Direct Current and Transient Analysis 153
>> plot (RL, IL)
>> grid on
>> xlabel(‘Load Resistance RL’), ylabel(‘Current IL’)
>> title(‘IL vs. RL’);
>> subplot(2,2,2)
>> plot(RL,VL)
>> grid on
>> xlabel(‘Load Resistance RL’), ylabel(‘Voltage VL’)
>> title(‘VL vs. RL’);
>> subplot(2,2,3)
>> P = IL.*VL;
>> plot(RL,P)
>> grid on
>> xlabel(‘Load Resistance RL’), ylabel(‘Power of RL’)
>> title(‘Power vs. RL’);
>> subplot(2,2,4)
>> plot(RL,IL,’o’,RL,VL,’+’,RL,P,’*’)
>> grid on
>> xlabel(‘Load Resistance RL’), ylabel(‘IL,VL,P=IL*VL’)
>> title(‘IL, VL, P vs. RL’);
>> legend(‘IL’,’VL’,’P’);
>> Imax = max(IL);
>> Vmax = max(VL);
>> Pmax = max(P);
>> Result = [Imax Vmax Pmax];
>> disp(‘ ******* R E S U L T S *******”)
>> disp(‘************************************************’)
>> disp(‘The maximum values for IL, VL, and PL for 0<RL<10 Ohms are:’);
>> disp (Result) % the plots are shown in Figure 2.51.
>> disp(‘ amps volts watts’)
>> disp(‘************************************************’)
**************************** R E S U L T S ********************
*****************************************************************
The maximum values for IL, VL, and PL for 0<RL<10 Ohms are:
2.0000 6.6667 5.0000
amps volts watts
*****************************************************************Clearly the results indicate that PRL-max ≠ IL-max (^) VL-max {5 W ≠ (2 A) (6.6667 V)} and
PRL-max = 5 W< IL-max (^) VL-max = Z 6.6667 W; the resulting plots are shown in Figure 2.51.
FIGURE 2.50
Network of Example 2.12.
RL
Vs = 10 V
5 Ω
−
- IL
VL
RL = 0:0.25: 10