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The negative root is extraneous, since the resulting value of vab would not satisfy KVL; thus, vab 2:071 2 4:289 V and i 2 2:071 4:142 A

CHAP. 1]

178 180 182

Notice that, because the resistance R2 is a function of current, the circuit is not linear and the superposition theorem cannot be applied.

For the circuit of Fig. 1-15, nd vab if (a) k 0 and theorems to simplify the circuit prior to solution.

24 25

(b) k 0:01.

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Solved Problem 1.3 Determine the effect of the following transactions on capital. (a) (b) (c) (d) Solution: (a) (b) (c) (d) pense. No effect only the asset and liability are affected. No effect same reason. Decrease in capital capital is withdrawn. Decrease in capital supplies that are used represent an ex Bought machinery on account. Paid the above bill. Withdrew money for personal use. Inventory of supplies decreased by the end of the month.

Fig. 1-15 (a) For k 0, the current i can be determined immediately with Ohm s law: i 10 0:02 A 500

190 197

Since the output of the controlled current source ows through the parallel combination of two 100- resistors, we have vab 100i 100k100 100 0:02 100 100 100 V 100 100 1

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(b) With k 6 0, it is necessary to solve two simultaneous equations with unknowns i and vab . Around the left loop, KVL yields 0:01vab 500i 10 With i unknown, (1) becomes vab 5000i 0 Solving (2) and (3) simultaneously by Cramer s rule leads to 10 500 0 5000 50,000 111:1 V vab 450 0:01 500 1 5000 3 2

197 199

For the circuit of Fig. 1-15, use SPICE methods to solve for vab if (b) k 0:05.

(a) The SPICE netlist code for k 0:001 follows:

Note!

26

Prb.1_5.CIR Vs 1 0 DC 10V R1 1 2 500ohm E 2 0 (3,0) 0.001 ; Last entry is value of k F 0 3 Vs 100 R2 3 0 100ohm RL 3 0 100ohm .DC Vs 10 10 1 .PRINT DC V(3) .END

(b) Edit <Prb1_5.CIR> to set k 0:05, execute the code, and poll the output le to nd vab V 3 200 V.

203 205

For the circuit of Fig. 1-16, nd iL by the method of node voltages if (a) 0:9 and (b) 0.

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Fig. 1-16 (a) With v2 and vab as unknowns and summing currents at node c, we obtain v2 vs v2 v2 vab i 0 R1 R2 R3 But i vs v2 R1 1

Substituting (2) into (1) and rearranging gives 1 1 1 1 1 v v v R1 R2 R3 2 R3 ab R1 s Now, summation of currents at node a gives vab v2 v i ab 0 R3 RL Substituting (2) into (4) and rearranging yields 1 1 1 v v2 v R3 R1 R3 RL ab R1 s Substitution of given values into (3) and (5) and application of Cramer s rule nally yield 2:1 0:1vs 0:1 0:9vs 1:9vs vab 2:1 1 2:21 0:8597vs 0:1 1:1 and by Ohm s law, iL vab 0:8597vs 0:08597vs RL 10 A

27

Only sales on account are recorded in the sales journal; cash sales are recorded in the receipts journal.

(b) With the given values (including 0 substituted into (3) and (5), Cramer s rule is used to nd 3 vs 1 0 vs vab 3 1 2:3 0:4348vs 1 1:1

CHAP. 1]

215 218

If V1 10 V, V2 15 V, R1 4 , and R2 6 in the circuit of Fig. 1-17, nd the Thevenin equivalent for the network to the left of terminals a; b.

R1 + V1 _

28

Fig. 1-17 With terminals a; b open-circuited, only loop current I ows. V1 IR1 V2 IR2 so that I V1 V2 10 15 0:5 A 4 6 R1 R2 Then, by KVL,

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