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circuits, electric circuits, hayt, solution manual
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Electric circuit analysis 8th^ edition by Hayt & Kemmerly chapter3 solutions. Prepared by Ranz. Exercise 1 ……………………………………………………………………………………………… Node is a point between two elements. Branch is a section between two nodes. Hence here, the number of elements and branches are essentially same. Nodes = 5 Elements = Branches = 7 a) 5 b) 7 c) 7
2nd option
KCL, total current entering a node = total current leaving a node Current supplied by source is the current passing through R1. Hence we can use KCL. 7+3 = i2 + 1 ; i2 = 9 A
KCL, total current entering a node = total current leaving a node Ic = 150 x Ib = 150 x 100 uA = 15 mA Ie = Ic + Ib = 15 mA + 0.1 mA = 15.1 mA
Vx = 2 mA x 4.7 kohm = 9.4 V I3 = - 5 x Vx = - 5 x 9.4 = - 47 A
Let us take the node between the two resistors connected to the source. KCL states that current entering a node should be equal to current leaving a node. Since both resistors are connected to source, current flowing through both should be same. By KCL I - I + ix =0 ; ix = Vx = ix x 0 ; Vx =0 V RESULT Proved in solution
a) By KVL, 9 + 4 + vx =0 ; vx = - 13 V ix = vx/ 7 = - 13/7 = - 1.86 A b) By KVL, 2 - 7 + vx =0 ; vx = 5 V ix = vx/ 8 = 5/8 = 0.625 A RESULT a) - 13 V ; - 1.86 A b) 5 V ; 0.625 A
Let current flowing through circuit be i. Va = - 5i 8Va = - 40i By KVL, 4.5 = 2i + 8Va +5i = 7i - 40 i = - 33i i = - 4.5/33 = - 0.136 A Let us number the devices from left to right as 1,2,3,4. Power absorbed by 1 : 4.5 x 0.136 = 0.6136 W 2 : 0.136 x 0.136 x 2 = 0.0369 W 3: - 40 x 0.136 x 0.136 = - 0.743 W 4 : 0.136 x 0.136 x 5 = 0.0925 W
