[wadenelson]

Since voltage drop testing is probably the single most important electrical diagnostic skill you can have... I think about it a lot.

Including different ways of explaining it for those who haven't "got" it yet.

So here's another.

Wires, contacts (in switches and relays) and connectors all have resistance.

All new and shiny, it's a very LOW resistance.

(As a reference point, 1000 feet of 14AWG stranded automotive wire has approximately 2.5 ohms)

Once contacts get worn or burned, connectors get corroded, and half the strands in a wire broken from flexing back and forth..... that resistance can increase considerably.

But it's STILL too low to measure effectively with an ohmmeter. Realistically, an ohmmeter can't measure anything ACCURATELY below, say, 10 ohms.

How hard you press the test leads to whatever you're measuring can cause your reading to vary by several ohms.

A corroded connector, or a ground connection to an auto or motorcycle frame might only add 1/10 of an ohm, .1 ohms, to a circuit.

That doesn't seem like much, but it can be enough to dim a headlight, cause a starter to click instead of spin, or cause all sorts of problems with ECM's and sensors.

So the way you FIND those corroded connectors, burned contacts, bad grounds, loose connections, etc is not by using an ohmmeter.

It's by using Ohms LAW.

Volts = Amps * Resistance.

If you have a 10 amp headlight, and a corroded headlight switch that adds .1 ohms of resistance...

I * R = the amount of voltage LOST passing through that switch, so

10 amps * .1 ohm = 1 volt.

So instead of the headlight getting a full 13.8V while the engine is running, it's only getting 12.8 volts. And as a result it's only about half as bright!

The key thing here is that a voltmeter CAN very accurately measure voltages, even down to millivolts (thousandths of a volt).

A one volt "drop" is easy to measure, unlike a .1ohm resistance.

So you turn the headlight on, and compare the voltage at the battery (alternator) with the voltage actually reaching the light.

Current HAS TO BE FLOWING! V=IR only works when I is non-zero! In other words you have to have the headlight ON, and illuminated.

You can't measure a voltage drop to a non-energized circuit!

Let's say you find the headlight is only getting 11.8 volts with the engine off, but the battery measures 12.6 volts.

You've got a voltage drop of .8 volts.

So you stick the black lead on the battery (-) terminal, and "walk" the red lead of your meter along the outputs of the (in this order)

battery + post
main fuse
ignition switch
dimmer switch
headlight connector
headlight socket

Say your readings at these points are

12.61 volts
12.60 volts
12.55 volts some drop is expected, acceptable...
12.49 volts
11.9 volts ***
11.87 volts

*** THERE IT IS! Between the dimmer switch and the headlight connector you've lost .6 volts. That's a lot!

You probably need to pull the headlight connector apart, clean/wire brush the contacts, and put it back together.

Your headlight will be noticeably brighter!

So what is a "bad ground." If you've got a wire bolted or screwed to the frame of your bike, whether to provide a taillight ground or whatever, and that is all rusty, corroded, loose, or even...bolted on top of fresh paint instead of bare, bright metal....chances are yo'uve got a "bad ground" with anywhere from .01 to .2 ohms of unwanted resistance.

A voltage drop occurs when current flows through an unwanted, un-planned for, or unexpected resistance in a circuit.

Now get that meter out and go find 'em!

 

[wadenelson]

OK so here's the quiz. Today I bought a 25 foot extension cord that says it can carry 13 A. The wire inside is listed at 16 gauge. Assume you loaded it up with a full course of Christmas lights carryig the full 13 A. If you had 110 V going into it how much voltage wiould the Christmas lights actually get.

In other words what is the expected voltage drop through this cord.


As a sidenote recognize that most home outlets are 15A. That means you could overload this cord before the circuit breaker would pop and melt the cord or start a fire if you plugged in one too many strings of Christmas lights

 

[Critter]

drop of 1.9% so you would have around 107v. The calculator I used wouldn't allow 25' but would take 20.

 

[wadenelson]

OK using an online voltage drop calculator that's one way to do it it's the easy way. Now let's do it the hard way.

16 AWG stranded wire has approximately 4.08 ohms per thousand feet

so .102 ohms for 25'

V= IR = 13 * .102 = 1.36V

110-1.36 = 108.6V

Good job Critter!

 

[WintrSol]

OK using an online voltage drop calculator that's one way to do it it's the easy way. Now let's do it the hard way.

16 AWG stranded wire has approximately 4.08 ohms per thousand feet

so .102 ohms for 25'

Actually, that would be .204 Ohms for 50', since there are two 25' conductors.

V= IR = 13 * .102 = 1.36V

110-1.36 = 108.6V

Good job Critter!

So, double the losses. Also, many household power sources are actually 117VAC, so that could be 117-2.72 = 114.38VAC; if your power is actually 110VAC, you have 107.38VAC.

 

[wadenelson]

Actually, that would be .204 Ohms for 50', since there are two 25' conductors.

Good catch, WintrSol! Thank you.

SOME of you are going...so my Christmas lights see 107v instead of 110V. Big deal!

You're right. The elves will never notice the difference.

But on a headlight circuit on a motorcycle, the difference between 13.8V and 12.8V might be 50% loss of brightness. In the 12V world a 1V voltage drop, due to dirty contacts, a bad ground, too thin wire.... can be huge.

Especially with ECM's reading sensors. And starter circuits.

Where you actually see voltage drop problems with extension cords isn't Xmas lighting, but on jobsites, where someone runs a compressor on the end of a 100' cord, and THEN tries to run Skilsaws, etc. while the compressor is running.

 

[hogcowboy]

Or why your standard backup generator can't run your refrigerator, freezer, TV, ect.