zen barcode ssrs The current supplied by the 60-V source is the sum of the three loop currents, I1 I2 I3 6 A. in Software

Creator QR in Software The current supplied by the 60-V source is the sum of the three loop currents, I1 I2 I3 6 A.

The current supplied by the 60-V source is the sum of the three loop currents, I1 I2 I3 6 A.
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Write the mesh current matrix equation for the network of Fig. 4-21 by inspection, and solve for the currents.
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Fig. 4-21 2 7 4 5 0 Solving, 25 5 0 0 7 5 25 19 4 5 19 4 700 536 1:31 A I1 50 4 6 6 0 4 5 19 4 3 32 3 2 25 0 I1 4 54 I2 5 4 25 5 I3 50 6
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ANALYSIS METHODS
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Similarly, I2 N2 1700 3:17 A 536 R I3 N3 5600 10:45 A 536 R
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Solve Problem 4.7 by the node voltage method.
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The circuit has been redrawn in Fig. 4-22, with two principal nodes numbered 1 and 2 and the third chosen as the reference node. By KCL, the net current out of node 1 must equal zero.
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Fig. 4-22 V1 V1 25 V1 V2 0 2 10 5 Similarly, at node 2, V2 V1 V2 V2 50 0 10 4 2 Putting the two equations in matrix form, 2 32 3 1 1 1 1 V1 5 6 2 5 10 10 76 7 6 76 7 4 54 5 1 1 1 1 V2 25 10 10 4 2 The determinant of coe cients and the numerator determinants are 0:80 0:10 0:670 0:10 0:85 5 0:10 0:80 1:75 N1 N2 25 0:10 0:85 From these, V1 1:75 2:61 V 0:670 V2 19:5 29:1 V 0:670
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In terms of these voltages, the currents in Fig. 4-21 are determined as follows: I1 V1 1:31 A 2 I2 V1 V2 3:17 A 10 I3 V2 50 10:45 A 2
For the network shown in Fig. 4-23, nd Vs which makes I0 7:5 mA.
The node voltage method will be used and the matrix form of the equations written by inspection.
ANALYSIS METHODS
[CHAP. 4
Fig. 4-23 2 1 1 1 6 20 7 4 6 4 1 4 Solving for V2 , 32 3 2 3 1 Vs =20 V1 7 4 76 7 6 76 7 6 7 5 1 1 1 54 5 4 V2 0 4 6 6
0:443 Vs =20 0:250 0 V2 0:443 0:250 0:0638Vs 0:250 0:583 7:5 10 3 I0 V2 0:0638Vs 6 6
Then from which Vs 0:705 V.
In the network shown in Fig. 4-24, nd the current in the 10- resistor.
Fig. 4-24 The nodal equations in matrix form are written by inspection. 2 1 1 6 5 10 6 4 1 5 32 3 2 3 1 2 V1 7 5 76 7 6 76 7 6 7 5 1 1 54 5 4 V2 6 5 2 2 0:20 6 0:70 1:18 V V1  0:30 0:20 0:20 0:70
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ANALYSIS METHODS
Then, I V1 =10 0:118 A.
Find the voltage Vab in the network shown in Fig. 4-25.
Fig. 4-25 The two closed loops are independent, and no current can pass through the connecting branch. I1 2 A Vab Vax Vxy Vyb 30 3A 10 I1 5 5 I2 4 3 V I2
For the ladder network of Fig. 4-26, obtain the transfer resistance as expressed by the ratio of Vin to I4 .
Fig. 4-26 By inspection, the network equation is 2 15 6 5 6 4 0 0 5 20 5 0 32 3 2 3 I1 Vin 0 0 5 0 76 I2 7 6 0 7 76 7 6 7 20 5 54 I3 5 4 0 5 I4 0 5 5 RL N4 125Vin
R 5125RL 18 750
N Vin I4 4 A R 41RL 150 and Rtransfer;14 Vin 41RL 150  I4
Obtain a Thevenin equivalent for the circuit of Fig. 4-26 to the left of terminals ab.
The short-circuit current Is:c: is obtained from the three-mesh circuit shown in Fig. 4-27.
ANALYSIS METHODS
[CHAP. 4
Fig. 4-27 2 15 5 0 32 I1 3 2 Vin 3
7 7 6 6 76 4 5 20 5 54 I2 5 4 0 5 0 Is:c: 0 5 15 5 20 Vin 0 5 Vin Is:c: R 150 The open-circuit voltage Vo:c: is the voltage across the 5- resistor indicated in Fig. 4-28.
Fig. 4-28 2 15 5 6 4 5 20 0 5 32 3 2 3 I1 Vin 0 76 7 6 7 5 54 I2 5 4 0 5 20 0 I3 25Vin Vin A 5125 205
I3
Then, the Thevenin source V 0 Vo:c: I3 5 Vin =41, and RTh Vo:c: 150  Is:c: 41 With RL connected to terminals ab, the output
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