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max power

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INHT,

hi, how should i go about solving this question?

Using the node method, find the voltage across Rx in terms of Rx.

Use E^2/Rx and find the term for power across Rx in terms of Rx

Find the first derivative of the power with respect to Rx and solve for its zero value.

When Rx = 18 ohms, the maximum power is obtained.

Ratch
 
collin55,

How many watts is dissipated in Rx when Rx = 18R ??????

Look at the graph in the attachment. Or compute it from equation (3) which gives 10368 watts.

Ratch
 
The circuit is very complex until you realise one thing.
Firstly, remove the 6R, 18v and 12R.
Maximum power in Rx is achieved when Rx is equal to 3R.
To see if this is true, simply change Rx to 2R or 4R and work out the watts dissipated.
Another way to look at it is this: When Rx is zero, the watts dissipated is zero, when Rx is very high, the watts dissipated is very small.

Make Rx = 3R and see what happens:
The voltage across Rx will be 18v.
Now put-back the 6R, 18v supply and 12R load resistor.
The voltage across Rx is 18v. This means no current will flow though the 6R resistor because it has 18v on each end.
Current through Rx = 36v/6R = 6 amp
Wattage dissipated by 3R = V2/3 = 108watts or I2xR = 62 x 3 = 108watts
We have turned a seemingly complex question into a very simple answer.

[MODNOTE]Removed unnecessary comments.[/MODNOTE]
 
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colin55,

Sorry I did not answer sooner, but I had places to go and things to do.

Now, if you looked at my analysis, you would have seen immediately from my node equations that I was considering the sources as current sources, not voltage sources like you did. Since the sources were marked with current direction, and were missing voltage polarity signs, I believe that was a good assumption.

But OK, let's assume they are voltage sources.

The circuit is very complex until you realise one thing.

Actually, I don't think it is complex at all. What thing should I realize?

Firstly, remove the 6R, 18v and 12R.
Maximum power in Rx is achieved when Rx is equal to 3R.
To see if this is true, simply change Rx to 2R or 4R and work out the watts dissipated.
Another way to look at it is this: When Rx is zero, the watts dissipated is zero, when Rx is very high, the watts dissipated is very small.

For the partial circuit you specify, that is true.

Make Rx = 3R and see what happens:
The voltage across Rx will be 18v.
Now put-back the 6R, 18v supply and 12R load resistor.
The voltage across Rx is 18v. This means no current will flow though the 6R resistor because it has 18v on each end.
Current through Rx = 36v/6R = 6 amp
Wattage dissipated by 3R = V2/3 = 108watts or I2xR = 62 x 3 = 108watts
We have turned a seemingly complex question into a very simple answer.

By fallacious reasoning, you have obtained the wrong answer to a slightly difficult problem. The goal is to find the resistance for maximum dissipation, not the dissipation at Rx=3. Since you have not performed any differentiation of the power with respect to Rx, you have no idea what the resistance for max power dissipation is. As you can see in the attachment, it occurs at Rx=2 for a dissipation of 112.5 watts, not 108 watts. The voltage increases across Rx as its resistance increases from zero, but after Rx=2, the power drops even though the voltage across the resistor still increases. The reasoning you expound above does not take that into effect.

See this review on Spot The Mistake:
**broken link removed** .

See the attachment which proves your mistake. I am surprised no one caught your error after all this time.

Ratch

[MODNOTE]Removed unnecessary comments.[/MODNOTE]
 
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Let us use your answer.
Current via 36v supply through 3R and 2R = 36/5 = 7.2A
Wattage dissipated by 2R = 7.2 x 7.2 x 2 = 103.68watts.
Voltage across 2R = 7.2 x 2 = 14.4v
Current delivered by 18v supply 18 - 14.4v through 6R = 3.6/6 = 0.6A
Wattage disspiated by 2R = .72watts.

However the 18v supply will increase the voltage across the 2R slightly and the 36v supply will not deliver 103.68watts.

Thus 2R does not dissipate 112 watts.

[MODNOTE]Removed unnecessary comments.[/MODNOTE]
 
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colin55,


Current via 36v supply through 3R and 2R = 36/5 = 7.2A

Tilt! Wrong! Incorrect application of superposition. You failed to take into account that the 6R resistor is in parallel with Rx, so that your calculated current is incorrect. The voltage across Rx from the 36 volt source is 12 volts.

Current delivered by 18v supply 18 - 14.4v through 6R = 3.6/6 = 0.6A

Again you fail to take into account that 3R is in parallel with Rx. The voltage across Rx from the 18 volt source is 3 volts.

So by applying the superposition theorem we get a voltage across Rx of 12 volts + 3 volts = 15 volts. That is the same as equation (2) of my attachment. 15 volts across Rx=2 gives a dissipation of 112.5 watts. The current from the 36 volt source is 6 amp, and the current from the 18 volt source is 1.5 amps for a total of 7.5 amps. 7.5 amps through R2 gives a dissipation of 112.5 watts.

Ratch

[MODNOTE]Removed unnecessary comments.[/MODNOTE]
 
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All this dense maths is a bit beyond me, but I think the problem is a bit obtuse - it states units are in V,A,Ω but it's not clear whether the sources are to be regarded as constant current or constant voltage.
 
I suggest you go and set up the circuit and see what happens.
You are trying to turn a simple circuit into something very complex.
Take one part of the circuit at a time.
Don't try adding in the 6R because it has 18v in series with it.

Again you fail to take into account that 3R is in parallel with Rx. The voltage across Rx from the 18 volt source is 3 volts.
This concept is entirely wrong. I don't know how you come to this reasoning. The 3R has 36v in series with it.
 
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whether the sources are to be regarded as constant current or constant voltage.
Unless otherwise stated, supply voltages are fixed DC voltages capable of supplying the current required to solve the problem.
 
colin55,

I suggest you go and set up the circuit and see what happens.

Why should I? My calculations were done carefully and are correct.

You are trying to turn a simple circuit into something very complex.

Nothing complex about it. Just an application of basic principles.

Take one part of the circuit at a time.

Isn't that what I did in my last post? First the 36 volt source, and then the 18 volt source. Didn't you follow that? Did you look at any of my attachments? If not, you are going off half-cocked.

Don't try adding in the 6R because it has 18v in series with it.

You have to do so because the voltage source has a resistance of zero. Voltage sources have a resistance of zero, current sources have a resistance of infinity. That principle is "staple goods". The Thevenin and Norton theorems make extensive use of that method.

This concept is entirely wrong. I don't know how you come to this reasoning. The 3R has 36v in series with it.

You need to get up to speed on how the superposition theorem works. Again, the 36 volts source is assumed to have zero internal resistance.

For the moment forget about dividing up the circuit. Look at my attachment. I calculated the voltage and wattage across Rx directly. Doesn't that mean anything to you? A direct calculation of voltage and power done without any mistakes you can find?

Ratch
 
No mistakes??? Just faulty reasoning.

Firstly you say Rx must be 18R
Then you say the dissipation is 10368 watts.
Now you say Rx is 2R and the dissipation is 112Watts.

How many more guesses are you going to provide???????????

A voltage-source has zero impedance when it is a voltage source. You cannot say the 6R is in parallel with the 2R when it has a voltage in series with it - and not account for the voltage.
 
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Throbscottle,

All this dense maths is a bit beyond me, but I think the problem is a bit obtuse

I find your lack of understanding and incomprehension--disturbing. The problem is straightforward and direct. The concepts are simple and so is the math. You won't get far in the electrical field unless you obtain more education.

it states units are in V,A,Ω but it's not clear whether the sources are to be regarded as constant current or constant voltage.

Those are standard MKS units, but that is irrelevant to the definition of a voltage/current source. You should know that voltage sources are assumed to have zero internal resistance and current sources are assumed to have infinite resistance.

Ratch
 
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hi Ratch,

LTSpice simulation of the problem shows [as you have already clearly proved] that the Wpeak occurs when the resistance is 18R.


Eric
 
colin55,

Firstly you say Rx must be 18R
The you say the dissipation is 10368 watts.

That was when I assumed those sources were current sources. Didn't you read my post about that?

Now you say Rx is 2R and the dissipation is 112Watts.

112.5 watts to be exact, if it is assumed the sources are voltage sources.

How many more guesses are you going to provide???????????

Guesses? What guesses? Both answers were the result of carefully considered calcuations.

A voltage-source has zero impedance when it is a voltage source. You cannot say the 6R is in series with the 2R when it has a voltage in series with it - and not account for the voltage.

Yes, you have to get up to speed on superposition. You take one source at a time and calculate the voltage or current of a element. You zero out all the voltage sources you are not using. Check the figures of my post #11 and see.

In any case, you need to review this whole thread carefully to refresh in your mind what we both avered, because you seem to be confused as to what was being presented.

Ratch
 
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ericgibbs,

LTSpice simulation of the problem shows [as you have already clearly proved] that the Wpeak occurs when the resistance is 18R.

Thanks Eric, I needed that. Now if you would be so kind as to present a confirmation of the problem answer using constant voltage sources, then I can go to bed.

Ratch
 
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