P 5.6-5 Determine the maximum power that can be absorbed by a resistor, R, connected to terminals a–b of the circuit shown in Figure P 5.6-5. Specify the required value of R.

Figure P 5.6-5

We know the maximum power will be when the load resistor is the same as the Thevenin resistance, so find the Thevenin resistor by deactivating the current source. The circuit now looks like

So we start by adding the 20 Ω and 120 Ω resistor in series so we have a 140 Ω resistor on the left. On the right we have a 10 Ω and 50 Ω resistor in series so we have a 60 Ω resistor. Now we add the 140 Ω and 60 Ω resistor in parallel.

So our circuit becomes

So we have two resistors in series and our Thevenin Resistance is

So R is

To find the maximum power we need

So we need to find the current in R, so adding R and now identifying mesh currents we have

Writing out Mesh equations: We have one super mesh and one normal mesh. The super mesh provides the two equations

and

Plug into MATLAB and the results are:

So max Power is

MATLAB Code and Results follow:

%Program to solve PH 320 Homework Problem P5.6-5

%version 2019-03-09 DW Donovan

clear all;

RR = [0 1 -1;

-(10 + 50) (20 + 10) (120 + 50);

(10 + 8 + 50 + 50) -(10) -(50)];

VV = [20 0 0]';

II = RR\VV;

i1 = II(1);

i2 = II(2);

i3 = II(3);

Pmax = i1.^2*50;

Ans ={['i1 = ' num2str(i1) ' A'];

['i2 = ' num2str(i2) ' A'];

['i3 = ' num2str(i3) ' A'];

['Max Power P = ' num2str(Pmax) ' W']};

Ans

%{

Ans =

'i1 = 0.2 A'

'i2 = 17.06 A'

'i3 = -2.94 A'

'Max Power P = 2 W'

%}

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This page last updated on March 9, 2019