Homework Solutions for Electric Circuits, 7^th edition, by Nilsson and Riedel

PROBLEM 4.10

(NOTE: Items in Green were added to facilitate the solution.)

a) Use the Node-Voltage method to find the branch currents i_a - i_e in the circuit shown.

My choice of reference node and node voltage variable has been included on the original diagram.

To write KCL at the two top essential nodes, we need to determine expressions for the voltages v₄ and v₂ in terms of the node ages.

v_a = - v₄ + 44
v₄ = 44 - v_a

v_b = v₂ - 2
v₂ = v_b + 2

Now write the KCL equations.

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Summing the currents into node a:

Simplifying gives

12 v_b - 12 v_a - 2 v_a + 132 - 3 v_a = 0
17 v_a - 12 v_b = 132 (Equation A)

Summing the currents into node a:

Simplifying gives

6 v_b - 6 v_a + 2 v_b + 3 v_b + 6 = 0
6 v_a - 11 v_b = 6 (Equation B)

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Solving equations a and b gives

v_a = 12 V
v_b = 6 V

Now solve for the five specified currents

i_a = (44 - v_a) / 4
i_a = (44 - 12) / 4
i_a = 8 A

i_b = v_a / 6
i_b = 12 / 6
i_b = 2 A

i_c = (v_a - v_b) / 1
i_c = (12 - 6) / 1
i_c = 6 A

i_d = v_b / 3
i_d = 6 / 3
i_d = 2 A

i_e = (v_b + 2) / 2
i_e = (6 + 2) / 2
i_e = 4 A

b) Find the total power developed in the circuit.

In my opinion, this question is a bit ambiguous. One can consider the power "developed" by by a resistor to be negative, in which case the total power developed must equal zero. However, I do not believe this is the intent of the question. I feel certain that the authors of the problem wish the sum of positive powers developed.

Since the resistors all absorb power (thus power absorbed by each is negative, we only need to consider the two sources.

We know that P_dev = vi if the arrow enters the negative voltage reference (one of the few situations where the equation is positive with the arrow entering the negative). Thus

P_44V = 44 (i_a) = 44 (8)
P_44V = 352 W

P_2V = 2 (i_e) = 2 (4)
P_2V = 8 W

Pdev = P_44V + P_44V
Pdev = 360 W

This page last updated at 5:57 PM on Thu, Sep 8, 2005