I00 where I0þI00¼I. Use superposition to obtain these currents. Ans. 1.2 A, 15.0 A, 16.2 A
4.31 Obtain the currentI in the network shown in Fig. 4.44. Ans: 12 A
4.32 Obtain the The´venin and Norton equivalents for the network shown in Fig. 4.45.
Ans: V0¼30 V;I0¼5 A;R0¼6
4.33 Find the maximum power that the active network to the left of terminalsabcan deliver to the adjustable resistorRin Fig. 4-46. Ans. 8.44 W
4.34 Under no-load condition a dc generator has a terminal voltage of 120 V. When delivering its rated current of 40 A, the terminal voltage drops to 112 V. Find the The´venin and Norton equivalents.
Ans: V0¼120 V;I0¼600 A;R0¼0:2
4.35 The network of Problem 4.14 has been redrawn in Fig. 4-47 and terminalsaandbadded. Reduce the network to the left of terminalsabby a The´venin or Norton equivalent circuit and solve for the currentI.
Ans: 8:77 A
Fig. 4-43
Fig. 4-44
Fig. 4-45
Fig. 4-46
4.36 Node Voltage Method. In the circuit of Fig. 4-48 write three node equations for nodes A, B, and C, with node D as the reference, and find the node voltages.
Ans:
Node A: 5VA2VB3VC¼30
Node B: VAþ6VB3VC¼0 from whichVA¼17;VB¼9;VC¼12:33 all in V Node C: VA2VBþ3VC¼2
8>
<
>:
4.37 In the circuit of Fig. 4-48 note that the current through the 3- resistor is 3 A giving rise to VB¼9 V. Apply KVL around the mesh on the upper part of the circuit to find currentI coming out of the voltage source, then findVAandVC. Ans: I¼1=3 A;VA¼17 V;VC¼37=3 V
4.38 Superposition. In the circuit of Fig. 4-48 find contribution of each source toVA,VB,VC, and show that they add up to values found in Problems 4.36 and 4.37.
Ans. (All in V)
4.39 In the circuit of Fig. 4-48 remove the 2-A current source and then find the voltageVo:c:between the open- circuited nodes C and D. Ans: Vo:c:¼3 V
4.40 Use the values forVC andVo:c:obtained in Problems 4.36 and 4.39 to find the The´venin equivalent of the circuit of Fig. 4-48 seen by the 2-A current source. Ans: VTh¼3 V;RTh¼14=3
4.41 In the circuit of Fig. 4-48 remove the 2-A current source and set the other two sources to zero, reducing the circuit to a source-free resistive circuit. FindR, the equivalent resistance seen from terminals CD, and note that the answer is equal to the The´venin resistance obtained in Problem 4.40. Ans: R¼14=3
Fig. 4-47
Fig. 4-48
Contribution of the voltage source: VA¼3 VB¼0 VC¼ 1 Contribution of the 1 A current source: VA¼6 VB¼3 VC¼4 Contribution of the 2 A current source: VA¼8 VB¼6 VC¼28=3 Contribution of all sources: VA¼17 VB¼9 VC¼37=3
4.42 Find The´venin equivalent of the circuit of Fig. 4-49 seen from terminals AB.
ans: VTh¼12 V;RTh¼17
4.43 Loop Current Method. In the circuit of Fig. 4-50 write three loop equations usingI1,I2, andI3. Then find the currents.
Ans:
Loop 1: 4I1þ2I2þI3¼3
Loop 2: 2I1þ5I2I3¼2 From whichI1¼32=51;I2¼9=51;I3¼7=51 all in A Loop 3: I1þ2I2þ2I3¼0
8>
<
>:
4.44 Superposition. In the circuit of Fig. 4-50 find the contribution of each source toI1,I2,I3, and show that they add up to values found in Problem 4.43.
Ans. (All in A)
4.45 Node Voltage Method. In the circuit of Fig. 4-51 write three node equations for nodes A, B, and C, with node D as the reference, and find the node voltages.
Ans:
Node A: 9VA7VB 2VC¼42
Node B: 3VAþ8VB 2VC¼9 From whichVA¼9;VB¼5;VC¼2 all in V Node C: 3VA7VBþ31VC¼0
8>
<
>:
From the source on the left: I1¼36=51 I2¼ 9=51 I3¼27=51 From the source on the right: I1¼ 4=51 I2¼18=51 I3¼ 20=51 From both sources: I1¼32=51 I2¼9=51 I3¼7=51
Fig. 4-49
Fig. 4-50
Fig. 4-51
4.46 Loop Current Method. In the circuit of Fig. 4-51 write two loop equations usingI1andI2as loop currents, then find the currents and node voltages.
Ans: Loop 1: 4I1I2 ¼2
Loop 2: I1þ2I2¼3 from which, I1¼1 A; I2 ¼2 A
VA¼9 V; VB¼5 V; VC¼2 V
4.47 Superposition. In the circuit of Fig. 4-51 find the contribution of each source toVA,VB,VC, and show that they add up to values found in Problem 4.45.
Ans. (all in V)
4.48 Verify that the circuit of Fig. 4-52(a) is equivalent to the circuit of Fig. 4-51.
Ans. Move node B in Fig. 4-51 to the outside of the loop.
4.49 FindVAandVBin the circuit of Fig. 4-52(b). Ans: VA¼9;VB¼5, both in V
4.50 Show that the three terminal circuits enclosed in the dashed boundaries of Fig. 4-52(a) and (b) are equivalent (i.e., in terms of their relation to other circuits). Hint: Use the linearity and superposition properties, along with the results of Problems 4.48 and 4.49.
From the current source: VA¼7:429 VB¼3:143 VC¼1:429 From the voltage source: VA¼1:571 VB¼1:857 VC¼0:571 From both sources: VA¼9 VB¼5 VC¼2
Fig. 4-52
64