I need to measure the resistance of some motorcycle alternator coils They should be about 1.5ohm I have a cheap DVM, which is of little use at this value Is there a simple way to do this keeping in mind that I have Zero electoral knowledge. Thanks
The coils will have higher "resistance" (reactance) when AC current is applied, if you can find a wall wart with AC output (possibly a dial-up modem wall-wart) with approx' 12V output - hook it up with a dial or indicator bulb in series, a gross difference in brightness between any two identical windings would indicate a problem. You can also make use of your meter on its AC voltage range for a more precise indication.
If you really do need an actual DC resistance measurement, the easiest way is to apply a regulated (and known) current to the winding and then measure the voltage developed, from there its a simple Ohm's law calculation: V/A=R.
This topic comes up frequently on News:rec.motorcycles.tech so the regulars have had plenty of practice giving advice.
You don't need to be registered to vote to measure resistance.
Sorry, couldn't er... resist.
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I bet you voted for Obama!
use Ohms law.
pass a known current like 100mA through the coil (eg use a 120 ohm resistor and a 12V supply) and measure the current using your DMM
+12V ---[120]---[coil]---[DMM]---0V then disconnect the DMM and set the circuit up again and measure the voltage across the coil.(view with fixed pitch font - eg cut and paster into notepad)
+12V ----[120]---[coil]---------- 0V | | `--DMM--'then divide current by voltage eg 150mV/100mA=1.50 ohms.
For the DC measurement, if you have not a regulated current source, and if your DVM can not measure current, try the following:
- buy a resistor between 1.5ohm to 15ohm, let's say R;
- put it in serial with the coil;
- put some DC voltage across the two resistors;
- measure the voltage around the coil (Vc) and around the resistor (Vr)
- compute coil resistance: Rc = (Vc)/(Vr)*R
Warning: depending the DC voltage and the value of R, you have to correctly select the power of R. Do not put too much DV voltage. For example, if R = 4.7ohm and if you apply 5V, then power =
4.7*(5/(1.5+4.7))^2 = 3 wattsOlivier
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