Circuit troubleshooting - simple Voltage dividers

[AcousticBruce]

When I looked at this I said... hmm R1 must be shorted. Makes sense to me.
I cant see any other way.

The book says the most probable failure is R2 is open?!? How is this? If R1 is fine, shouldn't it be a little less than 24v? Doesn't the 3.3kΩ take some of the voltage away?


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How is this 24V???? The ground of the current still has to travel through resisters to get to the voltmeter ground.

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[audioguru]

A modern voltmeter has an input resistance of 10M. Then its input current is so small that it doesn't affect the circuit.

 

[BeeBop]

When I looked at this I said... hmm R1 must be shorted. Makes sense to me.
I cant see any other way.

The book says the most probable failure is R2 is open?!? How is this? If R1 is fine, shouldn't it be a little less than 24v? Doesn't the 3.3kΩ take some of the voltage away?

If R2 goes OC then you no longer have a path, so you will measure 24V or the entire potential across R2. With an open circuit, R1 will not drop anything.
Read the part of the book under 'Step 1: Analysis.' both sides of R1 are at the same potential, since no current is flowing.

 

[AcousticBruce]

In picture 1 if R2 was open, that would mean, the entire circuit to the right of the VM would basically not be involved. Am I right?

So that being said. R1 the voltage source and the VM would be the only thing involved. So wouldnt the 3.3k resister reduce the 24v? I have hooked a LED to a 9V and it burned out rather quick... then I hooked a resister i think 370Ω in series and the voltage was reduced for the LED. So 1 resister should reduce voltage right?

 

[BeeBop]

In picture 1 if R2 was open, that would mean, the entire circuit to the right of the VM would basically not be involved. Am I right?

So that being said. R1 the voltage source and the VM would be the only thing involved. So wouldnt the 3.3k resister reduce the 24v? I have hooked a LED to a 9V and it burned out rather quick... then I hooked a resister i think 370Ω in series and the voltage was reduced for the LED. So 1 resister should reduce voltage right?

You are correct; the entire cct to the right of the VM would not be involved.

The current in a series circuit is the same throughout. The series resistor limits the current through the LED, and would do the same thing for the meter.

 

[AcousticBruce]

so is the book wrong based on what I have showed you? Is the second picture even possible?

 

[BeeBop]

so is the book wrong based on what I have showed you?

Um, no, I don't think so. How do you mean?

 

[AcousticBruce]

Look at the second picture... if R2 is open... it still shouldnt measure 24v. The Vs is 24V and resisters are in the path.

 

[BeeBop]

Look at the second picture... if R2 is open... it still shouldnt measure 24v. The Vs is 24V and resisters are in the path.

The truth is that an open component drops the entire voltage... Here is the answer:

A modern voltmeter has an input resistance of 10M. Then its input current is so small that it doesn't affect the circuit.

 

[AcousticBruce]

A modern voltmeter has an input resistance of 10M. Then its input current is so small that it doesn't affect the circuit.

I understand this, but this is not where I am confused.

If R2 is open... then how can it read 24V (in picture one) ? Can you not drop voltage with one resister?

 

[audioguru]

then how can it read 24V (in picture one) ? Can you not drop voltage with one resister?

Read the text. The text says that R1 is shorted so the meter is reading the 24V power supply voltage.

 

[AcousticBruce]

The text does not say that. Where do you see that?

The only way I find the reading on picture 2 to be accurate is if R1 is shorted AND the negative side of the VM is connected DIRECTLY to ground.

But the book clearly shows a reading on 24V and the voltage source itself is 24V... is the whole dang circuit shorted?


And AudioGuru I just dont understand what the loading effect of the VM has to do with my question... am I clearly missing a foundation? Slap me if needed.

 

[audioguru]

The meter has a very high input resistance so its current is almost nothing.
It actually reads 23.99999999999999999999999999999999999999999999999V.

 

[AcousticBruce]

AudioGuru I still dont know what the loading effect has to do with my question?

Maybe I am bad at phrasing my questions.

In picture 2
-The posistive side of VM has to go through a 3.3kΩ resister.
-the negative side of the VM goes through R3 R4 R5 and R6 (2.7k total)

So in my current way of thinking the fact that the VM is measuring in the middle of the circuit over a OPEN resister and the voltage is still traveling through a 3.3k and 2.7k (total parallel). So since the voltage source is 24V and the VM is hooked in the midst of a 3.3 on positive and a total of 2.7k on negative side than the VM should read a different amount, regardless of how new or old the VM is (the loading effect).

 

[BeeBop]

If R2 is open... then how can it read 24V (in picture one) ? Can you not drop voltage with one resister?

If there is one resistor in a series cct, then it will drop all the voltage. If that resistor goes open then the entire voltage of the supply will still be across that resistor.

 

[kchriste]

AudioGuru I still dont know what the loading effect has to do with my question?

The loading effect is due to the fact that the meter acts like a 10MΩ resistor in the circuit. Measurement affects behavior in circuits also.
See the real circuit here:

 

[AcousticBruce]

AudioGuru I still dont know what the loading effect has to do with my question?

Maybe I am bad at phrasing my questions.

In picture 2
-The posistive side of VM has to go through a 3.3kΩ resister.
-the negative side of the VM goes through R3 R4 R5 and R6 (2.7k total)

So in my current way of thinking the fact that the VM is measuring in the middle of the circuit over a OPEN resister and the voltage is still traveling through a 3.3k and 2.7k (total parallel). So since the voltage source is 24V and the VM is hooked in the midst of a 3.3 on positive and a total of 2.7k on negative side than the VM should read a different amount, regardless of how new or old the VM is (the loading effect).

 

[kchriste]

You are correct and are saying the same thing AudioGuru, I and others have said. What is confusing you in the book is the difference between ideal components and real world components. Your book assumes that the "meter" is an ideal component and thus has infinite resistance. If that was the case, then yes, the meter would read exactly 24V because there would be no current flow in the resistors. You need current flow in a resistor to have a voltage drop. Did you look at my real world diagram above?

 

[BeeBop]

AudioGuru I still dont know what the loading effect has to do with my question?

Maybe I am bad at phrasing my questions.

In picture 2
-The posistive side of VM has to go through a 3.3kΩ resister.
-the negative side of the VM goes through R3 R4 R5 and R6 (2.7k total)

So in my current way of thinking the fact that the VM is measuring in the middle of the circuit over a OPEN resister and the voltage is still traveling through a 3.3k and 2.7k (total parallel). So since the voltage source is 24V and the VM is hooked in the midst of a 3.3 on positive and a total of 2.7k on negative side than the VM should read a different amount, regardless of how new or old the VM is (the loading effect).

So if you do the calculations on the cct kchriste posted, how much voltage does the meter drop?

 

[AcousticBruce]

First of all sorry about the double post.

The loading effect is due to the fact that the meter acts like a 10MΩ resistor in the circuit. Measurement affects behavior in circuits also.
See the real circuit here:

Thank you so much. I finally snapped out of it. What i should have done is calculated the ( 10M / (3.3k + 10M + (15k^-1+4.7k^-1+(10k+2.2k)^-1)^-1 ) * 24 ~= 24V

yup yup... audio you were right... i just was thinking of the open resister as still open instead on it NOW being a 10M+

I made the circuit on a breadboard and the VM said 9V then I plugged a 10Ω
resister in parallel and it dropped to like 5V.... a light turned on over my head at that moment!

Thanks again
Thanks kchriste AudioGuru and Beebop for your patients