# Ohm reading when there is a short in a condenser motor

1. Dec 7, 2018

### fourthindiana

Preface to thread: The multimeter is set to read resistance in all the scenarios I ask about in this message.

The word "COMP" in the ladder diagram stands for compressor motor.

The three letters underneath the capacitor are DRC, which stands for Dual Run Capacitor.
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In the ladder diagram that I attached in the above picture, the start winding of the condenser motor is shorted to ground. The word "COMP" in the ladder diagram stands for compressor motor. About a week ago, an acquaintance of mine named Billy told me about what would happen in the situation portrayed in this ladder diagram. On the same day that Billy told me about this, I wrote down what he said to the best of my recollection. I wrote the following: "In the diagram, the condenser motor is shorted to ground, and the fact that the condenser is shorted to ground causes there to be resistance to ground at the contactor to ground."

Now I'm not so sure what Billy meant by "resistance to ground at the contactor to ground."

Billy told me that in the situation portrayed in the ladder diagram, my multimeter (set to Ohms) would have a resistance reading greater than zero Ohms but not an OL ( OL= open line). Billy told me that my multimeter would have a far different resistance reading if there was no short on the condenser motor or anywhere else than I would get if there was a short in the condenser motor. I'm not able to get in touch with Billy to get him to clarify this. I was hoping that you people could help me understand this.

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In the ladder diagram in the attached photograph, is the multimeter going to measure the resistance from everything to the right of where the red lead attaches to the contactor all the way to the ground?

In the ladder diagram in the attached photograph, is the multimeter only going to measure the resistance from where the red lead attaches to the contactor to the wire going from the contactor to the condenser motor to the start winding of the condenser motor while not measuring the resistance of anything in the compressor motor?

The answer cannot be yes to both of the above questions.

In the ladder diagram in the attached photograph, would the resistance measured by the multimeter be higher if there was no short to ground in the condenser motor?

Last edited: Dec 7, 2018
2. Dec 7, 2018

### jim hardy

Actually the answer can be "yes" to both questions. And it is.
That's because a multimeter when set to a Resistance scale actually measures not resistance but conductance.
It applies a modest voltage, typically 1.5 volts from its internal battery, measures the current that flows into the conductance of whatever you're testing.
Zero current means no conductance which is infinite resistance .
"Some" current means there's some conductance and the needle moves to a position on its scale in proportion to how much current flows.

But the marks underneath the needle don't say "current" they say "ohms".
The instrument designer made that meter scale show the result of an Ohm's law calculation that turns measured current to calculated resistance.
That's why ∞ ohms is on the left, that's zero current,
and zero ohms is on the right, that's some current.

In high end multimeters like a Simpson 260 , on the RX1 scale it might be as much as ¼ amp at zero ohms and that's why they use a "D" cell not a "AA".

....................
NOW, to answer your question (at last) :
In your diagram, the multimeter will read oops make that measure the sum of all the conductances to ground - zero for the compressor and some for the condenser motor.
And "some" + "zero" = just "some" .
But the scale underneath the needle is printed so it indicates resistance instead of conductance(it doesn't indicate current either, for that matter).
That's why the scale is non-linear - there's some algebra to get from current to resistance and that algebra is done for you by whoever printed the meter face..

Arithmetically, resistance is equal to $\frac{1}{Conductance}$ .
Conductances in parallel add , and that's sometimes handy.
Conductance is a legitimate electrical unit, it has name "Mho" which is "Ohm" spelt backward,
and they're just one anothers' inverse. ¼ Ohm = 4 Mhos ; ½ Mho = 2 Ohms.
We use Ohms a lot more often that we use Mhos , so just be aware they exist. You'll run across them sometime and when you do i hope you'll remember this conversation
...................................................

So Billy was right.
Put your finger where the red meter wire touches the black motor wires. Now move your finger along the black wires - if your finger can get to ground without having to jump a gap, the meter will "see" it and the needle will move.
If there's no path for current to get back to the meter, which in your diagram would be through ground, then the needle won't move.

The more the conductance the more current flows and the farther right the needle moves.

any help ?
In troubleshooting It's important to have an idea how your test equipment works - that way it won't fool you.

old jim

3. Dec 8, 2018

### fourthindiana

First of all, excellent post, jim hardy. Your post is very informative, but I still don't fully understand this.

I don't disagree with you, but I still don't understand how the answer can be yes to both questions.

I don't understand this sentence. To be more specific, I don't understand the meaning of "the current that flows into the conductance". I believe that conductance means how much current will flow through a path when one accounts for the resistance. To me, it sounds like the meaning of what you're saying goes something like the following: "It applies a modest voltage, typically 1.5 volts from its internal battery, measures the current that flows into the measurement of how well a given wire will transmit current." Please help me understand.

You say that the needle won't move. The only problem is that I don't definitely know that the default position of the needle is at infinite resistance. The photograph that you provided of an analog multimeter showed the needle at infinite resistance, but that might be because the multimeter was measuring something with infinite resistance at the time that the photograph was taken as opposed to infinite resistance being the default position of the needle on an analog multimeter.

If there was no short in the diagram at all, would the multimeter (set to measure resistance) read OL/Infinite resistance?

Last edited: Dec 8, 2018
4. Dec 8, 2018

### jim hardy

It's good that you question. It shows desire to understand the basics.

So much of understanding hangs on terminology that it's worth working on our electrical vocabulary.

We agree that "current" is flow of charges , usually along the inside of a wire.
"Resistance" is opposition to that flow, think of a restricting orifice . Ohms is a measure of the degree of restriction, analogous to smallness of the orifice.
"Conductance" is welcomeness to that flow, analogous to the largeness of the orifice.
It's that simple - is my restricting orifice half closed or half open ?

"I believe that conductance means how much current will flow through a path when one accounts for the resistance"
sounds to me correct.
You are understandably resisting the unfamiliar concept of conductance ( and believe it or not i actually didn't intend that pun..)

How much current will flow is found by Ohm's Law , $Amps = \frac{Volts}{Ohms}$

If your condenser fan happened to be 15 ohms to ground
and you applied 1.5 volts to ground
current of $\frac{1.5volts}{15 ohms} = {0.1amp}$ will flow
and an ohmmeter will indicate 15 ohms.

If your compressor suddenly becomes also 15 ohms to ground,
an additional 0.1 amps will flow for a total of 0.2 amps
and an ohmmeter will indicate $\frac{1.5volts}{0.2amps} = 7.5 ohms$

Point being - resistances in parallel do not add.
15 + 15 does not equal 7.5.

Conductances in parallel by contrast DO add.
15 ohms is $\frac{1}{15} = {0 .066667 mhos}.$
so let's check that
Conductance of the condenser motor = ${0 .066667 mhos}.$
+ Conductance of compressor motor = ${0 .066667 mhos}.$
.......................................................________________
Add conductances of both motors $= 0.1333333 mhos$

and $\frac{1}{0.133333 mhos} {= 7.5 ohms}$

.............................................................................................

so
???

Yes it will .
It will read the parallel combination of whatever resistances to ground are present.
To achieve that, it measures the sum of all currents flowing through however many conductances are present
then calculates and indicates equivalent resistance just as if there were only one path.

The reason i went into such detail was this statement
So - now you can tell Billy that you understand - a shorted condenser motor is one conductance to ground. But there may be others.

.
Ahhh i see. You're not accustomed to analog meters. I grew up with them and forget that they're increasingly uncommon.
Look up D'Arsonval meter.
They are a form of electromagnet, you push current through a coil to move the needle . A fine coil spring returns it to zero when current flow ceases..
In that picture up above , the meter needle is actually pointing slightly left of zero which is beyond the infinity mark on ohms scale.
That is its rest position ..
That's also where it would point when measuring an infinite resistance, because $\frac {1.5volts}{ ∞ ohms} = {Zero current}$ and the needle won't budge..
That particular meter should have been adjusted a little better sothe needle's rest position is right over the scale's zero marks.

Analog meters have a zero adjustment that should be tweaked to make sure the needle is on zero at rest .
That's because of two effects
1. The coil spring changes its length slightly with temperature .
2. When the meter is laying down horizontal versus standing up vertical, the weight of the needle itself has a small effect.

Here's another picture showing the zero adjust . Note it's not perfectly zeroed either.

old jim

5. Dec 8, 2018

### jim hardy

Yes it will.

Yes it will.
There is no conductance to ground in the compressor for it to measure. .
So the only one it can measure is the one in the condenser motor.

6. Dec 8, 2018

### fourthindiana

jim hardy,

Your post #4 on this thread is another excellent post on this thread.

My understanding of this is that the default position of the needle on an analog multimeter is at infinite resistance.

When one lead of a multimeter is attached to a circuit and the other lead of the multimeter goes to ground, I don't fully understand how a multimeter could apply voltage from the multimeter's internal battery to measure conductance through a circuit. When one lead of a multimeter is attached to a circuit and the other lead of the multimeter goes to ground, is it the case that a multimeter measures conductance in a circuit by applying voltage from its internal battery to a circuit and then the multimeter measures the current flow that leaves the multimeter? If not, how does a multimeter measure conductance in a circuit when one lead of a multimeter is attached to the circuit and the other lead of the multimeter goes to ground?

7. Dec 8, 2018

### jim hardy

If it seems i'm nitpicking semantics
it's because i find that
excruciating attention to detail in wording is the price of clear and precise communication

and we can have fun along the way !

8. Dec 8, 2018

### jim hardy

Yes !!! !

The rheostat adjusts maximum current , which is necessary as the battery ages ..
That one you set with meter leads shorted , zero ohms between them., so needle is over "0" on right end of Ohms scale.

When there's no path for current , which is zero Mhos which is ∞ Ohms,
needle returns to its rest position which should be over "∞" on left end of ohms scale.
You adjust rest position by very slightly rotating the coil spring mount.

That slotted screw head in photo rotates the coil spring mount by a few degrees .

old jim

Edit - credit for that fine image - it's at http://electriciantraining.tpub.com/14188/css/Figure-3-12-Simple-Ohmmeter-Circuit-94.htm

9. Dec 8, 2018

### fourthindiana

Okay. Thank you for your responses. I think I fully understand this now.

10. Dec 8, 2018

### jim hardy

11. Dec 9, 2018

### Svein

12. Dec 9, 2018