# Compute the useable power in a simple circuit with resistive vs. reactive loads

Gold Member

## Main Question or Discussion Point

This is not a homework question, but I am interested to calculate the power that can be extracted to perform useful work in two similar circuits. I have been told that it should be the same but this is not what I get.

Consider a voltage source as a fixed voltage source in series with an internal resistance $R_\text{source}$. This voltage source is connected in series with a load. If the load is purely resistive, then the total power generated by the source is worth $P=I^2R_\text{total}$ where $R_\text{total}=R_\text{source} + R_\text{load}$. In impedance matching $R_\text{load}=R_\text{source}$ and it turns out that the total power generated is split equally into the power source as Joule heat and into the load as Joule heat too. The Joule heat in the load is considered a useful work here. It is worth $I^2R_\text{source}$ (remember, I assumed impedance matching.)

So far so good. I have been told that if the load was not purely resistive, for example purely reactive, then the useful power would still be the one calculated above. But this is not what I get. Let's assume a purely reactive load, so $Z_\text{load} =i\Im{Z_\text{load}}$. In impedance matching $|Z_\text{load}|=R_\text{load}$ and so $\Im{Z_\text{load}} = R_\text{load}$. The total power generated is $P=Z_\text{total}I^2=(R_\text{load}+i\Im{Z_\text{load}})I^2=(R_\text{load}+iR_\text{load})I^2$. The real part of that expression is what gets dissipated as Joule heat and is exactly the same as before, and is also lost through the voltage source. The modulus or absolute value of that expression is worth $\sqrt{2}R_\text{load}I^2$. If this is equal to the generated power then it means only $(\sqrt{2}-1)R_\text{load}I^2$ is left for the load. If that square root of 2 was a 2, we would get the same as in the first case, but we don't. I can't see anything wrong with my reasoning. I do not see how to reach the same value as before.

I would appreciate comments and a pointer on where I go wrong.

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anorlunda
Mentor
I have been told that if the load was not purely resistive, for example purely reactive, then the useful power would still be the one calculated above.
That's wrong.

That's wrong too. The maximum power transfer theorem is meant for restive elements.

The real power dissipated in a purely reactive load is exactly zero. Therefore, the "useful" power is also zero, if useful is defined as the load power.

Baluncore
2019 Award
Your first problem is the two or three missing circuit diagrams and the confusion it causes.

Secondly you specify a fixed voltage source Vs, but don't specify DC or AC supply.
XL of an inductor is zero at DC. For AC, it generates a phase shift of current relative to voltage.

By impedance matching you think Rsrc = Rload, which dissipates maximum power in the load.

Then you replace the load with an inductor and make XL = Rs which is impossible, because they are orthogonal, like apples and oranges.

W = V²/(Rsrc+Rsrc); is not equal to W = V²/(Rsrc+zero).

For Vs = AC, with an inductive load, the current is phase shifted relative to voltage. But voltage does not occur in your math, so phase is hidden.

Confused.

Both Anorlunda and Baluncore are correct. When matching impedances you need to cancel the reactances between Zsrc and Zload. If you are matching a complex impedance to a real source such as a coax cable you need to introduce a matching network that cancels the load's reactance. If you are matching a complex impedance to a complex impedance, the matching network must still cancel the reactances of both the source and load reactances.

Here's an example: Suppose we want to match 52 ohms to 2000 ohms. To do that we need to add a reactance to the load. That reactance can be either inductive or capacitive but let's choose inductive. To find the value of that inductance we use the formula X = sqrt(Zlarger * Zsmaller - Zsmaller^2) or sqrt(2000 * 52 - 52^2) = 318.27. We add an inductor of reactance 318 ohms in series with the 52 ohm load. Now in order to cancel the reactance we have to convert the series values to parallel values. This can be done with these formulas. Rp = (Rs^2 + Xs^2) / Rs and Xp = (Rs^2 + Xs^2) / Xs in which Rp and Xp are the parallel values and Rs and Xs are the series values. The values I get are Rp = 2000 and Xp = 326.77. Now, in order to cancel the reactance we need to add a capacitor with a reactance of -326.77 ohms across the series combination of Rs = 52 and Xs = 318 leaving us with a purely resistive coupling.

• fluidistic and davenn
Gold Member
What I had in mind was a thermoelectric generator as voltage source, under a fixed temperature difference, hence fixed voltage. The derivations of the usable power in the literature all assume a purely resistive load, yielding the expression of the first case I described. But in pracrice nobody uses a thermoelectric generator to heat up a resistor. Usually one wants to harvest the power with a capacitor for example and by using a MPPT device to maximize the power produced and thus harvested. But I found no derivations that assume a reactive load, hence my try to see what we get. I was told that it doesn't make any difference if the load is reactive.

Is the thermoelectric generator a DC source? If so, it makes no sense to talk about reactances. All capacitors will be open and all inductors shorts. What is an MPPT device? If your thermoelectric generator is DC and the load reactance is inductive, it wouldn't make any difference.

anorlunda
Mentor
You're using the wrong equations. Thermoelectric devices generate DC. Rather than AC impedance, you should be using instantaneous differential equations. $L\frac{dI}{dt}$ and $C\frac{dV}{dt}$ do you know how to solve differential equations?

Gold Member
You're using the wrong equations. Thermoelectric devices generate DC. Rather than AC impedance, you should be using instantaneous differential equations. $L\frac{dI}{dt}$ and $C\frac{dV}{dt}$ do you know how to solve differential equations?
Yes I am familiar with DE's and basics RC, RLC, etc circuits. The problem here is that I'm not even sure about the circuit itself.
The MPPT is essentially an apparatus that implements an algorithm to track the source's impedance and adjusts the load's impedance accordingly, so that we're as close as possible to impedance matching.

My original question is whether the calculations and results we find in the literature about the power generated by a thermoelectric generator are still valid if we use a reactive load (e.g. using a very complex circuit so complicated that nobody bother to even write down the DE's that govern it, but which has a capacitor which we want to charge). I was told that yes, that the power generated result I get when I consider a purely resistive load will not change if the load changes, as long as I am still in the condition of impedance matching. This is what I wanted to check but failed so far.

anorlunda
Mentor
You can not deliver real power to a pure reactive load from any source, period.

• fluidistic
Baluncore
2019 Award
The MPPT is essentially an apparatus that implements an algorithm to track the source's impedance and adjusts the load's impedance accordingly, so that we're as close as possible to impedance matching.
Not quite. A Maximum Power Point Tracking circuit takes current from the voltage supply so as not to pull down the voltage too much, to maximise the energy extracted. The source can be non-linear so the impedance is not fixed.

The circuit required will be determined by; What range of voltage and charge do you expect your thermoelectric transducer to generate? What is the source of heat? Will the generator produce a steady flow of charge, or will it alternate over what cycle?

If you transferred charge to a capacitor, through an inductor, you could extract energy. But it would need a perfect diode which is an intelligent switch to block the charge flowing back again. The MPPT controller would become a switching converter. Each module would require a controller, which would lead to high costs and inefficiency.

Would a Peltier module not be simpler and more economic ?

• Joshy
Gold Member
Not quite. A Maximum Power Point Tracking circuit takes current from the voltage supply so as not to pull down the voltage too much, to maximise the energy extracted. The source can be non-linear so the impedance is not fixed.

The circuit required will be determined by; What range of voltage and charge do you expect your thermoelectric transducer to generate? What is the source of heat? Will the generator produce a steady flow of charge, or will it alternate over what cycle?

If you transferred charge to a capacitor, through an inductor, you could extract energy. But it would need a perfect diode which is an intelligent switch to block the charge flowing back again. The MPPT controller would become a switching converter. Each module would require a controller, which would lead to high costs and inefficiency.

Would a Peltier module not be simpler and more economic ?
I am just pondering about theoretical calculations, it isn't a real case. I do not understand the point of all these questions. I already specified that I consider a fixed delta T, why would the source of the temperature difference matter?
A thermoelectric generator is essentially a Peltier module. If you do not power the Peltier module but place it under a temperature difference, it will start to generate a voltage proportional to the temperature difference (and thus a current if the circuit is closed). Some companies sell thermoelectric generators and Peltier modules at different prices only because of very tiny differences in them that allows a TEG to work at slightly higher temperatures, for example. So they would use, say gold over copper for some parts. But that's not all companies and quite generally a Peltier module is no different at all physically than a TEG. So a Peltier module is just a TEG run "in reverse" where you inject a current and you produce a temperature difference.

Some commercial MPPT's adjust the impedance every X seconds, so the impedance of the load is indeed fixed for "macroscopic" timescales, though it indeed adapts to non linear sources as well. But this is beside the point.

Anyway I just talked to a friend and he explained a lot to me. He roughly said that if the load is purely reactive then it cannot produce what I call useful power. For that, it would need to have a resistive part. And that there are circuits made for that purpose, i.e. if one wants to charge a capacitor then somewhere there will be an inductor to kill off any reactive load. In the end the load can be simplified to a simple resistor like the 1st case I described here, so this is why people in the literature solely focus on that ideal case when they want to calculate the power produced by a thermoelectric generator. He also said that if I use the charge in the capacitor and let it build up again, over and over, then I am in kind of AC case and that the real power is I^2 times the real part (hence the resistive part) of the load's impedance. So he essentially "solved" my question.

• Tom.G
If you have a MPPT controlling the load on you TEC, then you can ignore what sort of load the MPPT sees when investigating the TEC side of the circuit. The MPPT algorithms only care about maximizing the power extracted from the TEC.

You can either dig into the guts of the MPPT to solve this (which is probably impossible), or just assume that it does it's job properly. If the MPPT is working, it will look like a resistor whose value changes to maximize the power from the TEC (within some limited range. perhaps). So then the load resistance seen by the TEC is dependent only on the TEC and environmental parameters.

The output of the MPPT will depend on the details of it's implementation, there isn't a universal answer for that.

• Tom.G
Baluncore
2019 Award
Which is why I asked you to identify the context.

It demonstrates that you are quite new to the field.

...left for the load. If that square root of 2 was a 2, we would get the same as in the first case, but we don't. I can't see anything wrong with my reasoning. I do not see how to reach the same value as before.
I suggest that we first determine which analysis mode should be used in the relevant system, steady-state AC analysis, steady-state DC analysis, or transient analysis?

For any complex system that may actually include equipment such as MPPT and capacitor filters, we usually take some compromises and choose an analysis method that produces reasonably accurate results.

If you really believe and determine that steady-state AC analysis is appropriate, this is good news because the problem has been simplified so you don't need to spend time doing transient analysis involving differential equations.

Back to the steady-state AC analysis in #1, I think that the "sqrt(2)R(load)I^2" just represents apparent power (S=I^2*Z), not the true power, nor the sum of internal resistive power and load dissipation power. Therefore, you can not subtract the source internal resistive power from it and treat the subtraction result as the remaining power of the load.

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He also said that if I use the charge in the capacitor and let it build up again, over and over, then I am in kind of AC case and that the real power is I^2 times the real part (hence the resistive part) of the load's impedance. So he essentially "solved" my question.
As a reminder, if there is a rectifier circuit or the like that generates a rapidly fluctuating DC voltage, then charge the filter capacitor and supply current to the resistive load at the same time, that is, the output of the rectifier drives the CR load in parallel, then we cannot use I^2*R for calculating the real power if the I is the root mean square value of the total output current C and R from the rectifier. However, as long as the I only represents the current of R, then the I^2*R is certainly still work.

In case that in this circuit only the total current of R and C is available, I can only represent the average power by taking the integrated average of VI over time or one period (if the waveform is periodic), and just use either computer simulation or power meter to solve practical problem.

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Baluncore