Current through a circuit when the key is open

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When the key in the circuit is open, no current flows, leading to an infinite resistance scenario according to Ohm's law. The internal resistance of the battery (0.6 ohm) is only relevant when current is present, which is not the case when the switch is open. Some participants debated whether the internal resistance should be considered part of the circuit, ultimately concluding that it should not when the switch is open. The discussion highlighted the ambiguity in the question's wording, leading to varied interpretations of circuit resistance under these conditions. Overall, the consensus is that the resistance of the circuit is effectively infinite when the switch is open.
Darshit Sharma
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Homework Statement
Consider the circuit diagram given below and answer the questions.
Relevant Equations
Ohm's law. V= IR
1708007546008.png



The equivalent resistance of the parallel combination is 2.4 ohm.

This implies that if the key is closed the total resistance of the circuit = R + r = 2.4 + 0.6 = 3 ohm

However, if the key is open, no current flows through the circuit and thus, by Ohm's law (R = V/I), as I tends to 0, R tends to infinity.
Thus, if the external resistance of the circuit is infinite, even if we add up the internal resistance of the battery ( 0.6 ohm), the resistance would still be infinity.

However, some of my fellow batchmates were fooling around with their reasoning as only the internal resistance of the battery would be taken into account and the final answer would be 0.6 ohm. They didn't provide a complete reasoning though.

Thus, the final question of this thread is what will be the resistance of the circuit, as the question questions and why?
 
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Hi,

So it comes down to: is the 0.6 ##\Omega## part of the circuit or not ?
I would say it is not.

I even find it questionable to say it's part of the circuit when the switch is closed, but this internal resistance must be taken into account to calculate the current when the switch is closed.

The reason for this is that one can for example investigate this 'circuit' with some other setup that doesn't include this particular battery.

Darshit Sharma said:
However, if the key is open, no current flows through the circuit and thus, by Ohm's law (R = V/I), as I tends to 0, R tends to infinity.
No, because then V=0 too.

##\ ##
 
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BvU said:
No, because then V=0 too.

##\ ##

Sir, please elaborate.


Screenshot 2024-02-18 040057.png
Screenshot 2024-02-18 040107.png



I am unable to make out why V =0.
 
BvU said:
Hi,

So it comes down to: is the 0.6 ##\Omega## part of the circuit or not ?
I would say it is not.


##\ ##
So sir final answer should be infinite or zero or 2.4 or 0.6?
 
When no current flows through the battery, it has effectively zero internal resistance. This is the case when the key is open.
 
Mister T said:
When no current flows through the battery, it has effectively zero internal resistance. This is the case when the key is open.
Ohk but V is still there?
 
Darshit Sharma said:
Ohk but V is still there?
Yes. You still have the full 6 V across the battery.
 
Darshit Sharma said:
Sir, please elaborate.
...
I am unable to make out why V =0.

If the switch is open, the voltage over the circuit is zero

It's all about what you consider to be part of "the circuit" and what not. My subjective and somewhat inconsistent take is that "the circuit" under consideration does not include the battery if it is not connected.

It's more about communication than about physics.

##\ ##
 
The question asks for the 'resistance of the circuit'.

When the key is in the open position, the 'gap' is a resistor of infinite resistance. This makes the overall circuit resistance = infinity.

IMO.
 
  • #10
BvU said:
If the switch is open, the voltage over the circuit is zero
Maybe there is some convention of which I am unaware. But we have an oval labelled "A" on the schematic. It has one terminal facing the bottom leg of the circuit and one terminal facing the leg on the top.

To me, the natural reading is that "the circuit" is everything between those two terminals. The oval labelled A is an unspecified sensor that is attempting to measure the circuit, trying to treat it as a voltage source in series with a resistor (Thevenin equivalent) or a current source in parallel with a resistor (Norton equivalent).

The wording of the question and the layout of the diagram seems to beg for the Thevenin equivalent interpretation.

With this interpretation, the voltage "over the circuit" can be whatever the sensor drives it to be.
 
  • #11
I know I'm not going to win this. To my worn-out brainwashed mind 'the circuit' is what is between A and B on top and the remainder is plumbing.
1708285568887.png

Usually A (black, with a circle around it) stands for an Ammeter (and I haven't yet seen any of those 'attempting to do' anything or 'trying to treat' something.

As I said, it's more about communication than anything else. All in good spirit.

Consultation with an authority proves me formally wrong. Somehow I don't feel inclined to repent.

And I'd be surprised if 'Thevenin' means anything for the OP

(by the way, the exercise quality is lacking IMHO)

##\ ##
 
  • #12
I think the intent of the question's author was to probe the student's understanding that internal resistance is present only when current flows through the battery.
 
  • #13
BvU said:
I know I'm not going to win this. To my worn-out brainwashed mind 'the circuit' is what is between A and B on top and the remainder is plumbing.
I agree with you that the electrical resistance of anything requires two points A and B between the resistance is to be determined. I disagree with your labeling A and B as you have because it is not very informative. The resistance between A and B going clockwise or counterclockwise? Why not choose the poles of the open switch? What if the switch were closed and put the leads of an ohmmeter at points A and B? The fact that the problem's author didn't supply points A and B makes me agree with you that the quality of this exercise is subpar.
Mister T said:
I think the intent of the question's author was to probe the student's understanding that internal resistance is present only when current flows through the battery.
I found this problem here as part of a preparatory exam. Unfortunately, no answers are provided to help us deduce the author's intent.
 
  • #14
I looked at some other questions from the preparatory exam and found these two representatives of the quality thereof :headbang:

Screen Shot 2024-02-18 at 3.26.20 PM.png

Screen Shot 2024-02-18 at 3.26.59 PM.png
 
  • #15
BvU said:
I know I'm not going to win this.
I did not want to make this an argument. It is just a discussion. No winners or losers. Just interested parties.

It is plausible that the "circuit" is indeed intended to be the pair of resistors in parallel and that the ammeter, switch, voltage source along with its embedded resistor are intended to perform the function of an ohmmeter.

The question then becomes: What reading should this "ohmmeter" display?

The answer then hinges on what assumptions the designer used when he calibrated the gauge. If the designer assumed an ideal 6 volt source without internal resistance then one reading would be expected. If the designer assumed an ideal 6 volt source with a 0.6 ohm internal resistance then a different (and more accurate) reading would be expected.

We need to guess how stupid the designer was.

Yes, I agree that it is a poor question that forces the student to read the mind of the person who posed it.
 
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  • #16
kuruman said:
I found this problem here as part of a preparatory exam. Unfortunately, no answers are provided to help us deduce the author's intent.
Terribly worded question. Mistakes in grammar. Doesn't tell you if you're supposed to read the ammeter when the key is open or closed (it's open in the diagram). Same for the current through the 4-ohm resistor.
 
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  • #17
jbriggs444 said:
We need to guess how stupid the designer was
My guess is that it was a committee of "experts". I taught a mini-course in MCAT physics prep. The questions in the book were an embarrassment. I had to weed out about half of them. After the first time I didn't want to teach it again but the pay was absurdly high.
 
  • #18
In my opinion, there is no “circuit” until there is at least one closed loop. For this reason, I think the resistance when the switch is open is zero.
 
  • #19
PhDeezNutz said:
In my opinion, there is no “circuit” until there is at least one closed loop. For this reason, I think the resistance when the switch is open is zero.
The resistance of a short circuit is zero.

If you think that an open circuit is an oxymoron then you should characterize the resistance of an open circuit as "undefined", not "zero".
 
  • #20
jbriggs444 said:
The resistance of a short circuit is zero.

If you think that an open circuit is an oxymoron then you should characterize the resistance of an open circuit as "undefined", not "zero".

Fair. Resistance is a property independent of current (provided the voltage isn't too high and actually rips it apart changing it's properties). So even if there is no voltage drop across the element its resistance is the same.

Rethinking this I would say the resistance when the switch is open is ##0.6 \Omega##
 
  • #21
PhDeezNutz said:
Rethinking this I would say the resistance when the switch is open is ##0.6 \Omega##
Again, the resistance between what point and what other point? Ideally, one defines and measures resistance using a four-point measurement. Here we are given no points! That's my point.
 
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  • #22
PhDeezNutz said:
Resistance is a property independent of current
That's Ohm's Law. It's an approximation that fails miserably in many cases.
 
  • #23
Half of the things bounced off my head; sorry to say that and for the delay. I was busy taking some similar quality exams.

So, after all the final verdict is the question, as always, is framed wrong.

However, the answer given is 2.4 ohm. I don't know what they have assumed and why.

PhDeezNutz said:
Fair. Resistance is a property independent of current (provided the voltage isn't too high and actually rips it apart changing its properties). So even if there is no voltage drop across the element its resistance is the same.

Rethinking this I would say the resistance when the switch is open is ##0.6 \Omega##
What about that infinite ohm resistor (open switch), Sir?

kuruman said:
Again, the resistance between what point and what other point? Ideally, one defines and measures resistance using a four-point measurement. Here we are given no points! That's my point.

I feel this is the sole reason for variable answers.



Anyways, what are some points of variable answers?
Which set of points should I consider to expect different answers?

And Resistance is independent of current ---- ok
but Current is dependent on Resistance ---- right? (Just confirming)


And if resistance is independent of current.....then how do we get infinite resistance across that open switch like all that I was doing R = V/I and since I = 0, R = infinity, was wrong?


I have attached the solution booklet.
 

Attachments

  • #24
Here is a screenshot of this awful question for those still interested in it. I am not.


Screen Shot 2024-02-28 at 6.02.31 PM.png
 
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  • #25
Darshit Sharma said:
And Resistance is independent of current ---- ok
I haven't read much of the thread, so sorry if I'm repeating another reply to this question. In an ideal circuit, the answer is yes. But in a real circuit, the resistor heats up more with more current through it, which increases its resistance a bit. The concept to search on is thermal coefficient of resistance. You don't need to worry much about it in your introductory circuit classes in school, but you most definitely need to take it into consideration in your more advanced circuit analysis classes and in real life circuit design.
 
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  • #26
Darshit Sharma said:
However, the answer given is 2.4 ohm. I don't know what they have assumed and why.
A 6 ohm and a 4 ohm resistor combined in parallel is equivalent to 2.4 ohms.

And if resistance is independent of current.....then how do we get infinite resistance across that open switch like all that I was doing R = V/I and since I = 0, R = infinity, was wrong?

Because when I = 0, V = 0 also.
 
  • #27
Mister T said:
A 6 ohm and a 4 ohm resistor combined in parallel is equivalent to 2.4 ohms.
But that 2.4 ohms should be in series with that infinite resistance open switch....so infinite resistance + 2.4 = infinite resistance
 
  • #28
Mister T said:
Because when I = 0, V = 0 also.
So 0/0 = undefined? But I've read somewhere that the resistance of an open switch is infinite



Just confirming once -- Is the answer 2.4 ohms to (i) part correct? 🤔


I am sorry to everyone here for both the lack of understanding on my part and the "awful" -ness of the question. One day I'll find a nice question for sure.
 
  • #29
Darshit Sharma said:
So 0/0 = undefined? But I've read somewhere that resistance of an open switch is infinite
Sorry for the confusion. I thought you were talking about an ohmic resistor but obviously you weren't. You were talking about a switch. In this case you have a nonzero voltage across the switch, but you have zero current through the switch. And the switch effctively offers infinite resistance.

But a switch is a nonohmic device, and the relation R = V/I does not apply to a switch.
 
  • #30
Darshit Sharma said:
So 0/0 = undefined? But I've read somewhere that the resistance of an open switch is infinite
In this case, with the switch open there would be 6 volts across the switch and no current flowing. That is not ##\frac{0}{0}##. The resistance of the open switch would be ##\frac{6}{0}##.

Whether you call that "undefined" or "infinite" depends in large part on whether you are a mathematician or an engineer.
 
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