# Can Power Be Measured as V^2/R Using a Bench Meter?

• Phillipdanbury
In summary: Three modules:1) 15.4 vdc, 30 amp max2) 15.4 vdc, 10 amp max3) 12 vdc, 5.8 amp maxI don't expect to reach the temperature difference where I will see these max values. One or two watts per module is probably realistic.The performance of a voltage source can be summarized by a graph showing how output voltage varies as load current is varied. At one extreme you have a point where load resistance is very high (an open circuit), and at the other extreme you have a point where load resistance is very low, practically zero Ohms (i.e., a short-circuit). Useful operation lies in between these extremes.
Phillipdanbury
Hello all,

I'm using peltier modules (also called thermoelectric modules) for a student project. If you're not familiar, they are small semiconductor devices that when heated on one side and cooled on another, power flows due to the Seebeck effect.

I'd like to measure power coming from individual modules. Then, when I have many modules wired in series, I'd like to measure total power of the device.

In one paper regarding a similar device, the authors compute P = I^2 * R. The authors don't say how they measure current. (Paper: A 500 W low-temperature thermoelectric generator/ Design and experimental study)

In another paper, the authors say they measure "open circuit voltages" and "short-circuit" current to derive power. I understand the open circuit voltage part, but I don't know what short circuit current is. (Paper: Ring-structured thermoelectric module).

Can I measure voltage only and compute P = V^2 / R?

If so, does the internal resistance of my bench meter (HP 34401A multimeter) count as the resistance in my circuit or do I need to add another resistor in series? Why do other researchers prefer other methods? Is the "short circuit" method preferable?

If anyone has a reference about measuring power of a device without a load, I would be very grateful. I would like to present my device at conferences and so a reference would be great.

If you know the resistance, yes you can measure the voltage and calculate power. Presumably the resistance used will be the resistance used in your actual circuit. You should be aware that there is an optimal value for the resistance to get maximum power out and you may have to try various values to find it.

So, I should measure the resistance across my single module and use that resistance value when computing power for that single module? Will I also add the internal resistance of the meter from the spec sheet? If not, why not?

Phillipdanbury said:
Can I measure voltage only and compute P = V^2 / R?
Providing that R is the resistance of an external rheostat that can handle the power, and R is known, then yes.

skeptic2 said:

What do I use as a load? A simple resistor? I think Oxygen is saying the same thing below. Any good references for this method? Even a textbook chapter?

NascentOxygen said:
Providing that R is the resistance of an external rheostat that can handle the power, and R is known, then yes.

A single module is probably a low power device, just a watt or two? So the cheapest way to test a module would be to buy 3 or 4 resistors, each 1 or 2 watts power rated, and arrange them series/parallel to give yourself 7 or 8 different resistance values for load testing.

What voltage and current is each module rated?

Phillipdanbury said:
In another paper, the authors say they measure "open circuit voltages" and "short-circuit" current to derive power. I understand the open circuit voltage part, but I don't know what short circuit current is.
The performance of a voltage source can be summarized by a graph showing how output voltage varies as load current is varied. At one extreme you have a point where load resistance is very high (an open circuit), and at the other extreme you have a point where load resistance is very low, practically zero Ohms (i.e., a short-circuit). Useful operation lies in between these extremes.

NascentOxygen said:
A single module is probably a low power device, just a watt or two? So the cheapest way to test a module would be to buy 3 or 4 resistors, each 1 or 2 watts power rated, and arrange them series/parallel to give yourself 7 or 8 different resistance values for load testing.

What voltage and current is each module rated?

Three modules:

1) 15.4 vdc, 30 amp max
2) 15.4 vdc, 10 amp max
3) 12 vdc, 5.8 amp max

I don't expect to reach the temperature difference where I will see these max values. One or two watts per module is probably realistic.

I picture them to be something like these: http://www.ebay.com/itm/12V-60W-TEC1-12706-Heatsink-Thermoelectric-Peltier-Cooling-Plate-Module-BE/122122181201

An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

NascentOxygen said:
I picture them to be something like these: http://www.ebay.com/itm/12V-60W-TEC1-12706-Heatsink-Thermoelectric-Peltier-Cooling-Plate-Module-BE/122122181201

Yes, those are the modules I am using.

tech99 said:
An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

Thanks for clarifying this. This is what I was doing originally before, but it didn't seem right when I thought about it.

Connecting a second meter to measure short circuit current along with open circuit voltage seems to change the reading. Considering my values are so small, maybe I'm better off calculating P = V^2 / R rather than measuring short circuit current? Would either of you mind posting ridiculously specific directions for how to set up my meter and resistors? I have a whole book of resistors. I can make any resistance in combination.

The ammeter and then the voltmeter are connected one after the other and not at the same time.

tech99 said:
The ammeter and then the voltmeter are connected one after the other and not at the same time.

My test rig applies heat from a hot water bath to one side and cold from a refrigerated bath to the other side. There's about 0.2 C variation from moment to moment. Because of this, by the time I disconnect vdc and connect adc, my delta T may change. Perhaps I'm better off with P = V^2/ R for this reason?

Phillipdanbury said:
...by the time I disconnect vdc and connect adc, my delta T may change.
1) Connect ONE lead of the ammeter to a peltier output terminal.
2) Connect the volmeter across the peltier output.
4) Connect the OTHER ammeter lead to the appropiate peltier output terminal. (Maybe us a pushbutton switch for this)

No need to disconnect the voltmeter. When the ammeter is connected, it will be a short circuit load for the peltier module.

jim hardy
tech99 said:
An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

I'm afraid this isn't correct. For instance suppose we have a 10 volt source with a 1 ohm internal resistance. If we put an ammeter directly across the source we get a current of 10 amps. Measuring open circuit voltage gives us 10 volts. Multiplying the two readings gives a power rating of 100 watts.

However the maximum power that can be obtained from this source occurs when a load of 1 ohm is connected. Then the voltage across a 1 ohm load is 5 volts and the power in the load is E^2/R or 25 watts, a significant difference from 100 watts.

Last edited:
Needless to say, I am very confused.

If there is a most appropriate way to measure total power from the generators, I can try to get additional hardware to do this. Please let me know. If any of you have resources to help me read and understand your comments further, I'm happy to do some additional reading on the subject.

Probably the easiest way to determine the power available from the modules (which by the way will vary depending on the resistance of your load) is to either connect your load or connect a power resistor, of a resistance and power rating similar to what you expect your load to be, across the Peltier modules and measure the voltage across it. The power will be E^2/R. Alternatively you could insert an ammeter in series with your load and find power by I^2*R.

Phillipdanbury said:
Needless to say, I am very confused.

If there is a most appropriate way to measure total power from the generators,

You need to focus on what's actually going on. Skeptic gave a hint, i'll try to add some detail to his hint. Then you'll appreciate Tom.G's suggestion.

In EE there is a theorem, "Transfer of power is maximum when impedance of load equals impedance of source". (for DC that is)
There's another theorem , "Any circuit can be represented by an ideal voltage source in series with an impedance." That representation is called "Thevenin Equivalent Circuit".

To find the Thevenin equivalent for a DC circuit you gather two measurements:
1. Measure open circuit voltage, write that number down as Vthevenin
2. Measure short circuit current

Then you divide result of (1) by result of (2) and write that number down as Rthevenin

Then you draw a circuit with a battery of voltage Vthevenin in series with a resistor of value Rthevenin .

That is the "Thevenin Equivalent Circuit". It will behave like the device under test. Well, so long as things stay simple ie neither Rthevenin nor Vthevenin is dependent on the other.

Tom.G's instructions tell you how to find "Thevenin Equivalent Circuit" for your modules. You should do it once for each one.

Now to maximum power transfer, as mentioned by skeptic :

When Rload is zero, obviously Rthevenin gets all the power.

With some algebra you could solve for power to load as f(Rload , Vthevenin and Rthevenin ).

You'll find efficiency is only 50% at point of max power transfer so most machines we don't run that heavily loaded. Radio antennas we do, though, to optimize signal transfer.

Anyhow, finding the Thevenin equivalent for your peltiers will give you an idea of the power available to a load. Then you could check how close to an ideal Thevenin circuit is each peltier gizmo by connecting a couple different Rload's to it and plotting results. Try 1X, ½X and 2X Rthevenin.
Chances are you'll find it somewhat nonlinear .
You can fit any three data points with a quadratic curve, and a least squares quadratic fit to four or five would make a nice slide for a presentation ...
you could teach beginners the interplay between internal resistance , power transfer and efficiency .

hope this helps,
have fun

old jim

Last edited:
Tom.G
skeptic2 said:
I'm afraid this isn't correct. For instance suppose we have a 10 volt source with a 1 ohm internal resistance. If we put an ammeter directly across the source we get a current of 10 amps. Measuring open circuit voltage gives us 10 volts. Multiplying the two readings gives a power rating of 100 watts.

However the maximum power that can be obtained from this source occurs when a load of 1 ohm is connected. Then the voltage across a 1 ohm load is 5 volts and the power in the load is E^2/R or 25 watts, a significant difference from 100 watts.
I agree with you, but for devices which are non linear, such as solar cells, which tend to act as a current source, the rough measurement seems to be a useful guide.

@Phillipdanbury
As @skeptic2 pointed out in post #16, the actual available power from the peltier module will be 1/4 of the value found from the open-circuit voltage and short-circuit current. Even that is an approximate value, not exact, because of nonlinear behavior and self heating of the module.

The suggestion of changing the load resistance and measuring the power at each value is a good one.
The power peak is somewhere near 1/2 the open-circuit voltage and 1/2 the short-circuit current.
For best accuracy and and fewer measurement errors, measure and write down both the current thru the load and the voltage across the load for different load values.
Then multiply the voltage by the current to yield Watts of power for each load. (To determine load resistance, divide the load voltage by the load current.)
The power should increase, reach a peak, then decrease for the various loads.
You may want to try more closely spaced load values around the peak to 'home in' on it.

For a more precise measurement you will need to stabilize the temperature across the module.

Please keep us up-to-date on your results, that way we can all learn!

Hello all,

Thank you all for these immensely helpful suggestions. I appreciate the explanations and background. I will begin collecting some data (data that I will keep this time).

I'll report back, probably in spring.

Thank you all again.

## 1. What is the purpose of using a meter R for V^2/R measurement?

The purpose of using a meter R for V^2/R measurement is to determine the resistance (R) of a circuit by measuring the voltage (V) and current (I) and using the formula R = V^2/I. This is commonly used in electrical and electronic circuits to ensure proper functioning and troubleshoot any potential issues.

## 2. How do you use a meter R for V^2/R measurement?

To use a meter R for V^2/R measurement, first connect the meter to the circuit by placing the positive lead on one side of the component being measured and the negative lead on the other side. Then, turn on the meter and select the appropriate setting for resistance measurement. Finally, take the necessary voltage and current readings and use the formula R = V^2/I to calculate the resistance.

## 3. Are there any precautions to take when using a meter R for V^2/R measurement?

Yes, there are a few precautions to take when using a meter R for V^2/R measurement. Make sure the circuit is turned off and disconnected from any power source before connecting the meter. Also, ensure that the meter is set to the appropriate range for the expected resistance value to avoid damaging the meter. Lastly, be careful not to touch any live wires or components while taking measurements.

## 4. Can a meter R for V^2/R measurement be used for both AC and DC circuits?

Yes, a meter R for V^2/R measurement can be used for both AC and DC circuits. However, it is important to select the correct mode on the meter for the type of current being measured. For example, if measuring AC circuits, the meter should be set to AC mode and if measuring DC circuits, the meter should be set to DC mode.

## 5. What are some common sources of error when using a meter R for V^2/R measurement?

Some common sources of error when using a meter R for V^2/R measurement include using the wrong meter setting, not properly connecting the leads to the circuit, and not accounting for any internal resistance in the meter. It is also important to ensure that the components being measured are not heating up or changing in resistance during the measurement process.

Replies
10
Views
3K
Replies
9
Views
4K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
40
Views
10K
Replies
14
Views
5K
Replies
13
Views
2K
Replies
24
Views
3K
Replies
3
Views
993
Replies
19
Views
2K