Resistance

[tartansailor]

Armed with a digital multimeter, but being electronically challenged, I was wondering if someone could shed some light on what is an acceptable resistance through a fuse circuit panel on my CD-25, and what is not acceptable?
The plan is to measure the resistance from the hot buss bar to the wire exiting the panel going to each of the various lights, etc.

Also what voltage would be considered acceptable at the terminal ends of each positive wire, cable?

Thanks in advance for any help you may wish to offer.

Dick

[Kurt]

first I suggest you measure the impedance of the meter leads themselves. You will find about .1 ohm. Then do your measuring. You shouldn't find more than another .1 ohm from the buss bar to the wiring exiting the panel. As far a voltage goes, it depends on the gauge and length of the subject wire. I believe the West cataloge contains a chart that addresses voltage drop and distance for various gauge wire.

[Oswego John]

Dick,,

Kurt is correct. With good, clean, bright connections, protected by No-Korode or other similar anti oxidant, the difference in potential and resistance from the positive buss bar to the beginning of the branch circuit is barely negligible.

You have to figure in and allow for the distance round trip between the panel and the fixture. A good rule of thumb to use is that the DC voltage drop at the fixture should be less than 10% of the voltage at the panel. That means that if the panel voltage is 12VDC, the voltage at the fixture ideally should be higher than 11VDC.

You are dealing with a series resistance circuit. The feed wire is resistance, the filament is resistance as well as the return wire going to the negative buss is resistance. The smaller CMA (circular mil area) of the wire, the higher is it's resistance. The higher the total resistance of the circuit, the lower will be the working voltage at the lamp. Too small the size of wire used and/or too far a distance between the panel and the fixture results in the high resistance acting as a rheostat, thus resulting in a dimmer bulb than is normal. Extremely high resistance results in producing excessive wire heat. When the copper wire heats up, it becomes more resistant to current flow and a vicious cycle takes part. This can result in the plastic insulation melting on the conductors and an ensuing fire. If luck has it's way, the positive and negative wires will short out and the fuse will blow or the breaker kick out.

So it is good practice to don't skimp on wire size, and to use the properly rated overcurrent protection, whether fuse or circuit breaker.

This principle applies to other electrical uses, ie bilge pump and galley appliances. I remember that once, a guy asked me why his steaming light was so dim? He wired his mast lights with this tiny annunciator wire, about the size of a telephone conductor wire. He might have gotten away with it if the light was in his cockpit. It was the distance that killed him. The wire ran from the panel to his console, through a switch, then to the base of the mast, through ratty connections, up to the top of the mast, through the lamp(s) [shared negative common ], back down the mast and back over to the panel. No way, Jose.

You must include the total load plus the total distance in determining the wire size for a properly functioning device.

O J

[Mark Yashinsky]

What is the problem you are trying to diagnose?

Light circuits are one thing, but you also said "etc". If you are working on the starter circuit or other high draw application, acceptable number can be different.

[tartansailor]

Kurt,
Thanks for the 0.1 ohm #. Readings varied from zero F (open circuit) to as high as 8.7 ohms, with the preponderance being around 4.3 ohms. Therefore a new panel is in order.
I wonder if circuit breakers would be more reliable than fuses in a salt air environment?
I know I'm a clod; thought impedance only referred to AC circuits.

OJ
Thanks for the insight. Already made the decision to rewire as the present wiring is not tinned! In order to get a 3% or less voltage drop and get the whole wiring bundle through the hinge hole at the base of the mast we need a 12-5 multi cable. Unfortunately West Marine only sells a 14-5.

Mark,
All the circuits in question are lighting. It now looks like a LED conversion would be optimum, but again there is uncertainty as to LEDs reliability in a salt air environment; and they are not cheap.

Dick

[Kurt]

Do you really need a new panel? did you try loosening/retightening any connections under terminal screws? You should also remove and clean any fuses and associated contacts...then see if the impedance drops down to a more acceptable reading. Spray everything with a little Boeshield.
I also question why you feel a need for 12 gauge wire. I believe my mast is wired with 16 gauge. The steaming and deck light don't draw much more than 1 amp each. Assuming both are illuminated at the same time, the ground return will be conducting 2 amps which is still a very light load. Round trip on my CD27 from the breaker panel up the mast and back is 50 feet. 16 gauge wire is adequate for a 50 foot run on a 2 amp load.

[Larry DeMers]

A more practical test would be to operate the circuit under test, using some cabin lights or whatever to provide 3-5 amps draw.
Next, carefully measure the voltage from ground to the DC voltage input of the panel. Then measure the voltage from ground to the output of the circuit and compare the two readings.
You want to see what the loss is across the contact . If the difference is larger than 100 mV (.1v), then the contact is poor, and needs cleaning or replacement. This holds true for All dc contacts (switches, fuses, relays etc).

The consequence of ignoring the voltage drop is that current flowing thru this resistance will disipate heat as a byproduct. Enough heat, and the contact will fuse closed, and then it could get interesting! Interesting is bad here.

Update that panel. It's fun to do if you apply your mind a bit, and write down all ideas, research parts available from Blue Sea or the other vendors, get your wire gauges right, and then methodically improve the reliability, safety and working ease of your boat. it's worth real money in the end.

I use DC magnetic circuit breakers that are rated to be used as a switch (their action is softer, and with much less spring tension due to the use of magnetics in the trip mechanism). They are handy to be used to activate say the starboard and port side buss that powers the reading lights (and the individual light would have it's own local switch to operate it) and fans in the cabin. Kill the breaker, and the whole buss is dead and can be worked on, while the opposite side of the boat has operating lights and power. Anyway, that is just one way to design your systems.

Your are right. *Impedance* refers exclusively to AC circuits. We are working with DC circuits only here, so resistance is the only component that matters. This simplifies the math a lot..ohms law is the only tool needed (I=E/R). Where with impedance, the math gets relatively complicated, with the addition of the frequency component.



Cheers,

Larry DeMers
s/v DeLaMer

Kurt,
Thanks for the 0.1 ohm #. Readings varied from zero F (open circuit) to as high as 8.7 ohms, with the preponderance being around 4.3 ohms. :( Therefore a new panel is in order.
I wonder if circuit breakers would be more reliable than fuses in a salt air environment?
I know I'm a clod; thought impedance only referred to AC circuits.

OJ
Thanks for the insight. Already made the decision to rewire as the present wiring is not tinned! :roll: In order to get a 3% or less voltage drop and get the whole wiring bundle through the hinge hole at the base of the mast we need a 12-5 multi cable. Unfortunately West Marine only sells a 14-5.

Mark,
All the circuits in question are lighting. It now looks like a LED conversion would be optimum, but again there is uncertainty as to LEDs reliability in a salt air environment; and they are not cheap.

Dick

[Mark Yashinsky]

were you saying to measure from ground to the battery side of the breakers in the panel (measure the voltage from ground to the DC voltage input of the panel) and from ground to the ground side of the circuit (measure the voltage from ground to the output of the circuit)? I think you want to measure to voltage ACROSS each section of the circuit that SHOULD have no resistance (switches, breakers, wire, etc). This way, you can set the voltmeter to a low range and have a couple of digits of sensitivity. Most meters only show 3 to 4 digits, and when you starting off tens of volts, its hard to see the little stuff. It would be real easy to put the probes on each side of the breaker on a low voltage scale. ANY voltage would be bad (heat). Do the same for lengths of wire (dont forget the connectors!). Again, any voltage showing up is resistance that you dont want.

[Julian]

These resistance readings sound mighty high. Are these just across the panel or is the load still attached? If so you may be reading the path through your bulbs, battery, etc.

[Julian]

Just a by the way,

V=IR (Volts = amps x ohms)

P=I^2 R (Watts = amps squared x ohms)

Power in watts is also heat energy so if a device has a resistance of 10 ohms on a 12 volt circuit (i.e.1.2 amps) it will generate 14.4 watts of heat or light or fire!

[tartansailor]

Mark, Julian,
Thanks for the input. I did the drill where contacting the meter's probes and pushing a button eliminates internal meter resistance.
Then one probe is placed against the ground buss and the other pressed against the hot wire going into the fused switch for a particular circuit. There is where the resistance readings were taken.
Unfortunately the repeatability is very poor, varying in units of ohms, not tenths.
This panel box has been exposed to a salt atmosphere for 31 years, so a good cleaning is in order, then to repeat the tests.

Dick

[Joe Mac Phee]

Dick,

I am late coming to this thread and so may be plowing old ground, I apologize.

In your first message, you said that you were going to measure resisitance from the "hot bus bar to the load connection". Later you say the measurement was from the "ground bus to the load connection".

I make the following assumptions.

The battery switch is turned off.
All other load switches are turned off.

If measurement was made from ground bus to load wire, you have measured the resistance of the load.

If you measured from the "hot bus" to the load wire, you measured the resistance of the switch/fuse assembly.

Since you are measuring with a digital meter which usually has a very high input impedance, you are essentially measuring a dry contact (very little current) which is sometimes difficult to do.

The technique suggested by Larry is probably more reliable and more representative. By measuring the circuit under load you get to your primary concern, i.e. voltage drop in the circuit.

I replaced the panel on my Cape Dory but it was for different reasons. I wanted circuit breakers instead of fuses and I needed more than the 5 circuits that were considered adequate in 1977.

Happy wiring, and do not touch anything that would jeopardize your brilliant future.

Joe Mac Phee

[Oswego John]

Joe,

I'm glad that you wrote that last post.

I realize that I already spoke on this subject but I'd like to amend one thing and to add a little bit more that might focus in on the subject a little sharper

There was a time that it was recommended that the voltage drop on a battery powered, series resistance circuit should be less than 10%. Using 12V DC as the working voltage, 10% of that would result in more than 10.8V measured at the circuit's return to negative.

In this day and age, considering the improvements in battery design and wiring methods, those figures aren't acceptable. It is more desirable, as well as reasonable, to strive for .05% voltage drop or .6 volts, leaving a net voltage of 11.4V DC

On a different note, I find that it is much more convenient and easier to test these panels and circuits with a clamp-on type inductance meter. (Radio Shack, Harbor Freight, Home Depot,etc.)

There is a simple diagram that I used as a graphic learning aid to shed light on the understanding of a simple DC circuit. If you have 30 seconds of idle time, do the following:

#1 Draw a circle freehand on any scrap of paper, ie: the back of an envelope, any size diameter you choose.

#2 Imagining the circle as the face of a clock, draw a line from 9 o'clock to 3 o'clock

#3 Midway across that line, draw another line from the center of the circle downward to 6 o'clock

#4 In the upper half of the circle, write the letter "E", which represents voltage. (Electromotive Force)

#5 In the lower left quarter, write the letter "I", which represents Amperes, or the amount of current flow.

#6 In the lower right quarter, write the letter "R", which represents resistance measured in Ohms.

This is a graphic to help explain what Julian previously wrote. To find what you are looking for, using mathematics, place your finger over it and use the remaining formula.

If you are looking for voltage, cover the "E" and use the remaining letters, "I" x "R"

To determine resistance, cover "R" and you will result in "E" over "I", or volts (12) divided by amperage

The same goes to find amperage, cover "I" and divide "E" by "R", or divide voltage(12) by resistance in ohms.

Try it, it works. And just to keep the ball rolling, if you want to find watts in the circuit, multiply the volts by the amps, or "E" x "I" or "I²" x "R"

Whew,
O J

[bottomscraper]

OJ,
Sorry to hijack this thread but I gotta know.

How do you do the superscript '2' like below?

Thanks

Rich

[quote="Oswego John"]

"I²" x "R"

O J[/quote]

[Oswego John]

Hi Rich,

A piece of cake.

Activate numeric pad.

While depressing Alt, hit 253 (numeric)

Have a healthy, safe, Happy New Year,
O J