I measured the voltage generated by my wind turbine as 0.91 V with a load of 180 Ohms. So theoretically, using ohms law, my wind turbine should produce an amperage of 0.05A. However, it produced 0.12A. I wanted to know what can account for this error?

  • 1
    \$\begingroup\$ How did you measure that current? \$\endgroup\$
    – uint128_t
    Feb 21, 2016 at 19:34
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    \$\begingroup\$ 0.91 volts divided by 180 ohms is 5 mA not 50 mA. \$\endgroup\$
    – Andy aka
    Feb 21, 2016 at 19:36
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    \$\begingroup\$ When I recently checked, Ohm's law was still holding, so there's something wrong with your measurement. (assuming the .05A is a typo and you meant .005A) Does the turbine put out DC, or AC? Are you measuring the voltage directly across the 180 Ohm load? Is it resistive? Have you looked at the output voltage with a scope? \$\endgroup\$
    – John D
    Feb 21, 2016 at 19:37
  • \$\begingroup\$ yes it should say 0.005A and it is DC and the load was a resistor \$\endgroup\$
    – user510
    Feb 21, 2016 at 21:18

2 Answers 2


Given a choice between Ohm's law and measurement error, I'm going to go with measurement error. After all, there is already a large math error on prominent display. 0.91V / 180 Ω = 0.00506 A making your initial calculation off by an order of magnitude.

Going with a typical mistake, I'll guess that you measured current across the load, thus short-circuiting the load and measuring at essentially 0 Volts. You are fortunate that your wind turbine only put out 0.12A, since a typical low-range ammeter fuse is 0.25A.


simulate this circuit – Schematic created using CircuitLab

  • Ammeters always go in SERIES with the load.
  • Voltmeters always go in PARALLEL with the load.
  • A multimeter needs to be hooked up correctly for the mode it is being used in.
  • \$\begingroup\$ my set up was alligator clips connected from the motor to the resistor and a different pair from the resistor to the multimeter. Is this wrong? \$\endgroup\$
    – user510
    Feb 21, 2016 at 21:20
  • \$\begingroup\$ A different pair sounds like you mean the first version, so yes, it's wrong, for current measurement. Correct, for voltage measurement. For current measurement a single lead from the motor to the meter, a single lead from the meter to the resistor, and a single lead from the resistor back to the other terminal of the motor - as in the second circuit. \$\endgroup\$
    – Ecnerwal
    Feb 22, 2016 at 2:27

Your generator also has some source impedance, you need to account for that in your calculations.

  • \$\begingroup\$ But the OP implied that he/she measured the output with the load attached, presumably across the load since there's no way to measure on the generator side of the source impedance. \$\endgroup\$
    – John D
    Feb 21, 2016 at 19:40

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