abcda: (2) 10.0 V – (6.0 N)I, – (2.0 N)I3 = 0%3Dbefcb: -(4.0 0)I, – 15.00 V + (6.0 N)I, – 10.0 V = 0(3)V + (6.0 N)I, - (4.0 N)I2= 0Solve equation (1) for I, and substitute into equation (2) (Enter your answer in 2.):10.0 V – (6.0 N)I, - (2.0 N)(I, + I2) = 0|(4) 10.0 V – (n)1, - (2.0 N)I, = 0Multiply each term in equation (3) by 4 and each term in equation (4) by 3 (Enter your answer in N.):(5)-100.0 V + (24.0 N)I, - (= 0(6)30.0 V – (24.0 N)I, – (6.0 N)I, = 0%3DAdd equation (6) to equation (5) to eliminate I, and find I, (Enter your first answer in V and your second answer in A.):V – (22.0 N)I, = 0%3DI2 =AUse this value of I, in equation (3) to find I, (in A):-25.0 V + (6.0 N)I, - (4.0 N)I2 = 0A%3DUse equation (1) to find I, (in A):I3 = I, + I, =%3D---Select---those indicated in the figure. The numerical values for the currents are correct.Finalize Because our values for I, and I, are negative, the directions of these currents areDespite the incorrect direction, we must continue to use these negative values in subsequent calculations because our equations were established with our original choice of direction. Find the currents I,, I,, and I, in the circuit shown in the figure. (Use Ɛ = 15.00 V.)%3DA circuit containing different branches.of4.0 NC6.0 N10.0 Vad2.0 NSOLUTIONConceptualize Imagine physically rearranging the circuit while keeping it electrically the same. Can you rearrange it so that it consists of simple series or parallel combinations of resistors? Youshould find that you cannot. (If the 10.0 V battery were removed and replaced by a wire from b to the 6.0 N resistor, the circuit would consist of only series and parallel combinations.)Categorize We cannot simplify the circuit by the rules associated with combining resistances in series and in parallel. Therefore, this problem is one in which we must use---Select---Analyze We arbitrarily choose the directions of the currents as labeled in the figure.According to the label, the current through the 6.0 Q resistor is in which direction in the figure?O to the rightO to the leftO downwardO upwardApply Kirchhoff's junction rule to junction c (Use the following as necessary: I, and I.):(1) I,+= 0We now have one equation with three unknowns: I,, I,, and I3. There are three loops in the circuit: abcda, befcb, and aefda. We need only two equations to determine the unknown currents. (Thethird loop equation would give no new information.) Let's choose to traverse these loops in the clockwise direction. Apply Kirchhoff's loop rule to loops abcda and befcb (Enter your answer in V.):abcda: (2)10.0 V-(6.0 0L - (2.00L = 0.

Ans 1) From the fig, apply junction law at point C ( The total incoming current = total outgoing…

abcda: (2) 10.0v - (6.0 0)/, - (2.0 0)/3 = 0 befcb: -(4.0 0)I, - 15.00 V + (6.0 n)I, - 10.0 V = 0 (3) V+ (6.0 0)I, - (4.0 0)I, = 0 Solve equation (1) for I, and substitute into equation (2) (Enter your answer in Q.): 10.0 V - (6.0 0)I, - (2.0 0)(I, + 12) = 0 (4) 10.0 v - ( - (2.0 0)I, = 0 Multiply each term in equation (3) by 4 and each term in equation (4) by 3 (Enter your answer in 0.): (5) -100.0 V + (24.0 0)I, - 0, = 0 (6) 30.0 V - (24.0 0)I, - (6.0 0)I, = 0 Add equation (6) to equation (5) to eliminate I, and find I, (Enter your first answer in V and your second answer in A.): v - (22.0 0)I, = 0 Use this value of I, in equation (3) to find I, (in A): -25.0 V + (6.0 0)I, - (4.0 0)1, = 0 I = Use equation (1) to find I, (in A): Finalize Because our values for I, and I, are negative, the directions of these currents are ---Select- Despite the incorrect direction, we must continue to use these negative values in subsequent calculations because our equations were established with our original choice of direction. 8 those indicated in the figure. The numerical values for the currents are correct.

abcda: (2) 10.0v - (6.0 0)/, - (2.0 0)/3 = 0 befcb: -(4.0 0)I, - 15.00 V + (6.0 n)I, - 10.0 V = 0 (3) V+ (6.0 0)I, - (4.0 0)I, = 0 Solve equation (1) for I, and substitute into equation (2) (Enter your answer in Q.): 10.0 V - (6.0 0)I, - (2.0 0)(I, + 12) = 0 (4) 10.0 v - ( - (2.0 0)I, = 0 Multiply each term in equation (3) by 4 and each term in equation (4) by 3 (Enter your answer in 0.): (5) -100.0 V + (24.0 0)I, - 0, = 0 (6) 30.0 V - (24.0 0)I, - (6.0 0)I, = 0 Add equation (6) to equation (5) to eliminate I, and find I, (Enter your first answer in V and your second answer in A.): v - (22.0 0)I, = 0 Use this value of I, in equation (3) to find I, (in A): -25.0 V + (6.0 0)I, - (4.0 0)1, = 0 I = Use equation (1) to find I, (in A): Finalize Because our values for I, and I, are negative, the directions of these currents are ---Select- Despite the incorrect direction, we must continue to use these negative values in subsequent calculations because our equations were established with our original choice of direction. 8 those indicated in the figure. The numerical values for the currents are correct.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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