Reconciling an apparent contradiction about Parallel Circuits

[fourthindiana]

I'm fairly sure that everything I wrote in the diagram in the attached photograph is consistent and accurate, but I don't understand how this can be. The rules involved seem to contradict each other. I hope you people can reconcile this for me.

It's my understanding that in the diagram in the attached photograph, the multimeter reading the voltage across the contactor would read zero voltage because the current would bypass the multimeter because the multimeter is 20,000 Ohms of resistance. The contactor is almost zero Ohms of resistance. I understand that the current at the contactor will take the path of least resistance and bypass the 20,000 Ohm multimeter by going through the contactor. However, why, then, would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the 10 Ohm resistor back to L2? It seems to me that the path of least resistance for the current from L1 to L2 would be from L1, through the closed contactor, and then through the 10 Ohm resistor, and then to L2.

If current will take the path of least resistance, why would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the contactor and the 10 Ohm resistor back to L2?

 

[Borek]

When you have resistances in parallel current goes through all of them. The larger the resistance, the lower the current, but it is there. It even goes trough the multimeter (are you sure it is just 20 kΩ?), although it is very, very low there. Note: contactor doesn't have 0 Ω resistance - its resistance is definitely very low, but not zero.

 

[fourthindiana]

When you have resistances in parallel current goes through all of them. The larger the resistance, the lower the current, but it is there. It even goes trough the multimeter (are you sure it is just 20 kΩ?), although it is very, very low there.

My HVAC instructor told me that my multimeter is 20,000 Ohms.

My Question: If current will take the path of least resistance, why would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the contactor and the 10 Ohm resistor back to L2?

Translation of your answer: Because in a parallel circuit, the current would go through the 50 Ohm resistor and the 10 Ohm resistor back to L2.

You didn't say why the current in a parallel circuit would not bypass the 50 Ohm resistor. You just said that the current would go through the 50 Ohm resistor. In the OP, I made it clear that I knew/know that the current will go through the 50 Ohm resistor. I appreciate your effort to help me understand, but you did not really answer the question.

You're saying that in a parallel circuit, the current goes through all the resistors. I'm basically asking why the current goes through all the resistors

Note: contactor doesn't have 0 Ω resistance - its resistance is definitely very low, but not zero.

Look at what I wrote in the OP: "The contactor is ALMOST zero Ohms of resistance." Since I said the contactor is almost zero Ohms of resistance instead of saying that the contactor is zero Ohms of resistance, doesn't that imply that I know that the resistance is an infinitesimally amount above zero?

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Edited to add this post script:

I just realized a possible significance of something you wrote: "The current even goes through the multimeter."

Ah--- this might be the key to solving this. If the current goes through the multimeter, then I don't think that there is a contradiction any more.

Is the resistance of the multimeter so high that the voltage is infinitessimaly small but higher than zero?

 

[Borek]

My Question: If current will take the path of least resistance

Answer: it will NOT take the path of the least resistance. It will flow through ALL resistances.

You start with a wrong assumption, no wonder you end with a contradiction.

 

[fourthindiana]

Answer: it will NOT take the path of the least resistance. It will flow through ALL resistances.

You start with a wrong assumption, no wonder you end with a contradiction.

Yeah. I think you probably did answer my question.

Please look at the edit I made at the end of post #3 of this thread and comment.

 

[Borek]

Is the resistance of the multimeter so high that the voltage is infinitessimaly small but higher than zero?

I suppose you mean current, not voltage, and it definitely is not infinitesimally small. If the multimeter has internal resistance of 20 kΩ you can expect current in μA or mA range (depending on the voltage, just apply the Ohm's law).

 

[fourthindiana]

I suppose you mean current, not voltage, and it definitely is not infinitesimally small. If the multimeter has internal resistance of 20 kΩ you can expect current in μA or mA range (depending on the voltage, just apply the Ohm's law).

I did/do mean to ask about the voltage.

You implied that the resistance of a multimeter would be much higher than 20,000 Ohms. I have a Fluke 116 multimeter. You are probably familiar with those. What do you think would be the resistance of a Fluke 116 Multimeter?

Assuming that the resistance of my multimeter would be whatever you would estimate, what would the voltage be at my multimeter in which the leads are across the contactor?

Wouldn't the voltage be just slightly above zero?

 

[Asymptotic]

My HVAC instructor told me that my multimeter is 20,000 Ohms.

Almost certainly the meter's voltage range is rated 20,000 ohms per volt.
It presents 20,000 ohms of resistance at the meter leads when dialed to the 1 volt full scale range, 200,000 ohms on the 10V range, 2 meg on the 100V range, and so on.

 

[fourthindiana]

Almost certainly the meter's voltage range is rated 20,000 ohms per volt.
It presents 20,000 ohms of resistance at the meter leads when dialed to the 1 volt full scale range, 200,000 ohms on the 10V range, 2 meg on the 100V range, and so on.

So in the situation presented in the diagram in the attached photograph, what would be the voltage reading on the multimeter across the contacts of the closed contactor?

 

[Asymptotic]

So in the situation presented in the diagram in the attached photograph, what would be the voltage reading on the multimeter across the contacts of the contactor?

What range is the meter on?

One approach is solve for the combined parallel resistances, then calculate current using Ohm's law.
Ignoring contact resistance for the time being, how much current flows through the load?

Actual values for contact resistance can be hard to come by on spec sheets, and falls within a relatively wide range, but for a set of largish contacts in good condition will be on the order of 50 microohms or less.

Lets say the contact in your circuit is 100 microohms ... what is the calculated voltage drop for the current you solved for before?

---------------------------
For example, voltage drop is 100 microvolts for 1 amp of current flowing through a 100 microohm switch resistance. Is the meter in question able to read this low a value?

 

[fourthindiana]

What range is the meter on?

I have a Fluke 116 multimeter. The Fluke 116 Multimeter does not have an adjustable range for voltage. I don't see what the range would matter anyway, since the voltage would be so low. It's not like the voltage could be too high.

One approach is solve for the combined parallel resistances, then calculate current using Ohm's law.
Ignoring contact resistance for the time being, how much current flows through the load?

I don't know how much current flows through the load. It's a hypothetical diagram. Please look at the diagram in the attached photograph.

Actual values for contact resistance can be hard to come by on spec sheets, and falls within a relatively wide range, but for a set of largish contacts in good condition will be on the order of 50 microohms or less.

Wouldn't the resistance of the contactor be so low as to be negligible?

Lets say the contact in your circuit is 100 microohms ... what is the calculated voltage drop for the current you solved for before?

I don't know how to calculate voltage drop. Heck, I don't even know how to calculate the voltage in the multimeter on the contactor. That's what I am asking you.

By Ohm's law, I am assuming that the current going through the 10 Ohm resistor is 23 Amps. And I am assuming that the current going through the 50 Ohm resistor is 4.6 Amps.

---------------------------

For example, voltage drop is 100 microvolts for 1 amp of current flowing through a 100 microohm switch resistance. Is the meter in question able to read this low a value?

I don't know.

 

[Asymptotic]

Been looking at Fluke 116 specs. Probably not a bad idea for you to dig thorough it as well.
Turns out meter input impedance is only 3,000 ohms when in auto-ranging mode.

I don't know how much current flows through the load. It's a hypothetical diagram. Please look at the diagram in the attached photograph.

It can be calculated. A short-cut for calculating the total resistance of two resistors in parallel is (R1*R2)/(R1+R2). For 10 and 50 ohms, this works out to 8.33 ohms.
Approximately 230V is across them (minus the unknown contact resistance), so since I=E/R the current is 230/8.33 ohms = 1.2 amps.

Consider a contact resistance of 100 microohms. Voltage drop is E=I*R, so with 1.2 amps through 100 microohms the voltage drop will be 120 microvolts (0.12 millivolts). Can the Fluke 116 read a voltage this low?

If it is set to the most sensitive (600 mV) AC voltage range, resolution is 0.1 mV. Measurement across a 100 microohm contact set will at best jiggle the display's least significant digit.

"Why measure voltage drop across a set of contacts at all then?" you may be asking. Because if they are in bad shape then voltage drop will be higher (sometimes, much higher - up to 230V, in this example, if they've blown completely open), and such tests are a useful troubleshooting tool.

Fluke 116 Manual
https://www.myflukestore.com/pdfs/c...er/116/manual/fluke_116_multimeter_manual.pdf

 

[fourthindiana]

Been looking at Fluke 116 specs. Probably not a bad idea for you to dig thorough it as well.
Turns out meter input impedance is only 3,000 ohms when in auto-ranging mode.

I don't know what auto-ranging mode is. As I said before, there is not adjustable voltage ranges on a Fluke 116 Multimeter. There is just a voltage setting for alternating current, and a different voltage setting for direct current, and that's it.

It can be calculated. A short-cut for calculating the total resistance of two resistors in parallel is (R1*R2)/(R1+R2). For 10 and 50 ohms, this works out to 8.33 ohms.
Approximately 230V is across them (minus the unknown contact resistance), so since I=E/R the current is 230/8.33 ohms = 1.2 amps.

The (R1*R2)/(R1 + R2) formula is only valid for when there are two resistors. I think that the resistance of the multimeter across the contacts makes the (R1*R2)/(R1+R2) formula invalid.

Consider a contact resistance of 100 microohms. Voltage drop is E=I*R, so with 1.2 amps through 100 microohms the voltage drop will be 120 microvolts (0.12 millivolts). Can the Fluke 116 read a voltage this low?

I don't know.

"Why measure voltage drop across a set of contacts at all then?" you may be asking. Because if they are in bad shape then voltage drop will be higher (sometimes, much higher - up to 230V, in this example, if they've blown completely open), and such tests are a useful troubleshooting tool.

For my hypothetical diagram in the attached photograph, let's assume that the contacts of the contactor are in excellent condition.

 

[Asymptotic]

I don't know what auto-ranging mode is. As I said before, there is not adjustable voltage ranges on a Fluke 116 Multimeter. There is just a voltage setting for alternating current, and a different voltage setting for direct current, and that's it.

Range selection is shown on page 9 in the manual.

The (R1*R2)/(R1 + R2) formula is only valid for when there are two resistors. I think that the resistance of the multimeter across the contacts makes the (R1*R2)/(R1+R2) formula invalid.

For my hypothetical diagram in the attached photograph, let's assume that the contacts of the contactor are in excellent condition.

Excellent condition doesn't mean zero ohms. All real world contacts have some (hopefully, very small) resistance.

Back to the original question.

It's my understanding that in the diagram in the attached photograph, the multimeter reading the voltage across the contactor would read zero voltage because the current would bypass the multimeter because the multimeter is 20,000 Ohms of resistance. The contactor is almost zero Ohms of resistance. I understand that the current at the contactor will take the path of least resistance and bypass the 20,000 Ohm multimeter by going through the contactor. However, why, then, would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the 10 Ohm resistor back to L2? It seems to me that the path of least resistance for the current from L1 to L2 would be from L1, through the closed contactor, and then through the 10 Ohm resistor, and then to L2.

Your understanding is wrong. Neither current path is bypassed. Current flows through both branches, it's just that more current flows through the lower resistance branch. For the parallel branch consisting of the meter and contact set, most of the current goes through the contact set, and only a little current flows through the voltmeter. For the 10 ohm and 50 ohm relay coil branch, most of the current flows through the 10 ohm coil, and the remainder flows through to 50 ohm coil.

Use Ohm's law to figure out how much current goes through each coil.

 

[phinds]

View attachment 235501 I'm fairly sure that everything I wrote in the diagram in the attached photograph is consistent and accurate, but I don't understand how this can be. The rules involved seem to contradict each other. I hope you people can reconcile this for me.

It's my understanding that in the diagram in the attached photograph, the multimeter reading the voltage across the contactor would read zero voltage because the current would bypass the multimeter because the multimeter is 20,000 Ohms of resistance. The contactor is almost zero Ohms of resistance. I understand that the current at the contactor will take the path of least resistance and bypass the 20,000 Ohm multimeter by going through the contactor. However, why, then, would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the 10 Ohm resistor back to L2? It seems to me that the path of least resistance for the current from L1 to L2 would be from L1, through the closed contactor, and then through the 10 Ohm resistor, and then to L2.

If current will take the path of least resistance, why would the current not bypass the 50 Ohm resistor in the parallel circuit and just only go through the contactor and the 10 Ohm resistor back to L2?

First off, the drawing is absurd. You have 115 volts on each side so the voltage across the elements is zero, yet you show it as 230 volts. Second, you have a variable capacitor shown apparently intended as a closed switch.

SO ... I have assumed that you mean that there is a voltage difference across the elements of 230 volts and I have redrawn the circuit that way, leaving out the closed switch (yes, it is not EXACTLY zero but we're going to assume that it is for now).

All this back and forth about what's doing what is just maddening to me. This is a staggeringly simple circuit. Here it is redrawn with your meter shown with a 20Kohm internal resistance in series with the galvanometer, and then analyzed:

There just isn't anything else to say about this circuit.

 

[fourthindiana]

First off, the drawing is absurd. You have 115 volts on each side so the voltage across the elements is zero, yet you show it as 230 volts. Second, you have a variable capacitor shown apparently intended as a closed switch.

I'm a student at a trade school, majoring in HVAC. I'm fairly sure that my instructor would agree with me that since each leg of power is 115 volts, the voltage across the 10-Ohm resistor is 230 volts, and the voltage across the 50-Ohm resistor is 230 volts. I think that when my instructor speaks of when the voltage is 230 volts, he means the voltage difference is 230 volts. The way that I drew the ladder diagram in the OP is similar to how ladder diagrams are drawn in HVAC textbooks.

There is no variable capacitor in the ladder diagram I drew. The switch is a normally closed contactor.

SO ... I have assumed that you mean that there is a voltage difference across the elements of 230 volts and I have redrawn the circuit that way, leaving out the closed switch (yes, it is not EXACTLY zero but we're going to assume that it is for now).

I don't know what the difference is between saying the voltage across the resistors is 230 volts and saying the voltage difference across the resistors is 230 volts.

All this back and forth about what's doing what is just maddening to me. This is a staggeringly simple circuit. Here it is redrawn with your meter shown with a 20Kohm internal resistance in series with the galvanometer, and then analyzed:

Please be patient. I'm a student at a trade school, majoring in HVAC, not an electrical engineer.

There just isn't anything else to say about this circuit.

There is plenty to say. I need clarification. I still don't understand.

Does the 11 mA going through the multimeter stand for 11.5 milliamps or 11.5 microamps.?

Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

 

[jim hardy]

I think that after your thread on shorted condenser fan motor you are now equipped to grasp this one using just your intuitiion

Please look at the edit I made at the end of post #3 of this thread and comment.

here's that edit

Edited to add this post script:

I just realized a possible significance of something you wrote: "The current even goes through the multimeter."

Ah--- this might be the key to solving this. If the current goes through the multimeter, then I don't think that there is a contradiction any more.

Is the resistance of the multimeter so high that the voltage is infinitessimaly small but higher than zero?

That's SOOOO doggone close...

For my hypothetical diagram in the attached photograph, let's assume that the contacts of the contactor are in excellent condition.

Okay.
Go back to that concept of :"conductance", which is the open-ness of a path to current flow.
The contactor is in good condition meaning it welcomes current flow.
Its contact has conductance of many Mhos(Siemens) , hundreds of them in fact if not thousands.
which means it has Resistance of just hundredths (if not thousandths) of an Ohm .
Remember Ohm's Law Volts = Current X Resistance
and reconsider your statement

Is the resistance of the multimeter so high that the voltage is infinitessimaly small but higher than zero?

No, the resistance of the multimeter so high contactor is so low that the voltage is infinitesimally small but higher than zero !

See ? You knew that already.
Much of learning is discovering what we already know.

old jim

 

[fourthindiana]

I think I am getting a better understanding of this.

I think that after your thread on shorted condenser fan motor you are now equipped to grasp this one using just your intuitiion

The information you told me on the thread about the shorted condenser fan motor certainly helped me understand this.

Okay.
Go back to that concept of :"conductance", which is the open-ness of a path to current flow.
The contactor is in good condition meaning it welcomes current flow.
Its contact has conductance of many Mhos(Siemens) , hundreds of them in fact if not thousands.
which means it has Resistance of just hundredths (if not thousandths) of an Ohm .
Remember Ohm's Law Volts = Current X Resistance
and reconsider your statement

No, the resistance of the multimeter so high contactor is so low that the voltage is infinitesimally small but higher than zero !

See ? You knew that already.
Much of learning is discovering what we already know.

old jim

According to what phinds told me, the current going through the multimeter across the contacts of the contactor is: Current= 230Volts/20,000-Ohms
Current = .0115 amps

You said that the resistance across the contactor would be hundredths (if not thousandths) of an Ohm.
I will estimate the resistance to be .05-Ohms.

Therefore, Voltage= .0115 amp X .05-Ohms
Voltage= .000575 volts
Is that a reasonable estimation and calculation of the voltage going through the multimeter across the contacts of the contactor?One thing I don't understand: If the multimeter reads only .000575 volts across the contacts of the contactor, why did phinds calculate the current going through the multimeter across the contacts of the contactor using the equation Current= 230 volts/20,000-Ohm? If the voltage at the multimeter going across the contacts of the contactor is only .000575volts, why was phinds able to put 230 volts into the calculation of the current instead of saying .000575 volts?

 

[phinds]

I'm a student at a trade school, majoring in HVAC. I'm fairly sure that my instructor would agree with me that since each leg of power is 115 volts, the voltage across the 10-Ohm resistor is 230 volts, and the voltage across the 50-Ohm resistor is 230 volts. I think that when my instructor speaks of when the voltage is 230 volts, he means the voltage difference is 230 volts. The way that I drew the ladder diagram in the OP is similar to how ladder diagrams are drawn in HVAC textbooks.

My bad. I'm use to dealing w/ DC and there was no indication that this was AC.

EDIT: and by the way, if the two sources are in phase then the voltage IS zero across the elements.

There is no variable capacitor in the ladder diagram I drew. The switch is a normally closed contactor.

the symbol you used was the symbol for a variable capacitor. As I said, I assumed this was what you were talking about when you said closed switch.

Does the 11 mA going through the multimeter stand for 11.5 milliamps or 11.5 microamps.?

Milliamps. It's simple arithmetic.

Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

If you put a voltmeter across a short it reads zero. The meter has a very high resistance and the short has very low. An ideal short has zero. In the real world I suppose the meter might have a microamp or two flowing through it. Because of the current difference, essentially nothing flows through the galvanometer so it reads zero.

 

[jim hardy]

If the voltage at the multimeter going across the contacts of the contactor is only .000575volts, why was phinds able to put 230 volts into the calculation of the current instead of saying .000575 volts?

Remember my quip about 'excruciating attention to detail in wording" ? I think there was a simple mis-communication. It happens.

He is right about the green one
and you are right about the red one.

old jim

 

[phinds]

Remember my quip about 'excruciating attention to detail in wording" ? I think there was a simple mis-communication. It happens.
View attachment 235520

He is right about the green one
and you are right about the red one.

old jim

Yeah, there are two different things going on here and my impatience got the better of me so I wasn't as careful as I should have been. My most recent post talks about the meter across the "contractor" (by which I assume he means the switch) and my previous post added the meter across the resistors.

 

[jim hardy]

Yep.
It'll feel good when we see "The Light Come On" for him.
I think he's progressing well.

old jim

 

[fourthindiana]

I'm still confused.

My bad. I'm use to dealing w/ DC and there was no indication that this was AC.

This is an AC circuit. I should have said that in the OP. My apologies.

EDIT: and by the way, if the two sources are in phase then the voltage IS zero across the elements.

That is totally in contradiction to what my HVAC instructor writes. Why does my HVAC instructor say otherwise? Is it because the voltage is zero across the elements but the voltage difference (whatever that means) is not zero across the elements?

the symbol you used was the symbol for a variable capacitor. As I said, I assumed this was what you were talking about when you said closed switch.

In HVAC, that is the symbol for a normally closed contactor. A contactor is a type of relay.

If you put a voltmeter across a short it reads zero. The meter has a very high resistance and the short has very low. An ideal short has zero. In the real world I suppose the meter might have a microamp or two flowing through it. Because of the current difference, essentially nothing flows through the galvanometer so it reads zero.

But I am asking about putting a voltmeter across a normally closed contactor, not a short. Why are you talking about shorts?

 

[fourthindiana]

Remember my quip about 'excruciating attention to detail in wording" ? I think there was a simple mis-communication. It happens.
View attachment 235520

He is right about the green one
and you are right about the red one.

old jim

But in phinds' diagram on the right on post #15 on this thread, phinds wrote: "V10= 230 V, so
I10= 230V/10-Ohm= 23 amp

V50= 230 V, so
I50= 230V/50-Ohm= 4.6 amp

Vm= 230 Volts, so
Im= 230/20,000= 11.5 milliamps"The Vm and I am parts of that diagram had to mean Voltage of the multimeter resistor and Current of the multimeter resistor.

Frankly, jim hardy, I'm not 100% sure what part of what phinds wrote that you are referring to when you say that phinds was referring to the 10-Ohm resistor.

Therefore, I still don't know the answer to the following question: Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

 

[fourthindiana]

Yep.
It'll feel good when we see "The Light Come On" for him.
I think he's progressing well.

old jim

jim hardy, I think that the key to my understanding this lies in the following questions: If the multimeter reads only .000575 volts across the contacts of the contactor, why did phinds calculate the current going through the multimeter across the contacts of the contactor using the equation Current= 230 volts/20,000-Ohm? If the voltage at the multimeter going across the contacts of the contactor is only .000575volts, why was phinds able to put 230 volts into the calculation of the current instead of saying .000575 volts?

When phinds did the calculation to find the current, phinds had to be referring to the multimeter because phinds wrote: Current= 230 volts/20,000-Ohm . The only resistor in my diagram with 20,000-Ohm of resistance is the multimeter.

 

[jim hardy]

Frankly, jim hardy, I'm not 100% sure what part of what phinds wrote that you are referring to when you say that phinds was referring to the 10-Ohm resistor.

no, i think he was referring to the METER CONNECTED ACROSS the 10Ω resistor , the meter i circled in green.in your photo
this one in his sketch from post 15
a picture is worth a thousand not-quite-precise words ?

it is with considerable trepidation i venture to say what somebody else meant

but i feel pretty safe I'm right on this one. Corrections welcome.

 

[Tom.G]

But I am asking about putting a voltmeter across a normally closed contactor, not a short. Why are you talking about shorts?

short circuit
noun : a connection of comparatively low resistance accidentally or intentionally made between points on a circuit between which the resistance is normally much greater
(from: https://www.merriam-webster.com/dictionary/short-circuit)

This is one of those cases where a term "short circuit", or "short", can have many subtle meanings depending on where it is used. A contactor, as you know, has a very low resistance thru its closed contacts. When it was stated there was a "short" across the meter, it was the bolded parts of the definition above that the writer had in mind. With a very low resistance connection (a "short") across the meter, the measured voltage would be extremely low... perhaps so low that the meter would read zero.

For the confusion between closed contacts and a variable capacitor, that is more common than it should be and I've seen it here on PF a few times. The drawing standards for Electronic schematics and Industrial Control wiring diagrams use different definitions for the same symbol.

The upper item, two parallel lines, in Industrial Control is an open contact, usually of a relay or contactor. In Electronics it is a Capacitor.
The lower item, with a diagonal line thru it, in Industrial Control is a closed contact. In Electronics (if there is an arrowhead on the diagonal) it is a Variable Capacitor.

EDIT:
And just to confuse things more, if the upper item has a circle around it, in Architectural drawings it is a wall outlet!

Hope this helps clear things up a little!

Cheers,
Tom

 

[jim hardy]

Back to preciseness in wording.

A closed contact is almost a short circuit...

you estimated

You said that the resistance across the contactor would be hundredths (if not thousandths) of an Ohm.
I will estimate the resistance to be .05-Ohms.

a pretty reasonable estimate.
Compared to the 10 Ω resistor that's what, 200X greater conductance ?

As Ohm's law becomes more natural to you , you too will become accustomed to the shortcuts we all take .
They're confusing to beginers.
Perhaps it's true - familiarity breeds perhaps not contempt, but at least inattention to detail.

old jim

 

[Asymptotic]

Apologies for getting the math (wildly) wrong; @phinds numbers are correct, and I updated the sketch to match.

Therefore, I still don't know the answer to the following question: Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

Lets view the question from another perspective. Fluke 116 meter specs for the default, auto-ranging voltmeter function "Auto-V LoZ" is a 600.0 V range with a resolution of 0.1 V. Per the manual, auto-ranging also "sets the Meter’s input impedance to approximately 3 kΩ to reduce the possibility of false readings due to ghost voltages."

We don't know what the contact resistance is (although we do know it ought to be very nearly zero ohms), the current through the contacts is 27.6 amps, and meter resolution is 0.1 volts. Solving Ohm's law for resistance (R=E/I), 0.1V/27.6 amps = 0.0036 ohms.

This particular meter will read contact voltage drop as no more than 0.1V when contact resistance is 0.0036 ohms (3.6 milliohms) or less. Chances are this is why your instructor claims it will read zero voltage. It depends on the characteristics of the contacts in question, but 3.6 milliohm (preferably, significantly less) would be necessary for a contact set capable of conducting this high a current. Why do I say this?

Power=Volts * Amps = I²R.
27.6² * 0.0036 ohms = 2.76 watts, and that translates to a fairly high amount of contact heating. For comparison, specs for a typical "ice-cube" relay with 6A rated contacts is 50 milliohms maximum contact resistance, which produces 1.8 watts (6² A * 0.05 ohms).

In your example circuit, if contact resistance was much higher than 0.0036 ohms they'd be dissipating even more heat, and would tend to be unreliable. For instance, power dissipation for a 0.05 ohm (50 milliohm) contact resistance at 27.6 amp is 38 watts. This is on par for a mid-range soldering iron heating element, and would fairly quickly burn up the contacts.

That is totally in contradiction to what my HVAC instructor writes. Why does my HVAC instructor say otherwise? Is it because the voltage is zero across the elements but the voltage difference (whatever that means) is not zero across the elements?

My assumption reading your sketch is of a typical U.S. residential circuit where L1 to neutral is 115V, L2 to neutral is 115V, and voltage between L1 and L2 is 230V. No neutral is shown in the sketch, and makes it somewhat ambiguous.

But I am asking about putting a voltmeter across a normally closed contactor, not a short. Why are you talking about shorts?

To add to the answers above, imagine what would happen if a closed contact set was wired across a 230V source. For the 0.0036 ohm contact set postulated above, I=E/R, 230V/0.0036Ω, or just shy of 64,000 amps. Chances are the source wouldn't be able to provide that much current, and circuit protection (breakers, and/or fuses) would clear, but probably not before the contacts were destroyed.

 

[fourthindiana]

I'm still confused.

Apologies for getting the math (wildly) wrong; @phinds numbers are correct, and I updated the sketch to match.

View attachment 235530

Lets view the question from another perspective. Fluke 116 meter specs for the default, auto-ranging voltmeter function "Auto-V LoZ" is a 600.0 V range with a resolution of 0.1 V. Per the manual, auto-ranging also "sets the Meter’s input impedance to approximately 3 kΩ to reduce the possibility of false readings due to ghost voltages."

We don't know what the contact resistance is (although we do know it ought to be very nearly zero ohms), the current through the contacts is 27.6 amps, and meter resolution is 0.1 volts. Solving Ohm's law for resistance (R=E/I), 0.1V/27.6 amps = 0.0036 ohms.

I know Ohm's Law. I know that R=E/I. However, I'm still confused by how your equation solving for resistance across the contacts. You say that the resistance across the contacts is 0.1 Volts/27.6 amps. You're saying that the voltage going through the contacts is 0.1 volts because the meter resolution of a Fluke 116 multimeter is 0.1 volts. But what does meter resolution mean? Why is the meter resolution the value we should plug in for the voltage when calculating the resistance going through the contactor?

This particular meter will read contact voltage drop as no more than 0.1V when contact resistance is 0.0036 ohms (3.6 milliohms) or less. Chances are this is why your instructor claims it will read zero voltage.

Here's my interpretation of what you're writing here: If the resistance of the contactor is .0036-Ohms or less, the voltage drop that would result from the resistance of the contactor would be no more than 0.1Volts.

But doesn't that just mean that if the resistance of the contactor is .0036-Ohms or less, that the voltage going through the contactor would be reduced by no more than 0.1V from what the voltage going through the contactor would be if the contactor literally had no resistance at all? If the contactor had no resistance at all, wouldn't there be 230 volts at the contactor? In other words, if the resistance of the contactor is .0036-Ohms or less, then couldn't the voltage at the contacts be 229.9 volts? I think I might not understand a key feature of electricity here.

It depends on the characteristics of the contacts in question, but 3.6 milliohm (preferably, significantly less) would be necessary for a contact set capable of conducting this high a current. Why do I say this?

Power=Volts * Amps = I²R.
27.6² * 0.0036 ohms = 2.76 watts, and that translates to a fairly high amount of contact heating. For comparison, specs for a typical "ice-cube" relay with 6A rated contacts is 50 milliohms maximum contact resistance, which produces 1.8 watts (6² A * 0.05 ohms).

In your example circuit, if contact resistance was much higher than 0.0036 ohms they'd be dissipating even more heat, and would tend to be unreliable. For instance, power dissipation for a 0.05 ohm (50 milliohm) contact resistance at 27.6 amp is 38 watts. This is on par for a mid-range soldering iron heating element, and would fairly quickly burn up the contacts.

I think I understand this part of your post.

My assumption reading your sketch is of a typical U.S. residential circuit where L1 to neutral is 115V, L2 to neutral is 115V, and voltage between L1 and L2 is 230V. No neutral is shown in the sketch, and makes it somewhat ambiguous.

Yes, I meant for this ladder diagram to be a ladder diagram of a typical US Residential circuit. There are two legs of power, L1 and L2. I don't see how my not having drawn a neutral in this ladder diagram makes the ladder diagram somewhat ambiguous. I think that most houses in the USA have two legs of power.

 

[Tom.G]

If the contactor had no resistance at all, wouldn't there be 230 volts at the contactor? In other words, if the resistance of the contactor is .0036-Ohms or less, then couldn't the voltage at the contacts be 229.9 volts? I think I might not understand a key feature of electricity here.

It sounds like you have got it either correct or very close!
Here are some details.

With the contacts closed:

1. If you put one meter lead on L2 and the other lead on L1, you will read the power line voltage, in this case 230V.
2. If you move the meter lead from L1 to the contactor terminal with the resistors you will read about 229.9V, the difference between the 230V and the 0.1 V voltage drop across the contacts.
3. If you move the meter lead from L2 to L1 you will read the voltage across the contacts, about 0.1V.
4. If the contacts Open, you will read almost 230V, the voltage across the contacts.

Comments:
Step 2 above is the voltage across the resistors.
Step 4 will actually be 229.4V because the 3000 Ohm meter allows 0.077A to flow. This 0.077A causes a 0.64V drop across the resistors.

Edit: added image

Cheers,
Tom

 

[CWatters]

If the contactor had no resistance at all, wouldn't there be 230 volts at the contactor? In other words, if the resistance of the contactor is .0036-Ohms or less, then couldn't the voltage at the contacts be 229.9 volts? I think I might not understand a key feature of electricity here.

You are confusing these two things..

a) the voltage on the right hand side of the contactor (about 229.9V) and
b) the voltage drop across the contactor (about 0.1V)

 

[Asymptotic]

I'm still confused by how your equation solving for resistance across the contacts. You say that the resistance across the contacts is 0.1 Volts/27.6 amps. You're saying that the voltage going through the contacts is 0.1 volts because the meter resolution of a Fluke 116 multimeter is 0.1 volts. But what does meter resolution mean? Why is the meter resolution the value we should plug in for the voltage when calculating the resistance going through the contactor?

Not quite. What I'm asking is, how much contact resistance is necessary at 27.6 amps (value of the current flowing through them) in order to develop a 0.1V voltage drop? Meter resolution is 0.1V, so this is the resistance necessary to make the meter display change from 0.0V to 0.1V.

One way to think about meter range and resolution is to compare it to a more familiar measuring device. For example, consider a yardstick.

This "meter's" range is 1 yard. It can't measure anything longer than a yard any more than a 600V meter can measure voltages higher than 600V.

Now to resolution. A yardstick with marks at only the origin point (0) and 1 yard mark has exceptionally poor resolution (the thing being measured either fits between the 0 and 1 yard marks, or it doesn't). Dividing it with marks at 1 foot interval increases resolution, but it still isn't very useful, so increasingly finer division marks are added. How much resolution is necessary depends on the application. A carpenter's yardstick may need a resolution no finer than 1/8th of an inch, while a machinist requires finer resolution, and his may go down to 1/64th inch.

In your case, if voltage drop measured by the Fluke 116 meter reads zero, it doesn't necessarily mean the amount of voltage drop is exactly zero, just that it may lie somewhere between zero, and the smallest division of your scale (0.1V).

 

[fourthindiana]

I have reviewed this thread a few times in the past few days.

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor. Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

What do you people think?

 

[fourthindiana]

Not quite. What I'm asking is, how much contact resistance is necessary at 27.6 amps (value of the current flowing through them) in order to develop a 0.1V voltage drop? Meter resolution is 0.1V, so this is the resistance necessary to make the meter display change from 0.0V to 0.1V.

One way to think about meter range and resolution is to compare it to a more familiar measuring device. For example, consider a yardstick.

This "meter's" range is 1 yard. It can't measure anything longer than a yard any more than a 600V meter can measure voltages higher than 600V.

Now to resolution. A yardstick with marks at only the origin point (0) and 1 yard mark has exceptionally poor resolution (the thing being measured either fits between the 0 and 1 yard marks, or it doesn't). Dividing it with marks at 1 foot interval increases resolution, but it still isn't very useful, so increasingly finer division marks are added. How much resolution is necessary depends on the application. A carpenter's yardstick may need a resolution no finer than 1/8th of an inch, while a machinist requires finer resolution, and his may go down to 1/64th inch.

In your case, if voltage drop measured by the Fluke 116 meter reads zero, it doesn't necessarily mean the amount of voltage drop is exactly zero, just that it may lie somewhere between zero, and the smallest division of your scale (0.1V).

I appreciate your efforts to help me understand this. However, let me be frank with you.

The only significance of the concepts of voltage drop and meter resolution that you've said so far is that the two concepts are a useful troubleshooting tool. You said that know voltage drop helps one determine what condition that the contacts are in. I've read your posts on this thread several times both to try to understand them and to try to understand how voltage drop and meter resolution relate to the OP of this thread. To me, voltage drop and meter resolution don't seem germane to the topic of this thread. All this stuff you're writing about voltage drop and meter resolution seems like much ado about nothing, at least to me.

 

[jim hardy]

What do you people think?

I think you're getting mighty close...

In the spirit of 'excruciating attention to detail in wording..'

I have reviewed this thread a few times in the past few days.

Great ! We take that as a compliment .

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

I'm not familiar with the term "Series Laws". Do you mean Kirchoff's Current Law ?

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

Yes ! That's the beginnings of circuit analysis.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor.

Well, not quite. It's in series with the parallel combination of the 10 and 50 ohm resistors.

I guess instructor has not yet covered parallel versus series connections ?

Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Amperage through the contactor is the SUM of the currents through the 10 and 50 ohm resistors. That's 27.6 amps as @Tom.G 's image in post 31.

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps 27.6 multiplied by .05-Ohms
Voltage at contactor= 1.15 1.38 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

Yes it could.Lastly
You have to be careful when saying "voltage AT (anyplace)."
That's because voltage is always taken as a DIFFERENCE between two places.
That's why voltmeters have two wires.

We get sloppy in terminology and it always confuses newbies.
As i said , Voltage is always a difference
L1 is shown as 115 volts. But 115 volts DIFFERENT from what ?
The answer is "115 volts different from circuit common, which in house wiring is almost always connected to "ground:" .
L2 is also 115 volts different from "ground" but it's always opposite direction from L!, so the difference between L1 and L2 is 230 volts.

If you place voltmeter's black lead on circuit ground and its red lead on L1 you'll measure 115 volts.
If you place voltmeter's black lead on circuit ground and its red lead on L2 you'll measure 115 volts.
If you place voltmeter's black lead on L2 and its red lead on L1 you'll measure 230 volts.
Try it - you need to make that intuitive , the thought process needs to be immediate and automatic.

Likewise - if you're going to say "Voltage AT the contactor.."
you must also say 'to where that difference is measured..' Red lead to contactor, Black lead where - to L1. to L2, to ground, or somewhere else ?
If you say instead "Voltage ACROSS the contactor " you've implied red lead to one side of contactor and black lead to other side of contactor.
Try it on your training simulator - that thinking needs to become an automatic habit.
That preciseness in communication will prevent others from misunderstanding what you mean.

When troubleshooting i always ask for a voltage measurement like this:
"What do you read from L2 to R10's side of the contactor?"
and my guys all know that i mean "Place voltmeter's black lead on "From"(L2) and red lead on "To"( resistor side of the contactor) and tell me what you read."
Always describe your voltage so it's unambiguous exactly what difference you are measuring.

It's a bad habit we pick up from reading tire pressures.
We forget that we're actually measuring the difference between pressures inside and outside the tire.

 

[CWatters]

I have reviewed this thread a few times in the past few days.

Today it occurred to me that I could estimate the voltage going through the contactor using the Series laws, Ohm's Law, and using the estimate .05-Ohms of resistance for the contactor.

In post #15 on this thread, phinds already used Ohm's Law to establish that the amperage going through the 10-Ohm resistor is 23 amps. Phinds calculated the amperage of the 10-ohm resistor by saying I= 230 volts/10ohms.

In the diagram in the photograph I attached to the OP, the contactor is in series with the 10-Ohm resistor. Since the amperages of all resistors are equal in a series circuit, doesn't that mean that the amperage going through the contactor is 23 amps?

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

What do you people think?

It took me awhile to find a data sheet for a contactor that specified the contact resistance. Hope this link works but it says <0.4 milli ohms (<0.0004 Ohms).

https://www.google.com/url?sa=t&sou...WMAR6BAgAEAE&usg=AOvVaw0N9YpKwqUzGdWJMvhNb-QG

Don't get too worried about what your tutor said. When he says "the voltage at the contactor is 0V" he means "the voltage drop across the contactor is approximately 0V because the resistance is low". The voltage drop you calculated is less than 1% of the supply voltage so its a reasonable approximation. What is the likely tolerance of the resistors in your circuit? That will probably have a far greater effect on the current than the contact resistance.Edit: Sorry I can't get that link to work. When Google finds a PDF for you how do you copy the link? I used "copy link address" and then paste it in but it doesn't work.

 

[jim hardy]

Edit: Sorry I can't get that link to work. When Google finds a PDF for you how do you copy the link? I used "copy link address" and then paste it in but it doesn't work.

here's where it sent me

https://www.tdk-electronics.tdk.com/download/2060308/ac1d068dad91b8308dd031bfb3d2f33e/high-voltage-contactor-pb.pdf

 

[Asymptotic]

I appreciate your efforts to help me understand this. However, let me be frank with you.

The only significance of the concepts of voltage drop and meter resolution that you've said so far is that the two concepts are a useful troubleshooting tool. You said that know voltage drop helps one determine what condition that the contacts are in. I've read your posts on this thread several times both to try to understand them and to try to understand how voltage drop and meter resolution relate to the OP of this thread. To me, voltage drop and meter resolution don't seem germane to the topic of this thread. All this stuff you're writing about voltage drop and meter resolution seems like much ado about nothing, at least to me.

The point about meter resolution is, if voltage across the contacts is less than the smallest value the meter can display, then the meter will read zero volts even though a small voltage may actually be present. To further complicate the issue, a Fluke 116 meter reading of 0.1V could represent an actual value between 0.050V to 0.149V - to know it more closely, a higher resolution meter must be used.

Therefore, I still don't know the answer to the following question: Why does my instructor say that the multimeter would measure zero voltage across the contacts of the contactor in the ladder diagram in the attached photograph?

This is how resolution ties into the OP. If contact resistance is lower than a certain value for a given current (the amount of resistance necessary to drop 0.1V), then the meter will read zero voltage across the contacts.

 

[DaveE]

First off, the drawing is absurd. You have 115 volts on each side so the voltage across the elements is zero, yet you show it as 230 volts. Second, you have a variable capacitor shown apparently intended as a closed switch.

1) You may want to consider phase? When you see contactors and 10ohm resistor with numbers like 115, 230, 380, 480, then you should consider the possibility that it is an AC circuit.

2) Variable capacitors should be drawn with an arrow through them (according to me) not a line; like a variable resistor. You are confusing the way schematics are drawn by an electrician with the way they are drawn by pcb designers. For example: https://www.nrc.gov/docs/ML1025/ML102530301.pdf

Really, it's ok the way he described it.

This is a staggeringly simple circuit.

Maybe it is to you, by why would he ask about it if he thought that? Are you trying to find faults with his question, show everyone that you know a lot about about electronics, or are you trying to help educate someone who is learning?
What really is the point of your post? Is your reply going to encourage people like the OP to ask similar questions in the future?

 

[fourthindiana]

The point about meter resolution is, if voltage across the contacts is less than the smallest value the meter can display, then the meter will read zero volts even though a small voltage may actually be present. To further complicate the issue, a Fluke 116 meter reading of 0.1V could represent an actual value between 0.050V to 0.149V - to know it more closely, a higher resolution meter must be used.This is how resolution ties into the OP. If contact resistance is lower than a certain value for a given current (the amount of resistance necessary to drop 0.1V), then the meter will read zero voltage across the contacts.

Good points. Now I see how voltage drop and meter resolution are relevant to the OP of this thread. My apologies for saying that the topics are not germane when meter resolution and voltage drop are germane.

 

[Averagesupernova]

I don't think that I need to mention this, but imagine an analog meter with a full scale reading of 10,000 volts. Does meter resolution mean something to you now when trying to measure a tenth of a volt on the 10 KV scale? Exaggeration often times helps me wrap my head around something.

 

[fourthindiana]

Excellent post, jim hardy.

I'm not familiar with the term "Series Laws". Do you mean Kirchoff's Current Law ?

I don't think that what I call the "series laws" are the same as Kirchoff's Current Law. To me, the "Series Laws" are the following three laws:
1# The total current in a series circuit= the current in the first resistor= the current in the second resistor= the current in the third resistor...
2# The total voltage in a circuit= the voltage of the first resistor + the voltage of the second resistor+ the voltage of the third resistor...
3# the law of resistance for series (which I think is similar to the series law regarding voltage but I don't know for sure)

Well, not quite. It's in series with the parallel combination of the 10 and 50 ohm resistors.

I guess instructor has not yet covered parallel versus series connections ?

My instructor has covered parallel versus series connections, but I just momentarily forgot about this stuff because it's been months since I last used it.

Amperage through the contactor is the SUM of the currents through the 10 and 50 ohm resistors. That's 27.6 amps as @Tom.G 's image in post 31.

Okay. Then the voltage through the contactor is 27.6 amps multiplied by .05-Ohms. Voltage through contactor= 1.38 volts

Lastly
You have to be careful when saying "voltage AT (anyplace)."
That's because voltage is always taken as a DIFFERENCE between two places.
That's why voltmeters have two wires.

I was planning on creating a new thread asking about this. I thought that maybe it was the case that voltage is the difference between two places. Now I don't have to create a new thread regarding this.

 

[jim hardy]

I was planning on creating a new thread asking about this. I thought that maybe it was the case that voltage is the difference between two places. Now I don't have to create a new thread regarding this.

There are LOTS of folks whose thinking on that subject is , for lack of a better word, "muddy" .

It has come up in a lot of threads before.
As @Averagesupernova said

Exaggeration often times helps me wrap my head around something.

here's the exaggeration i use to plant the concept. It's repeated in several old threads

Langauge gets so confusing... because we all take shotcutsI was taught in high school that voltage is potential difference
but what was potential ?
The concept of work done moving against a field just didnt seem to relate .

What clarified it for me was when concept of absolute potential clicked in place -

' the work done in bringing a unit of charge from infinity to wherever you are...'

i could just imagine myself traveling from Alpha Centauri to Miami Central High School with a bucketful of charge and tabulating the force encountered every centimeter along the way ... and adding up all those Force X Distance numbers to arrive at "Work Done" .

Nobody is going to go out to infinity and measure the force on a charge,
or hold a voltmeter lead there for you to make that measurement either,

so

we have to settle for measuring the difference in absolute potentials between two points closer to us, and we call that voltage,
We do not know the absolute potential of either point and cannot measure it
but their difference is readily measured by a voltmeter with two leads of finite length.

'Difference in absolute potentials' shortens to potential difference , aka 'voltage'.

So
In my world, voltage is always a potential difference between two points.
We get sloppy in our communication and do not always define those two points,
one of them is usually a 'circuit common' of some sort which is often called 'ground'
though it may be or may not be connected to Earth .

If that potential difference exists because of current flowing through a resistance, that's a voltage drop.
If that potential difference exists because of, let's say the chemistry going on inside a battery, that's not a voltage drop it's just a voltage (or maybe a voltage rise).

Sorry to be so verbose
but the price of clear communication is excruciating attention to language.

Hope mine was clear .. exaggeration is often a useful tool.

old jim

if you nail that concept now, early in your studies, you'll avoid having to "unlearn" a lot of wrong stuff later on.

Good Luck in your class !

old jim

 

[NTL2009]

Jim Hardy's post#36 covers it all very well. I almost wasn't going to join in, but I think there may be a few things I can add.

I congratulate you on your patience and persistence. It's often hard for some us to remember struggling with these concepts, they become 2nd nature after a while, and sometimes hard to explain to beginners. Adding to that, some of those helping in this thread are approaching it at a high level and in general terms (and confusing you a bit with that), while your instructor is looking at things in a bit too low and simple level, and making a bit too many assumptions/generalizations without explanations. You are getting caught in the middle. Hang in there, you are definitely making progress!

I think you have this straight now, but in the earlier posts, you seemed to be confused by the phrase "Current takes the path of least resistance". That does not mean the single path of the lowest resistance (though it seems to read that way). It means whatever path/paths it can take. And when you have parallel resistors, the path of least resistance is to take all the paths combined. I know some here don't like analogies at this level, but I'm not one of them. Consider two parallel lanes of traffic, one with a 50 mph speed limit, one with a 30 mph speed limit. You would never say that the way to get the most cars down those two lanes is to make them all take the 50 mph lane. Obviously, you get more cars by using both lanes to their limits. Currents divide this way too, taking all the parallel paths it can find.

From your post #34:

Therefore, the following equation will show the voltage going through the contactor:

Voltage at contactor= 23 amps multiplied by .05-Ohms
Voltage at contactor= 1.15 volts

Although my 1.15 volts seems like a relatively tiny amount of volts, my instructor said that the voltage at the contactor would be zero volts. Would the resistance going through the contactor possibly be substantially less than .05-Ohms?

It's useful to try to put some real world experience to these assumptions. A first pass guess of 0.05 Ohms for the contactor was reasonable - after all, 0.05 is a small number compared to the 10 and 50 Ohm values in that circuit. And you did the math right, assuming 23 amps ( I guess this was just with the 10 Ohm in the circuit?), then 23 Amps x 0.05 Ohms = 1.15 V. Now think about power, Volts * Amps. That would be ~ 26 watts at that contactor. Think of a 25 watt soldering iron, or a 25 watt filament bulb. I'd think that contactor would burn up if it was dissipating that much heat. So later on we learn that the contactor is spec'd at typically less than 0.4 milli-Ohms. So that would be 9.2 milli-Volts @ 23 Amps, so ~ 0.21 Watts - much better!

So now you realize that a contactor that has become dirty or pitted, and has had its resistance increase just slightly, can start burning up pretty fast in a circuit with tens of Amps.

So when your instructor says there is zero volts across the closed contacts (and listen to Jim Hardy - voltage is across a component, current is through a component), he is speaking loosely. It's shorthand for "very low, almost zero, not enough to be concerned about". And it is appropriate for dealing with circuits like this, you can't get bogged down in every detail. We know what someone means when they refer to a "12 Volt Battery" in a car, even though it is ~ 12.75 volts at rest, and about 13.8 V when the alternator is running, and maybe ~ 10V when cranking a cold engine. But we have to know when shorthand is OK, and when we need exact values.

This will all be 2nd nature to you too very soon. Get the basics down, everything builds on that.