Maximum power transfer

[K8MHZ]

I think you are on the right track for what Gar is looking for, but you need to think about the size of the capacitor a little more.

Steve

Yep, a 10H coil and a 10F cap aren't a wash at 60 Hz.

 

[rattus]

120106-2213 EST

This question is for those that do not already know the answer, but like a thought provoking challenge.

Given an ideal constant voltage source of 100 V and an internal resistance of 5 ohms what is the load resistance that will produce the maximum power dissipation in the load resistance? What is the current and power in the load at maximum power transfer? How do you solve this problem?

How was this related to Edison's concept of a parallel power distribution system?

.

First off, one must write the expression for power in the load. Any takers?

 

[K8MHZ]

120109-1115 EST

I expected that many reading this thread would know the answers to the questions of the two separate paragraphs of my post #1. It was not intended that those that knew the answer would answer, but rather that it would challenge others to think about the questions.

Has the discussion helped anyone? I do not know. But it has caused at least a little discussion.

Steve66 is the only one so far to have even made any mention that relates to my original second paragraph.

T. J. not every electrician needs to know the answers to the questions in this thread to perform their daily work. Does the knowledge help them? Maybe. A broad basic understanding of things you work with is probably useful. More specifically to maximum power transfer --- it is a very important concept relative to photo-voltaic energy sources. It is also very important to understanding what Edison did and how this resulted in the evolution of the electric power industry.

.

Yes, this thread has been useful. And will be even more useful if someone finds an error in my work....it's been years since I had to figure out this stuff.

Max power = 2000 watts
Current = 20 A

To get 2000 watts out of the load, 5 ohms must be resistive. So, a 'load' consisting of a 10H coil only, would need an appropriately sized cap to 'neutralize' the coil, along with 5 ohms of pure resistance to produce real power.

I figured the size of the cap to be .0000007 H.

 

[K8MHZ]

First off, one must write the expression for power in the load. Any takers?

Power in our load will still be V x A, as in order to get max power all the reactance will have to be cancelled out.

 

[Joethemechanic]

Let me ask this,

What form do you want your answer in?

Or is this an essay question?

 

[Joethemechanic]

If this was RF I would be saying you need a reactance at the load end big enough to absorb a full wave, but the fact that a 60 cycle full wave is like 3,000 miles long, well that makes for a big big something.

 

[K8MHZ]

If this was RF I would be saying you need a reactance at the load end big enough to absorb a full wave, but the fact that a 60 cycle full wave is like 3,000 miles long, well that makes for a big big something.

It's not the length of the wave the components react to, it's the amount of time they spend going from one point in oscillation to another.

 

[K8MHZ]

If this was RF I would be saying you need a reactance at the load end big enough to absorb a full wave, but the fact that a 60 cycle full wave is like 3,000 miles long, well that makes for a big big something.

Oh, and 60 Hz can be radio. That is about the frequency that the big ELF transmitters in Michigan use to use to send messages to submarines that are under water.

One of the antennas went from Michigan's UP into Wisconsin.

 

[K8MHZ]

I figured the size of the cap to be .0000007 H.

That 'H' is really an 'F'.

:dunce:

 

[Joethemechanic]

Wait

Am I over-complicating this?

Max power transfer takes place at unity.


Give me a clue here, I mean I'm a high school drop out millwright and rigger

I really haven't had to deal with any in depth electrical problems since,,,,,,,

9th grade electronics class in 1974

 

[K8MHZ]

Wait

Am I over-complicating this?

Max power transfer takes place at unity.


Give me a clue here, I mean I'm a high school drop out millwright and rigger

I really haven't had to deal with any in depth electrical problems since,,,,,,,

9th grade electronics class in 1974

If by 'unity' you mean the impedances of the source and load are equal, that is correct. BUT only true resistance makes power. 5 ohms of reactance won't make any watts. That's why things get more complicated when we add reactance.

 

[gar]

120109-1354 EST

K8MHZ:

Your capacitance is correct, 0.704 ufd.

Your power is not. From a power standpoint the AC circuit degenerates to the same as the DC circuit. 100 V / 10 ohms = 10 A, and power is 10^2 * 5 in the load. or 500 W. 1000 W is supplied from the voltage source and thus 50% efficiency of power transfer to the load.

For those unfamiliar with series resonant LC circuits the resonant frequency is f = 1/(2*Pi*(L*C)^-2). In an ideal series resonant circuit (meaning 0 losses) the impedance at resonance is 0, and if there were losses, then the impedance equals a series resistance equal to the losses. As soon as you move off resonance a reactive component becomes part of the impedance.

.

 

[Besoeker]

I figured the size of the cap to be .0000007 H.

Note your later correction from H to F.
Don't disagree with the figure.
0.00000070362 F

 

[K8MHZ]

120109-1354 EST

K8MHZ:

Your capacitance is correct, 0.704 ufd.

Your power is not. From a power standpoint the AC circuit degenerates to the same as the DC circuit. 100 V / 10 ohms = 10 A, and power is 10^2 * 5 in the load. or 500 W. 1000 W is supplied from the voltage source and thus 50% efficiency of power transfer to the load.

For those unfamiliar with series resonant LC circuits the resonant frequency is f = 1/(2*Pi*(L*C)^-2). In an ideal series resonant circuit (meaning 0 losses) the impedance at resonance is 0, and if there were losses, then the impedance equals a series resistance equal to the losses. As soon as you move off resonance a reactive component becomes part of the impedance.

.

Where did you get the 10 ohms from?

From Post #1

Given an ideal constant voltage source of 100 V and an internal resistance of 5 ohms

 

[wirenut1980]

Where did you get the 10 ohms from?

From Post #1

5 ohms in the source impedance and 5 ohms in he load. Use both to calculate the real current in the circuit.

 

[Joethemechanic]

BUT only true resistance makes power..

Ok I am thinking you are not talking about resistance in the way resistance is normally thought of.

In my head I can kinda see it. Like in a generator. I sort of see electrons in the windings of the armature sitting in their valence shells resisting the "push" of the magnetic field. If the electrons don't resist the "push" then no power transfer is taking place.

I know this from practice because if I open the circuit, and remove the electron's resistance to the field, my mechanical imput drops to only what is eaten up by bearing friction and windage.


In a lot of ways electricity is just a line shaft. Much easier to turn corners with though.

And we can add a counterweight to it just like a line shaft, something like a big synchronous cap

 

[K8MHZ]

5 ohms in the source impedance and 5 ohms in he load. Use both to calculate the real current in the circuit.

Gotcha, thanks!

 

[rattus]

My way:

My way:

I would solve this problem as follows:

I = Vs/(r + R)

Where Vs is the source voltage, r is the load resistance, and R is the source resistance., then find the power, P.

P = r x Vs^2/(r + R)^2

then find the derivative,

dP/dr and set it to zero.

Now solve for r.

A little thought tells us that if the source is reactive, then the load must also be reactive, the complex conjugate of the source impedance so the reactances offset each other.

 

[K8MHZ]

Ok I am thinking you are not talking about resistance in the way resistance is normally thought of.

In my head I can kinda see it. Like in a generator. I sort of see electrons in the windings of the armature sitting in their valence shells resisting the "push" of the magnetic field. If the electrons don't resist the "push" then no power transfer is taking place.

Not quite true.

There can be electron flow without power. Since it takes power to push the electrons no matter if they turn into watts or not, there is a difference between power in and power out in a reactive load. That difference can be expressed as a ratio and extrapolated into a power factor rating.

It's a tough concept to mentally distinguish between resistance (expressed in ohms and directly related to power) and impedance (also expressed in ohms but does not always result in power, yet still requires amps to overcome).

 

[K8MHZ]

I would solve this problem as follows:

I = Vs/(r + R)

Where Vs is the source voltage, r is the load resistance, and R is the source resistance., then find the power, P.

P = r x Vs^2/(r + R)^2

then find the derivative,

dP/dr and set it to zero.

Now solve for r.

A little thought tells us that if the source is reactive, then the load must also be reactive, the complex conjugate of the source impedance so the reactances offset each other.

Why would we have to solve for 'r' if we know 'R'?

Where was it mentioned that the source was reactive?