Do breakers add resistance to the circuit?

I am guessing very little. So I need some direction pls.

Or is it so minute, one doesn't have to worry about it?


Dnk....

Breaker contacts do add impedance to a circuit. Whether it is relatively insignificant depends on what you're calculating.
For voltage drop calcs, I wouldn't worry about it.
I have on a few occasions used a bit of engineering judgment to apply values for contact impedance to reduce calculated fault current values.

The latter.

Rememberthat Resistance = heating.
Breakers do get warm, the heating watts of which are often considered when computing the BTUs in A/C sizing of electrical equipment rooms.
I do have some data regarding the heating watts of some breakers stashed somewhere.

Doesn't the non-magnetic trip element essentially depend on I^2 x R, meaning there is some small amount of impedance or resistance in a heating element?

(Above is supposed to be I-squared x R)
Radar

A little off topic:
with the num lock on
If you hold down the alt key and enter 0178 you will get the "squared" (²) symbol when you let up the alt key
0216 will get you Ø, 0176 comes out degrees°, etc.

Here's the whole Table of ANSI characters

The current actually doesn't flow through the bimatalic element. The element is bonded in such a way that when it heats as to allow it to flex and bend in order to hit the trip bar and trip the breaker. The conductor heats the bimetalic element.
If you have a scrap 100a or larger breaker take it apart. It will be fun to look at.

SO Dave, you're saying the load current only passes thru the conductor which, with minimal resistance, causes some small degree of conductor heating which in turn heats the bimetalic strip, deforming it to the point of hitting the trip bar. Inverse time element - the time required to activate shortens as the degree of overcurrent increases.

Thanks for the tip electure. so what I meant to say previously is that P = I²R. Cool.

Radar

The short answer is: it depends.
Some breakers show almost no resistance when buzzed with an ohmeter others I have tested show 15 to 16 ohms. This is dependant on many factors; the design of the breaker, the ampacity, etc.

[This message has been edited by IanR (edited 01-05-2006).]

Measuring resistance across the line and load of a breaker does not measure resistance of the "bimetallic element. The resistance that is measured is that of the contacts themselves. Now, I know where this will lead.
Now I would expect the reply now to be "contact resistance is not good in that the contacts would heat." Contact resistance can vary all over the map. Each time the breaker is opened and closed it changes as the moving and stationary contacts rub together. And when measured with an ohmmeter the voltage used to measure the resistance is minimal when compared to that which the breaker is actually applied to. Under actual operating conditions with voltage applied one would have to actually measure the line to load voltage which represents the voltage drop due to resistance within the breaker. That voltage along with the current value could be used to calculate the resistance using ohm’s law.
Remember that voltage drop equals heating which affects the temperature rise of the breaker. Breakers do give off heat.
Should one elect to measure resistance across the line load of a breaker the results should never be used to determine the condition of the beaker unless the readings a very high of show and open circuit. The breaker should be exercised. Also in many cases what may look like a questionable condition often changes to an acceptable condition when placed into service over time.
The industry recognized testing of molded case circuit breakers is NEMA AB4.
The best thing that a person can do is simple take a breaker apart. It takes a lot if guessing as to what is happening inside away.

Makes sense Dave. Just doing the simple math on the values that Ian has measured on a breaker with, say, 100 amps of line current and applying I²R would imply generating around 150-KW of heat on each line, which just cannot be.

Radar

I agree, a breaker can not operate with 15 to 16 ohms as it would be melting down.