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Fig. 4-15, where it is evident that the resistors R1 ; R2 ; . . . ; Rn can be connected one at a time, and the resulting current and power readily obtained. If this were attempted in the original circuit using, for example, network reduction, the task would be very tedious and time-consuming.
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Fig. 4-15
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At times it is desired to obtain the maximum power transfer from an active network to an external load resistor RL . Assuming that the network is linear, it can be reduced to an equivalent circuit as in Fig. 4-16. Then I and so the power absorbed by the load is V 02 RL V 02 PL 0 2 4R 0 R RL " R 0 RL 1 R 0 RL  2 # V0 R 0 RL
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It is seen that PL attains its maximum value, V 02 =4R 0 , when RL R 0 , in which case the power in R 0 is also V 02 =4R 0 . Consequently, when the power transferred is a maximum, the e ciency is 50 percent.
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Fig. 4-16
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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 RL 0, in which case power delivered to the load is zero (i.e., at a minimum).
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4.1 Use branch currents in the network shown in Fig. 4-17 to nd the current supplied by the 60-V source.
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ANALYSIS METHODS
[CHAP. 4
Fig. 4-17 KVL and KCL give: I2 12 I3 6 I2 12 I4 12 60 I1 7 I2 12 I1 I2 I3 I4 Substituting (10) and (11) in (13), I1 I2 2I2 I2 4I2 Now (14) is substituted in (12): 60 I1 7 1 I1 12 10I1 4 or I1 6 A 14 10 11 12 13
Solve Problem 4.1 by the mesh current method.
Fig. 4-18 Applying KVL to each mesh (see Fig. 4-18) results in 60 7I1 12 I1 I2 0 12 I2 I1 6 I2 I3 0 6 I3 I2 12I3 Rearranging terms and putting the equations in matrix form, 60 19I1 12I2 12I1 18I2 6I3 0 6I2 18I3 0 Using Cramer s rule to nd I1 , 60 12 0 19 0 I1 18 6 12 0 6 18 0 12 0 18 6 17 280 2880 6 A 6 18 2 or 19 12 4 12 18 0 6 32 3 2 3 0 60 I1 6 54 I2 5 4 0 5 18 I3 0
CHAP. 4]
ANALYSIS METHODS
Solve the network of Problems 4.1 and 4.2 by the node voltage method.
With two principal nodes, only one equation is necessary. V1 60 V1 V1 V1 0 12 6 12 7 from which V1 18 V. Then, I1 60 V1 6A 7
See Fig. 4-19.
Fig. 4-19
In Problem 4.2, obtain Rinput;1 and use it to calculate I1 .
Rinput;1 R 2880 2880 10  288 11 18 6 6 18 I1 60 60 6A Rinput;1 10
Then
Obtain Rtransfer;12 and Rtransfer;13 for the network of Problem 4.2 and use them to calculate I2 and I3 .
The cofactor of the 1,2-element in R must include a negative sign: 12 6 12 1 1 2 0 18 216 Then, I2 60=13:33 4:50 A: 12 18 13 1 1 3 0 6 72 Then, I3 60=40 1:50 A: Rtransfer;13 R 2880 40  13 72 Rtransfer;12 R 2880 13:33  216 12
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.
ANALYSIS METHODS
[CHAP. 4
Fig. 4-20 2 19 4 7 7 Thus, 32 3 2 3 I1 60 7 7 13 7 54 I2 5 4 60 5 60 I3 7 19 2 3 19 7 7 6 7 R 4 7 13 7 5 2880 7 7 19
Notice that in Problem 4.2, too, R 2880, although the elements in the determinant were di erent. All valid sets of meshes or loops yield the same numerical value for R . The three numerator determinants are 60 7 7 N2 8642 7 4320 N3 4320 N1 60 13 60 7 19 Consequently, I1 N1 4320 1:5 A R 2880 I2 N2 3A R I3 N3 1:5 A R
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