# Measuring the current from a micro generator affects its output?

I am performing an experiment where I have built a scale model of a wind turbine, and am measuring the current it can produce with a micro generator. I am using a standard multi meter and the micro generator is DC. I do not know much about the topic, but whenever I start measuring the current, the turbine slows down significantly and reduces the current. By my understanding, the ammeter function should have very low resistance and allow current to flow through the circuit like normal.

I am also trying to calculate the power produced by the windmill by the equation P=IV, where the voltage is unaffected by my multi meter measurements. Should I ignore the windmill's readings as its current decreases and multiply by its maximum current values? How does a real windmill generator work compared to this?

## Answers and Replies

Bystander
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See "short circuit."

That makes sense, thank you. Should I therefore take the average, maximum current values the multimeter presents and multiply by the voltage to get the most accurate representation of it's power output?

Bystander
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multiply by the voltage
What is the voltage of a "short-circuited" generator? You might want to try a known resistance, "load," and vary that until you get a stable set of simultaneous voltage AND current measurements.

I measured the maximum currents by allowing the turbine to spin up to maximum velocity, then touching the multi meter to see the initial current measured as the generator "short circuits." I am taking the voltage values from an average of the measured voltage using the multi meter as a voltmeter. There is a light bulb attached to the circuit with a measured value of 72.6 ohms, but I have no idea how to create/vary a load. Is the method I mentioned before not acceptable?

Bystander
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"Settling times" on measurement instruments might not be quite as "instantaneous" as you believe;
measured value of 72.6 ohms,
is that a "cold bulb" resistance? Voltage across the bulb times current through the bulb equals power.

I believe it is the "cold bulb" resistance. However, when I try to measure the current through the bulb while the turbine is operating, it short circuits. Do I need to attach an extension with a small resister to add more?

I believe it is the "cold bulb" resistance. However, when I try to measure the current through the bulb while the turbine is operating, it short circuits. Do I need to attach an extension with a small resister to add more?
Usually, measuring the current should not effect the power output or voltage right?

Bystander
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measure the current through the bulb while the turbine is operating, it short circuits.
Measure the current by placing the meter in series with the bulb, NOT parallel; measure the voltage in parallel.

This might be a stupid question, but in order to do so I must strip the red wire slightly?

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Bystander
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I'll assume you have a single circuit, bulb plus generator? You'll need to insert a "current sensing resistor" in series with the bulb; this requires actually cutting the wire. Is there any problem with that? You can then measure the voltage drop across the current sensing resistor, calculate the current, I=Vsense/R, then look at the voltage drop across the bulb, Vbulb, and calculate the power, Pbulb=IVbulb. The current sensing resistor needs to be small enough that it doesn't appreciably reduce the current, but it needs to be large enough that it yields a measurable voltage.

• Asymptotic
Okay, thank you so much for your help!

• berkeman
Tom.G
Science Advisor
This video shows another way to measure the current. The demonstration has a resistor in the circuit but it is not needed for what you are doing; the resistor seems to be a carryover from an earlier experiment.

Cheers,
Tom

berkeman
Mentor
Should I therefore take the average, maximum current values the multimeter presents and multiply by the voltage to get the most accurate representation of it's power output?
Since you are experimenting with a wind generator, it's a good time for you to learn about the Maximum Power Point (MPP) and how important it is in power conversion from wind, solar, hydro, etc.

Are you familiar with the concept of the maximum energy transfer from a source to a load? And how matching the load resistance to the source resistance gives you maximum power transfer? For example, if the source resistance of a battery is 100 Ohms, then you get the maximum energy transfer (and max power transfer) if your load is 100 Ohms too. It the load resistance is higher than 100 Ohms, you get less current than the MPP, and if the load resistance is lower, you get more current but less voltage, and the power transfer is again less than at the MPP. (Remember that Power = V * I)

This is used in power generation like in your experimental setup, because if your load resistance (or "impedance") is not matched to the output resistance of the generator, you are wasting power. As you saw when you short-circuited the output of the generator, you got very little power out. And if you open-circuit the output of the generator, you get little or no power out.

So in your experiment, you will get the most power out by using a potentiometer as the load, and varying the resistance until you measure the most power (monitor both Vout and Iout using two DMMs -- the Vout DMM is across the pot, and the Iout DMM is in series with the wire to the top of the pot). Put the numbers into an Excel spreadsheet to start to get a feel for what the output resistance is for your generator, and what the MPP is.

Then, you can vary the wind in your setup so the generator spins faster, and find that new MPP. It will generally vary with the wind velocity. In real life, voltage converter circuits are used to match the load to the output impedance of the generator. Without these MPP converters, solar and wind generators would be much less efficient! Maximum Power Point Tracking (MPPT) at Wikipedia: https://en.wikipedia.org/wiki/Maximum_power_point_tracking

EDIT / ADD -- If you do use a potentiometer for the load of your generator in your experiments, be sure it has a power rating that is high enough so it will not get hot as you get close to the MPP. Maybe start with a 5 Watt, 500 Ohm potentiometer...

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• Asymptotic