Basic Ohms law question

Status
Not open for further replies.
T

TLE2221

Member
Hello,
I have a very easy, basic electricity question related to ohms law.
How do you use the following example as a teaching tool related to ohms law?
The other day a coworker installed a used breaker in panel. It was given to him from a communications contractor that needed to feed a temporary circuit. Single pole, 20 amp breaker, 120 volts. (QO120)
The panel was feed from a three phase service.
Upon metering the lugs in the panel, the expected voltage was found. 110 volts, All phases to neutral, 208 phase to phase. Everything was as normal.
Upon testing the breaker, no load, (he tested it before he hooked up the branch circuit) ((smart)). He found that the load side of the breaker read 88 volts. It was determined that the breaker came from a Navy base and it had signs of corrosion and a white residue on the load side terminal connections. I would imagine the breaker had major corrosion on the internal over current components. Just a guess, as I have not had any experience with this issue yet. Navy yard and salt water?
He pulled out the breaker and did a resistance test. It read 7 ohms I do believe, but I might have heard him wrong. Anyhow, it was somewhere around that. This was from the point where the breaker clips onto the panel bus to the point where the wire terminations are. He was reading through all the internal components.
I got to thinking about it as it is related to ohms law, and wanted to prove out all the numbers.
He had a 22 volt drop when he metered the load side of the breaker. Wire terminal to neutral
Applied Voltage of 110 volts,
VD of 22
Resistance of 7 ohms through the breaker.
Can I prove out these numbers using ohms law? We don?t have a complete circuit; we just have a resistor in series with the voltage.
I can obviously find what my current would pull, E/R = I but but it still won?t help me prove out the Voltage drop.
Just in thinking about all things stupid, if you would have hooked the terminal of the breaker directly to the neutral bar, It would have created a 15 amp load through the breaker and could have proven ohms law if you would have used an amp clamp. This would just be a basic series circuit with one resistor.
I am just trying to nail down some basic electrical fundamentals as I feel like this is a perfect example to learn from. I can do the ohms law math very easy, but am having trouble applying it to case as it relates to the voltage drop of 22 volts. It is a lot easier for me to understand all the electrical theory aspects when you can put them into real life examples.

Thanks!
 
I'm trying to think how you could have this voltage drop without any current flowing through this circuit? I guess when measuring with your meter, the meter is still drawing some current though the circuit and through the meter? This current I imagine would be very small.

For a 22V drop at 7ohms this means there would be about 3.14A of current in the circuit?
 
Pitt123 said:
For a 22V drop at 7ohms this means there would be about 3.14A of current in the circuit?
Click to expand...
A voltmeter works by placing a very large resistor in parallel with the test leads, and measures the current through that resistor. So the current in the leads will be very small, far far less than 3.14 amps.
 
TLE2221 said:
Just in thinking about all things stupid, if you would have hooked the terminal of the breaker directly to the neutral bar, It would have created a 15 amp load through the breaker and could have proven ohms law if you would have used an amp clamp. This would just be a basic series circuit with one resistor.
Click to expand...
Yes, you are right, that would be stupid. :cool:

Your math works, in that 110 divided by 7 gives you about 15 amps. But that 7 ohm resistance is not likely to remain constant. By the time you installed the breaker, connected the wire, and turned the breaker on, you might have knocked loose much of the corrosion products, reducing the internal resistance from 7 ohms to perhaps .007 ohms, and you would have an arc flash on your hands (quite literally)!
 
charlie b said:
A voltmeter works by placing a very large resistor in parallel with the test leads, and measures the current through that resistor. So the current in the leads will be very small, far far less than 3.14 amps.
Click to expand...
I think that was precisely his point. If there was that much of a voltage drop through the breaker with only DVM current in voltmeter mode, I would expect that the resistance would be a lot higher than 22 ohms. Contact resistance from corrosion, however, is a lot different from a factory made resistor. Its value may change a lot with the applied voltage and current pushed through it, so the 7 ohms and the 22 volt drop may not relate to each other. I think the solution is obvious, though. Throw that breaker away. :D
 
Absolutely Mr. Beck,
Not smart at all, but it?s good to think hypothetically at times.
I should have said all things stupid, and hypothetical!
 
Status
Not open for further replies.
Top