I'am having difficulty understanding the statements in the encircled part. What internal losses is it talking about? and how did it come up with its formula for calculating it? TIA.
I'am having difficulty understanding the statements in the encircled part. What internal losses is it talking about? and how did it come up with its formula for calculating it? TIA.
Poorly defined problem. No definite statement on which two points they are Theveninizing. Best to do this problem by node analysis. Only one undefined node, call it V.
The node equation is :
Easily solved for V
The power across 10 ohm resistor is:
The plot of the voltage across the resistor is:
And the plot of the power dissipated in the resistor is:
As you can see, when the R resistance becomes 20 ohms, no voltage is across the 10 ohm resistor.
When R = 0, 12 volts is across the 20 ohm resistor, so the wasted power is 7.2 watts. When R = 20 ohms, V = 8 volts and 4 volts is across resistor R = 20 ohms, which causes 0.8 watts wasted. 8 volts across the 20 ohm resistor causes 3.2 watts of wasted power. Added together we get 4.0 wasted watts when R = 20 ohms.
When R = 0, 12 volts is across the 20 ohm resistor, so the wasted power is 7.2 watts. When R = 20 ohms, V = 8 volts and 4 volts is across resistor R = 20 ohms, which causes 0.8 watts wasted. 8 volts across the 20 ohm resistor causes 3.2 watts of wasted power. Added together we get 4.0 wasted watts when R = 20 ohms.Ratch
Your last calculation is not correct. If R=20 ohm, using your equation to calculate V at the connection point
of the three resistors, we obtain V=7 V. Then
Pr1 = (12-7)^2/20 = 1.25 W
Pr2 = (7-8)^2/10 = 0.1 W
Pr3 = (7)^2/20 = 2.45 W
then the wasted power is Pr1+Pr3 = 3.7 W. Node analysis and current branch analysis of course must match each other.
By the way, when V=8 V, there is no power dissipated onto the load regardless its ohmic value
Your last calculation is not correct. If R=20 ohm, using your equation to calculate V at the connection point
of the three resistors, we obtain V=7 V. Then
Pr1 = (12-7)^2/20 = 1.25 W
Pr2 = (7-8)^2/10 = 0.1 W
Pr3 = (7)^2/20 = 2.45 W
then the wasted power is Pr1+Pr3 = 3.7 W. Node analysis and current branch analysis of course must match each other.
By the way, when V=8 V, there is no power dissipated onto the load regardless its ohmic value