Voltage Drop

[ranger58sb]

Originally Posted by Rusty123

Voltage drop is 1.24 volts, not 0.24. But if no current is flowing (as he claims), the resistance wouldn't matter. Not saying he doesn't have some bad connections somewhere, but there's gotta be current in this scenario, which needs to be eliminated to resolve.

OP said 11.96, and Tashtego typed 10.96. Latter could have been a simple typo...

-Chris

[Jim Cate]

Back to the chase:
If there is zero current flowing, there can be NO voltage drop, no matter how stuffed the wire or terminals are. Ergo, if he has measured the drop correctly, there IS current flowing. If it were my boat, I would want to track this stray current down ASAP. It may simply be an unknown but legitimate load, but could be leakage current, busy eating up some critical component of the vessel.

Cheers,

Jim

[Terra Nova]

12.2 volts is about the level of a battery which has sat, unattended, for months with no external loads.

[Rusty123]

Originally Posted by ranger42c

OP said 11.96, and Tashtego typed 10.96. Latter could have been a simple typo...

-Chris

Right, my mistake.

[Ex-Calif]

Originally Posted by Terra Nova

12.2 volts is about the level of a battery which has sat, unattended, for months with no external loads.

Which leads to the quality of the battery or the quality of the meter.

My meter as one step above a go/no-go gauge. Measuring a 1/4 volt voltage drop on my meter would be futile.

Was the battery tested with the cables disconnected?

I gotta believe if this is an on-going issue that a 1/2-1 amp load would have drained the battery in short order.

This all isn't adding up yet...

[Nicholson58]

Originally Posted by Ketchgould

What's the easiest way to trace down my problem?

Thanks

You could pull a new wire. If your readings are true on volts, you must be sucking off amps somewhere. Perhaps a partially compromised wire? I had enough loss in my pool shed this spring to pop fuses. Turns out the insulation was failing in a conduit and bleeding off power. With sea salt, etc. you could be leaking power.

[Ketchgould]

The wire is fifty years old. I am going to replace and see what that does.

[F51]

I'm in the camp that says if there is a voltage drop there has to be current flowing. Rusty made a good point though. If the meter ground wire moved to a different ground at the far end of the 27 foot wire, that may account for some or all of the difference. There should not be much resistance in 27 feet of 14ga, about .08 ohms as was stated. So the current would be about 3 amps if my math is correct. That's a lot of current to be unaware of.

[jeepbluetj]

Originally Posted by Ex-Calif

My meter as one step above a go/no-go gauge. Measuring a 1/4 volt voltage drop on my meter would be futile.
........

This all isn't adding up yet...

My cheaper than a beer harbor freight meters (either $4 or free depending on if there's a coupon) have always been pretty accurate to about a hundreth. (compared to a fluke). It's amazing how well they work, until they don't. But at $4, I have more than one.


And a voltage drop on a wire with no load other than a high-impedance meter? Yup. Somethings not adding up. There's a load, or it's amazingly bad. (Cause the 10mohm+ impedance of the meter is a load... just a tiny tiny tiny one)

[smac999]

a bad connection or crimp, corroded wire, or a wire with only one strand left would still show the full voltage with no load.

he definatlly has a load. in which case the voltage drop may or may not be normal. need a amp meter to figure that one out.

if this wire comes off another bus / switch ect the drop may be in the wires before this one. (which means the load could be in the main wires and not this circuit)

[btrayfors]

Guys,

EVERYTHING in a DC circuit has resistance: wires, terminals and connectors, switches and, of course, the load(s).

Therefore, by definition, there is a voltage drop of some magnitude across each component in the circuit: wires, terminals and connectors, switches and, of course, the load(s).

The magnitude of the voltage drop depends on several things including, of course, the amount of current flowing in the circuit. No doubt.

HOWEVER, it is perfectly possible to measure a voltage drop across a component with NO current flowing apart from the miniscule amount required for a quality digital voltmeter to function. In a badly corroded circuit or one otherwise compromised (loose connections, badly corroded contacts, voltage leak, etc.) this may be quite large.

But, the true voltage drop in the circuit can only be measured accurately when the usual or intended LOAD is switched in. Very often, this drop will be much larger than that measured with no load other than a voltmeter.

If you don't believe there's a measurable voltage drop with each component, try measuring the DC resistance across any one. EXAMPLE:

1. touch the leads on your multimeter together tightly and measure the resistance (ohms). My Fluke 189 reads about 0.15 ohms resistance.

2. now, touch the same leads to an unattached battery lug, measuring the resistance from the stud to various parts of the contact surface where the lug fits onto the battery terminal. My Fluke 189 reads anywhere from 2 to 8 ohms resistance on a used, but not badly corroded lug.

Any resistance by definition means voltage drop -- Ohms Law!

Loose connections -- a very common occurrence on a boat -- mean a lot of resistance. So, too, do badly installed ring terminals and other connectors.

RE: the OP's situation, I agree that unless the wire were very badly compromised there would be only a small voltage drop. But, there would be one: wires, even new wires, have some resistance.

Bill

[Rusty123]

Bill's post prompted me to dust off some of my old basic electronics theory (a whole lot easier with the internet, incidentally). While I was familiar with the difference between the operation of an "ideal voltmeter" (infinite resistance), and a real meter (high resistance), my brain had converted this to a mental thumb rule of "no practical difference in the real world".

But like most thumb rules, sometimes they don't apply, and I think this is one of those cases.

A real voltmeter (with a resistance of 1 - 20 Mohms) has little effect on a circuit when connected in parallel with other loads of much lower resistance, which is the usual application. But in the OP's case, there are no loads (at least in theory), so the current flow is entirely a result of the meter's insertion into the circuit. Which means that the meter has a major effect on circuit parameters, including incurring voltage drops wherever resistance exists (which, as Bill points out, is everywhere).

Another way to view the scenario is that the OP is measuring his battery voltage with two different lead sets - the one that came with the meter, and another with an extra 27 feet of likely resistive wire spliced in. It's not surprising that these two measurements are different.

Thus, I believe the OP's measurement of voltage differential in different locations would be more meaningful with the circuits under normal load, so that the meter thus is an insignificant contributor. Alternatively, the resistance of the wire in question could be measured directly, as others have pointed out.

Please correct me if I'm off in LaLa land here.

[jeepbluetj]

Rusty, you're not completely in lala land. The impedance of the meter _will_ affect the circuit. But generally in a completely negligible way.

I.e. Lets say we have 10000 feet of 14 gauge wire hooked up to a 12v battery (good wire, good connections, etc) and let's just say 12v for giggles.

The resistance of 10K feet of 14g wire is ~50 ohms ( 0.0025 ohms/ft x 10K feet x 2 conductors)

Lets assume a total crap DMM with an input impedance of around 1mohm (decent ones are 10-20x that, some can get to the gig ohm range)

Total resistance of the circuit when measuring the voltage is 1,000,050 ohms.
Current flow when measuring is 12 uA (micro amps)
Voltage drop over the wire is ~.6 mV
Voltage measured would be 11.9994 volts (assuming you had equipment that could do it)

Therefore, either that wire is absolutely horrible beyond belief (like just a few atoms of copper connected) - or there's a load on it somewhere. Or the meter has terrible input impedance.

[Rusty123]

Yeah, I should have clarified that I'm not implying that the OP's wire isn't bad, only that my instinct as to expected voltage was faulty.

To extend your thought process a couple more steps -- if the wire drops .25 volts, the meter drops the remaining 11.75 volts (12 volts nominal). Which means that the resistance of the wire is 2.1% of the resistance of the meter (.25/11.75). So if the meter resistance is 1 M ohm, then the wire resistance must be 21 K ohm, right?

[Ex-Calif]

Originally Posted by Rusty123

But in the OP's case, there are no loads (at least in theory), so the current flow is entirely a result of the meter's insertion into the circuit. Which means that the meter has a major effect on circuit parameters, including incurring voltage drops wherever resistance exists (which, as Bill points out, is everywhere).

I don't think OP has confirmed or denied that he has isolated the load end yet. I don't know if he is testing a completed or open circuit at this point.

Also he hasn't told us what circuit this is. 27 feet is a longish run but what's at the end.

It would be kinda funny if we are talking about a bad wire and he is flashing the Alternator field circuit with the engine off...

Rgds