Thevenin theorem application

Hello!

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.

Internal losses means the power which is not delivered to the resistor in question i.e. 10 ohm.

That means in case of R = 0 the power which is lost is the power which is dissipated across resistor 20 ohm that's why 7.2 watts.

But I don't know how he arrived at 12 watts for the second case.

mjuneja said:

But I don't know how he arrived at 12 watts for the second case.

Click to expand...

Yeah, this also the part that confuses me. I don't know if it is a typo or what.

As per my calculation it should be 3.7 watts which includes the dissipation across R as well as 20 ohm resistor.

You can calculate the dissipation across each resistor by using the superposition theorem.



If, instead, R1=20 ohm, R2=10 ohm, R3=20 ohm I obtained:

I1=0.25 A, I2=0.1 A, I3=0.35 A that is:

Pr1=20*(0.25)^2=1.25 W
Pr2=10*(0.1)^2=0.1 W
Pr3=20*(0.35)^2=2.45 W

So the internal losses are Pr1+Pr3=3.7 W as calculated by mjuneja, while the load dissipates 0.1 W (as correctly stated in the text of the problem)

paulmdrdo said:

View attachment 144387

Hello!

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.

Click to expand...

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.

Ratch

Ratch said:

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

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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

albbg said:

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

Click to expand...

Yes, you are correct. I used the wrong voltage for the power calculations. Thanks for the correction.

Ratch