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Find the maximum potential difference across a series circuit

  • Thread starter mhrob24
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  • #1
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Homework Statement:

Three capacitors
C1 = 10.0 µF, C2 = 16.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum potential difference across the series combination.

Homework Equations:

Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......

in parallel circuit: C(eq) = C(1) + C(2) + C(3) ........

C = Q/V
I'm not really sure what I need to find exactly. From what I'm seeing, I could give C1 the max potential difference of 125V because it has the lowest capacitance, and because V = Q/C, this means the capacitor with the highest potential difference across its plates will be the one with the lowest capacitance. Then I can just solve for Q, and once I have Q, I'd have everything I need in order to solve for the voltages between C2 and C3....but I'm not sure where to go from there. I usually see questions worded as find the potential differences "across each capacitor in the combination", not "across the series combination". If they were asking for the maximum potential DROP across the series I think I could do that by just finding the lowest voltage in the series (should be C3 since C3 has the highest capacitance out of the 3 capacitors), and then subtracting that from 125 which would give me to total drop in voltage across the series.
 

Answers and Replies

  • #2
collinsmark
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Homework Statement: Three capacitors
C1 = 10.0 µF, C2 = 16.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum potential difference across the series combination.
Homework Equations: Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......

in parallel circuit: C(eq) = C(1) + C(2) + C(3) ........

C = Q/V

I'm not really sure what I need to find exactly. From what I'm seeing, I could give C1 the max potential difference of 125V because it has the lowest capacitance, and because V = Q/C, this means the capacitor with the highest potential difference across its plates will be the one with the lowest capacitance. Then I can just solve for Q, and once I have Q, I'd have everything I need in order to solve for the voltages between C2 and C3....but I'm not sure where to go from there. I usually see questions worded as find the potential differences "across each capacitor in the combination", not "across the series combination". If they were asking for the maximum potential DROP across the series I think I could do that by just finding the lowest voltage in the series (should be C3 since C3 has the highest capacitance out of the 3 capacitors), and then subtracting that from 125 which would give me to total drop in voltage across the series.
I think you might be overthinking the question. Had the question be asking "what is the maximum charge [technically, 'charge separation'] of the three capacitors", that would have required some significant calculations. But that's not the question.

The way the question is phrased, the answer is quite straightforward. Let Kirchhoff come to the rescue.
 
  • #3
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I'm somewhat familiar with Kirchhoff's rules but only because I just googled it. This question came from an online HW platform and each set of questions are based off a specific chapter. In this case, its chapter 8 (capacitors) for me. Kirchhoff's rules aren't discussed until chapter 10, so this question should be assuming the reader doesn't know about that yet. At least i'd hope so.....

Would the answer simply be 125v? In a series, there will be voltage drops as the current travels across each capacitor in a series, so If 125v is the maximum potential any capacitor can have, then I guess the maximum potential difference across the whole series can't be any greater than 125V?
 
  • #4
collinsmark
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Would the answer simply be 125v?
That's not quite right. The maximum voltage (i.e., potential) across each component is 125 V. So if you have three such components in series, what is the maximum voltage (potential) across all three?
 
  • #5
collinsmark
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On further thought, this problem statement might have been phrased really, really poorly. The way the problem was phrased, the answer is quite simple. However, if the problem was phrased as such: "The three capacitors are connected in series with initial charge of zero on each capacitor. What is the maximum potential difference that can be applied across the series connection such none of the capacitors are at risk of damage?" that would lead to a very different answer.

That said, that's not the way the question was phrased. The way the question was phrased leads to a relatively simple answer.
 
  • #6
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From your previous response, if the maximum V that any of the 3 capacitors can have in this circuit is 125V, and this is a series, then the sum of the voltages across each capacitor will add up to the total supply voltage of the circuit? That would mean 125 * 3 = 375V for the maximum voltage across the entire series? I know its not right because I entered that in and it was still incorrect so I know I'm wrong but I don't know where.

EDIT: I don't even know how that would be possible for each of them to have a max of 125v since their will be a voltage drop across each capacitor. I'm lost. I guess I'm just not reading this question correctly or something if the answer is that easy.
 
  • #7
collinsmark
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From your previous response, if the maximum V that any of the 3 capacitors can have in this circuit is 125V, and this is a series, then the sum of the voltages across each capacitor will add up to the total supply voltage of the circuit? That would mean 125 * 3 = 375V for the maximum voltage across the entire series? I know its not right because I entered that in and it was still incorrect so I know I'm wrong but I don't know where.
Okay, then that tells me that the problem statement was actually phrased really, really poorly. The problem statement, as it's written, says nothing about the initial charge of any of the capacitors and makes no restrictions that each capacitor cannot be charged individually. But, as you've indicated, this doesn't seem to indicate the situation the problem statement is struggling to suggest.

So, let's make the assumption that one cannot charge capacitors individually, and that the initial charge on each capacitor is zero. That will lead to a different answer. Let's go with those assumptions, since that seems to be what is actually being asked.

Given that, what do you have so far?

[Hint: given the above restrictions, if there is a charge Q on one of the capacitors, what does that tell you about the charge on either of the other capacitors?]
 
  • #8
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Yeah sorry about that. The online HW site im using (Webassign) has a reputation around here of being notoriously terrible with the questions they ask, so thanks for sticking with me still.

Well, to start, I guess If the max V that any single capacitor here can have is 125V, that must mean the battery powering this circuit is 125V right? (assuming 1 battery is connecting these 3 capacitors in series). And the capacitor "C1" should be the one that has the max. voltage of 125V across its plates because it has the lowest capacitance right?

Am I going in the right direction with my thinking here?
 
  • #9
collinsmark
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Yeah sorry about that. The online HW site im using (Webassign) has a reputation around here of being notoriously terrible with the questions they ask, so thanks for sticking with me still.

Well, to start, I guess If the max V that any single capacitor here can have is 125V, that must mean the battery powering this circuit is 125V right? (assuming 1 battery is connecting these 3 capacitors in series). And the capacitor "C1" should be the one that has the max. voltage of 125V across its plates because it has the lowest capacitance right?

Am I going in the right direction with my thinking here?
Start with this: since the capacitors are in series, any current that passes through a given capacitor is the same as the current going through any of the other capacitors, right?

What does that tell you about the charge on a given capacitor compared to the charge on any of the other capacitors?

(Again, we're assuming that the initial charge on the capacitors is zero, before any current is applied.)
 
  • #10
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That all the capacitors in the series will have the same magnitude charge on each of their plates.

So initially, they will all have zero charge, and then will have charge Q on each plate. (+Q and -Q)
 
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  • #11
collinsmark
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That all the capacitors in the series will have the same magnitude charge on each of their plates.

So initially, they will all have zero charge, and then will have charge Q on each plate. (+Q and -Q)
Correct. Each capacitor has the same charge. That's the key to solving this problem. :wink:

So, what's the least amount of charge that will cause any one of the given capacitors to reach 125 V?
 
  • #12
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Hey, hope you’re still with me. I’ve been dealing with a cold so I got some early sleep yesterday.

I know the charge on each of the plates can’t be any greater than 1.25e-3 because capacitor C1 has a capacitance of 10.0e-6. Dividing this from Q will give us 125V ( V = Q/C) for that capacitor, so any other higher magnitude charge would give a higher voltage than 125V, thus causing damage. Since the other 2 capacitors in this circuit are of higher capacitance, this would mean the charge 1.25e-6 would allow each of the 3 capacitors to stay within the 125V range, thus not causing any damage to either of the capacitors.

Is that correct?

If it is, I think I know how to solve it from there. I would just use the charge
Q= 1.25e-3 and solve for V for the other two capacitors, then I would add up the voltages of the 3 capacitors and that would give me the maximum supply voltage for the circuit?.....I think?

EDIT: I was right! 245.78V

Thanks for the help @collinsmark
 
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