P 4.2-6 Simplify the circuit shown in Figure P 4.2-6 by replacing series and parallel resistors with equivalent resistors; then analyze the simplified circuit by writing and solving node equations.

(a) Determine the power supplied by each current source.

(b) Determine the power received by the 12-Ω resistor.

Figure P 4.2-6

There is the 40 Ω and the 10 Ω resistors in parallel they become effectively

So this 8 Ω is in series with the 12 Ω So we have a 20 Ω effective resistor. We also have a

60 Ω and a 120 Ω resistor in parallel.

So now we can redraw the equivalent circuit and set up the node equations.

We have also placed a ground point for reference, and labeled 3 node points. Now we can write out the 3 node equations.

At Node a

At Node b

At Node c

We can rewrite these equations into standard form and use MATLAB to solve for the three nodes, v_a, v_b, and v_c.

(a) Determine the power supplied by each current source.

Since the two current sources have common bottom terminals connected which we are calling ground, so the voltage across the 3 mA source is v_a, and the voltage across the 2 mA source is v_b. So we can plug these equations into MATLAB and solve.

and

(b) Determine the power received by the 12-Ω resistor.

The Power dissipated by the 12 Ω resistor can be found by using

Again put these equations in to MATLAB and let it do the calculations.

From MATLAB we get the answers

MATLAB code and results follow:

%Program to Solve PH 320 Homework Problem P.4.2-6

%version 2019-01-28 DW Donovan

clear all;

RR = [(1/20 + 1/20) (-1/20) (-1/20);

(-1/20) (1/20 + 1/10) (-1/10);

(-1/20) (-1/10) (1/20 + 1/10 + 1/40)];

II = (1e-3)*[3 2 0]';

VV = RR\II;

va = VV(1);

vb = VV(2);

vc = VV(3);

P3 = va*3;

P2 = vb*2;

P12 = -((va - vb)/20).^2*12;

Ans = {['Va = ' num2str(va) ' V'] ['P3 = ' num2str(P3) ' mW'];

['Vb = ' num2str(vb) ' V'] ['P2 = ' num2str(P2) ' mW'];

['Vc = ' num2str(vc) ' V'] ['P12 = ' num2str(P12) ' W']};

Ans

%{

Ans =

'Va = 0.244 V' 'P3 = 0.732 mW'

'Vb = 0.228 V' 'P2 = 0.456 mW'

'Vc = 0.2 V' 'P12 = -7.68e-06 W'

%}

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This page last updated on January 28, 2019