Power

Hello PeterR,

Power in an AC circuit depends on the waveform.

Many AC calculations assume the waveform is a Sine Wave from Mains Power source.

AC waveforms can be almost any shape: Square, triangular, Sine, and combinations plus variations of those basic AC waveforms.

Average power and maximum power are far different from each other.

Kind Regards....

I believe I wrote the question wrong, and did not explain better. Please have a look at the following link, and let me know if you agree.

Thanks for all your help

http://www.opamp-electronics.com/tutorials/maximum_power_transfer_theorem_1_10_11.htm

Hello again PeterR,

Good tutorial there, thanks.

In any AC Power transfer circuit, the maximum power transfer is when the source impedance equals the load Impedance, at the frequency being transferred - This is the same as for DC circuits.

Because in AC circuits, the DC resistances remain the same, but AC resistances (Impedances) alter with frequency, AC calculations can be extremely complex, particularly in cases of "unusual (non-Sinewave) waveforms" and/or multiple frequencies.

As you can see, on that web-linked page, DC examples and calculations only are given.

Does that answer the question you intended?

Kind Regards....

Thanks for all your help

Hello PeterR,

Maximum Power Transferred vs Source and Load Resistance/Impedance

In the diagram above, power is being transferred from the source at voltage V and fixed source resistance R_S, to a load with resistance R_L resulting in a current I. By Ohm's law, I is simply the source voltage divided by the total circuit resistance:

The power P_L dissipated in the load is the square of the current multiplied by the resistance:

We could calculate the value of R_L for which this expression is a maximum, but it is easier to calculate the value of R_L for which the denominator

is a minimum. The result will be the same in either case. Differentiating with respect to R_L:

For a maximum or minimum, the first derivative is zero, so

or

In practical resistive circuits, R_S and R_L are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:

This is positive for positive values of R_S and R_L, showing that the denominator is a minimum, and the power is therefore a maximum, when

R_S = R_L.

You can also demonstrate maximum power transfer graphically by plotting the power transferred (y-axis) against the load resistance (x-axis) for a given fixed source resistance. Hint: For convenience, set E = 1 volt, R_S = 1 ohm, and plot

P_L vs R_L for 0 <= R_L < ∞.

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Maximum Power Transfer vs Efficiency

The condition of maximum power transfer does not result in maximum efficiency. If we define the efficiency η as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from the circuit shown above that

Consider three particular cases:

• If R_load = R_source, then η = 0.5.
• If , then η = 1.
• If R_load = 0, then η = 0.

The efficiency is only 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity (though the total power level tends towards zero). When the load resistance is zero, all the power is consumed inside the source (the power dissipated in a short circuit is zero) so the efficiency is zero.

(adapted from Wiki)

Cheers!
-e

Hi,

"The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power."

This is correct for DC circuits or AC circuits supplying only resistive elements. In case of impedances, the maximum average (real) power will be transferred from the source to the load if the load impedance equals the conjugate of Thevenin impedance of the circuit.