Low resistance

Discussion in 'Electronic Basics' started by PW, Nov 14, 2008.

1. PWGuest

I need to measure the resistance of some motorcycle alternator coils
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

2. ian fieldGuest

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 so the regulars

3. ian fieldGuest

The childish PHucker troll escaped from .

4. JamieGuest

I bet you voted for Obama!

http://webpages.charter.net/jamie_5"

5. Jasen BettsGuest

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

+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.

6. Olivier ScalbertGuest

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 watts

Olivier