# Homework Help: Thévenin Circuit: Power Delivered vs Power Transferred?

1. Aug 3, 2014

### abhitubby

1. The problem statement, all variables and given/known data

Q8. Circuit analysis plays an important role in the analysis of systems designed to transfer power from a source to a load. Maximum power transfer can best be described with the aid of the circuit shown in Figure 8a.

A circuit is given (please refer to attached file).

a) A resistive network can always be replaced by its Thévenin equivalent.
Replace the network by its Thévenin equivalent and redraw the circuit.

b) Determine the value of the Thévenin resistance that permits maximum power delivery to RL.

c) For the circuit show in Figure 8b, find the value of RL that results in maximum power being transferred to RL.

d) For the circuit shown in Figure 8b, calculate the maximum power that can be delivered to RL.
When RL is adjusted for maximum power transfer, what percentage of the power delivered by the 360V source reaches RL.

2. Relevant equations

[1] Ohm's Law: V=IR
[2] Power=(I^2)*R
[3] Max Power= [(VTH)^2]/4RTH

3. The attempt at a solution

a) Employing Thévenin's theorem: I removed the load (resistor marked as RL; then I shorted voltage source (there are no circuit sources to leave open) and found RTH to be 160Ω (resistors in series). VTH is 240V (as there are no other sources).
Thus, I ended up with a Thévenin circuit with a 240V source, a 160Ω resistor and RL (all in series).

b) It is in this part and in part c) that I encounter the most confusion.
I know that when RTH is equal to RL (as seen from the load) maximum power is transferred to the load. So the answer to this section should be 160Ω. However, I am confused as both b) and c) seem to be the same. What is the difference between 'power delivery' and 'power transfer'? Is it that more power can be delivered to the load (ie. arrive at the load) than what is actually transferred to it (ie. be absorbed by the load)?

c) Please see confusion expressed in part b)

d) Still confused about the previous sections, I continued onward under the assumption that the last statement in italics was true (see part b). The circuit doesn't show a variable resistor but the question states that the voltage source is now 360V. (I went with it)
RL was adjusted for max power, so it is at 160Ω. The new equivalent resistance is then (1/160 + 1/120)^-1 + 40 which is roughly 108.57Ω.

Calculating power delivered to the load using formula [2] and substituting in I=V/R from Ohm's Law, I arrived at P=(V^2)/R. Plugging in 360V and 108.57Ω, I got roughly 1193 watts delivered.

Then calculating max power, I used formula [3] using the same voltage and resistance (360V, 108.57Ω) as before. and I get roughly 298 watts. Which gives me (298/1193)*100% = roughly 25% power absorbance.

(I am very unsure about including the load resistance to make a new equivalent resistance for the circuit, what is the purpose of doing this? I tried to follow my lecture slides, but they are not very clear)

If someone could shed some light on this and poke me in the right direction, I would be really appreciate it!
Peace

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2. Aug 3, 2014

### Staff: Mentor

The diagram shows a 240 V source. Where does the 360 V value come from?

Check again. When the 240 V supply is shorted are the two resistors really in series as "seen" from RL? And with RL removed, is the output voltage really the full 240 V?

Parts (b) and (c) are not the same. One asks about the value of Rth to maximize power delivered to the load, while the other asks about how to choose the load resistance RL to maximize power delivered to RL (presumably given some value of Rth that can't be changed).

So in one case you get to freely choose a value for Rth. In the other it is assume that Rth is a fixed value.

3. Aug 4, 2014

### abhitubby

This is a past exam question, and I've it copied out word for word (just did a recheck :tongue:)
I was also confused at that sudden value change, but I assumed that they had neglected to show the load resistance as variable on the diagram. Either that, or a it's typo (I went with 360V while trying to do the question).

No, they are not! I always forget to look at it from the point of view of the the load.
They would be parallel resistors, and so would have a combined value of (1/120 + 1/40)^-1 which is 30Ω.

When finding VTH, one goes through each source present in the circuit, shorting voltage sources, and leaving current sources open. In each case one finds a voltage (V', then V'' etc.) and the final VTH is simply the sum of all those voltages. In this case, after shorting the voltage source, I am left with two resistors in parallel, and an open load. I don't think I can calculate a voltage using Ohm's Law as there is no current source present. Would I need to somehow use the partition of voltage method? I have never come across such a situation before.

Alright, I think I get what you're poking at. For part (c) my answer would then be 30Ω as this is the Thévenin resistance of the circuit, and I know that maximum power is transferred to the circuit when RL is equal to RTH.

Then in part (b) I would use [2] and sub in VTH/(RL+RTH) for i, and show that if RTH was zero, then the maximum value of power would be possible.
(but then I noticed in [3] that value of RTH resulting in max power would be 1Ω. Are either of these methods correct? Or does one of them not apply in this situation? Or are both wrong?)

4. Aug 4, 2014

### Staff: Mentor

It's likely that the source was changed from 360 V to 240 V in order to "refresh" the question for a new exam, but the person who made the change failed to make the necessary changes in the rest of the question text. This is not an unusual occurrence, unfortunately. I would recommend using the 240 V value throughout, but make a note in your answer submission that you have done so.

You find the Thevenin resistance by suppressing the sources and determining the net resistance as "seen" by the load. But you find the Thevenin voltage by determining the voltage at the open-circuited load terminals with all the sources active (superposition can be used to sum their contributions as you've noted). In this case you can find the open-circuit voltage that's produced across the 120 Ω resistor due to the 240 V source.

Yes, that's correct.

Yes, that's true. If no power is lost in the delivery network (i.e., no power is dissipated in the Thevenin resistance) then that power is not wasted, and is available to deliver to the load).

How did you determine that? Surely if there is any Thevenin resistance then power will be dissipated by it.

5. Aug 7, 2014

### abhitubby

This is is quite likely. I'll remember that for the next time I come across such an issue

... Alright. I understand why the voltage across the load will be the same as the voltage across the 120Ω resistor. This can be found using partition of voltage. VTH=240V * (120/120+40) which is 180V.

I really don't know how that formula is brought about. It's derivation is unfortunately not shown in my lecture notes, it is simply stated. However, I follow what you're saying about dissipation. If this formula is correct, then wouldn't a RTH which is zero cause division by zero? Or, is this another case in science where zero is an approximation for a very tiny value?

Part (d)

Calculating power delivered to the load using formula [2] and substituting in I=V/R from Ohm's Law, I arrived at P=(V^2)/R. Plugging in 240V and 30Ω, I got roughly 1920 watts delivered.

Then calculating max power, I used the same formula but with combined RTH and RL (30+30=60Ω) using the same voltage as before. and I get 960 watts. Which gives me (298/1193)*100% = 50% power absorbance.

Is this the right method I am following?

Thank you very much.

6. Aug 7, 2014

### Staff: Mentor

You should be able to derive the formula for the power delivered to the load starting with the Thevenin equivalent circuit with the load connected. With that in hand you should be able to tell by inspection what has to happen with Rth in order to maximize the power in RL; you don't even need calculus to be able to see it.

I think you'll want to use the Thevenin voltage and Thevenin here. That's the point of taking the Thevenin equivalent: to simplify the source network down to an easily analyzed source and series resistance. You have Vth in series with Rth and RL.

You want to compare the power delivered to the load RL with the total power delivered by the source (or equivalently the power dissipated by both Rth and RL).