Potential difference between three capacitors in series

In summary: The charge on a capacitor is given byQ = CVNow, in your case you've calculated a net capacitance Ct and you have a voltage Vt = 12V placed across it. What is the charge on that capacitance?Next, what do you recall about how charges are distributed amongst capacitors in series?so I can work out the charge for C1 by 10x10-6 x 12 = 120uC however I don't know how to apply that in a way to find the potential difference.
  • #1
Ware182
7
0
Hi,
I'm having a problem with this circuit finding potential difference between capacitors. I know the answers by playing around on simulation software A=12v B=8v and C=4v however, I am struggling to use the formula provided Vc= (Vt/Ct)*C to find the answers. It works fine in a parallel circuit with components of the same value but I know that series is different.

1. Homework Statement

upload_2015-2-28_21-22-33.png


Using the formula Vc = (Vt / Ct) * C


  1. Calculate the voltage at A when SW1 is switched to 12V. 12V (v= I x R)
  2. Calculate the voltage at B when SW1 is switched to 12V. 8V
  3. Calculate the voltage at C when SW1 is switched to 12V. 4V
I have worked out Question A using ohms law- not the formula provided just to check against simulation software.
So far I understand that VC will be A, B, C
Vt is voltage total=12v
Ct is capacitance total 1/ct=1/c1+1/c2+1/c3 =3.3 recurring
vt/ct=3.6
Now I'm struggling to understand what I times 3.6 by to make either 4, 8 and 12 finishing the formula.

Can anyone help me please?
Thanks
 
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  • #2
Symmetry should give you your answer with hardly any calculation at all. All components in series share the same current, so if all the capacitors are the same size, what can you say about the charges that end up on them?
 
  • #3
Ware182 said:
vt/ct=3.6
Now I'm struggling to understand what I times 3.6 by to make either 4, 8 and 12 finishing the formula.
What does vt/ct yield? (what are the units?)
 
  • #4
Hi Ware182. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

I'm not sure what form of answer they are looking for. Do you know about the exponential rise of voltage in a series RC circuit?
 
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  • #5
VT/CT =3.6uF VT being voltage total of 12V, CT being Capacitance total of 3.3uF recurring

from the formula provided it states that I now times that answer by C (i'm guessing means capacitance) this does not give me the expected figures.

I'm thinking that the formula provided is incorrect for a series circuit, as it works perfectly on parallel.

Or is it me going wrong somewhere?
Thanks
 
  • #6
NascentOxygen said:
Hi Ware182. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

I'm not sure what form of answer they are looking for. Do you know about the exponential rise of voltage in a series RC circuit?

long story short- no i don't know about exponential rise of Voltage in RC circuits

however thank you all for your prompt replies!
 
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  • #7
Ware182 said:
VT/CT =3.6uF VT being voltage total of 12V, CT being Capacitance total of 3.3uF recurring

from the formula provided it states that I now times that answer by C (i'm guessing means capacitance) this does not give me the expected figures.

I'm thinking that the formula provided is incorrect for a series circuit, as it works perfectly on parallel.

Or is it me going wrong somewhere?
Thanks
I suspect that you may be mistaken in the formula that you're using.

For a capacitance C the relationship between the charge and voltage on it is given by

Q = CV

Now, in your case you've calculated a net capacitance Ct and you have a voltage Vt = 12V placed across it. What is the charge on that capacitance?

Next, what do you recall about how charges are distributed between capacitors in series?
 
  • #8
gneill said:
I suspect that you may be mistaken in the formula that you're using.

For a capacitance C the relationship between the charge and voltage on it is given by

Q = CV

Now, in your case you've calculated a net capacitance Ct and you have a voltage Vt = 12V placed across it. What is the charge on that capacitance?

Next, what do you recall about how charges are distributed between capacitors in series?

so I can work out the charge for C1 by 10x10-6 x 12 = 120uC however I don't know how to apply that in a way to find the potential difference.

I understand that between each capacitor on this circuit it will drop 4V and 40 micro coulombs

However I don't understand how to use this information to solve the equation.
 
  • #9
Ware182 said:
so I can work out the charge for C1 by 10x10-6 x 12 = 120uC however I don't know how to apply that in a way to find the potential difference.
No, the full 12V is not across every capacitor. The full 12V is across the total equivalent capacitance that you found, Ct. So you can find the charge on THAT capacitance.

Now, this is where my question about how charge is distributed amongst capacitors in series becomes critical. You need to know how this works! Since all components that are series connected always share the same current, any change in charge on capacitors in series must be identical. So if the charge on your equivalent capacitor Ct is Q, every one of the series capacitors that make up Ct has the same charge Q.

If you know the charge on a capacitor then you can find the voltage across that capacitor via Q = CV.

In this case you find the charge Q on Ct, then apply that charge to each of the capacitors that make up Ct to find their individual voltages.
 
  • #10
gneill said:
No, the full 12V is not across every capacitor. The full 12V is across the total equivalent capacitance that you found, Ct. So you can find the charge on THAT capacitance.

Now, this is where my question about how charge is distributed amongst capacitors in series becomes critical. You need to know how this works! Since all components that are series connected always share the same current, any change in charge on capacitors in series must be identical. So if the charge on your equivalent capacitor Ct is Q, every one of the series capacitors that make up Ct has the same charge Q.

If you know the charge on a capacitor then you can find the voltage across that capacitor via Q = CV.

In this case you find the charge Q on Ct, then apply that charge to each of the capacitors that make up Ct to find their individual voltages.

Ok, thank you, I will have a play and see what i can make of it.
 
  • #11
gneill said:
No, the full 12V is not across every capacitor. The full 12V is across the total equivalent capacitance that you found, Ct. So you can find the charge on THAT capacitance.

Now, this is where my question about how charge is distributed amongst capacitors in series becomes critical. You need to know how this works! Since all components that are series connected always share the same current, any change in charge on capacitors in series must be identical. So if the charge on your equivalent capacitor Ct is Q, every one of the series capacitors that make up Ct has the same charge Q.

If you know the charge on a capacitor then you can find the voltage across that capacitor via Q = CV.

In this case you find the charge Q on Ct, then apply that charge to each of the capacitors that make up Ct to find their individual voltages.

ok so I've spent all day on it and I think I've finally got it.

1/10 +1/10 + 1/10 = 0.3
1/0.3 = 3.3 recurring
Qt = 3.3 recurring * 12 = 40
40/10 = 4 (4V drop across each capacitor)
40/10 + 40/10 +40/10 = 12V (Applying Kirchoff's law to check)

can you confirm that this is the way you were trying to show me
 
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  • #12
Ware182 said:
can you confirm that this is the way you were trying to show me
Yes, that's the method.

Try to remember to attach units to results, even intermediate step results. That or work symbolically until the very end where you can plug in numbers. It's much easier to spot errors with variables than amongst a sea of numbers.
 
  • #13
Thanks for all your help. that's good advice I'll be sure to apply that.
 

1. What is potential difference between three capacitors in series?

Potential difference, also known as voltage, is the difference in electric potential between two points in a circuit. In a series circuit, the potential difference is divided between the three capacitors in proportion to their capacitance values.

2. How is potential difference calculated in a series circuit?

In a series circuit, potential difference can be calculated by adding up the individual potential differences across each component. In the case of three capacitors in series, the total potential difference would be the sum of the potential differences across each capacitor.

3. What happens to the potential difference if one capacitor in a series circuit is removed?

If one capacitor is removed from a series circuit, the potential difference across the remaining capacitors will increase. This is because the total potential difference is divided between a smaller number of components, resulting in a higher voltage across each remaining capacitor.

4. How does the placement of the capacitors affect the potential difference in a series circuit?

In a series circuit, the placement of the capacitors does not affect the potential difference. This is because the potential difference is determined by the total voltage of the circuit and the relative capacitance values of the capacitors, not their physical placement.

5. Can the potential difference across each capacitor in a series circuit be equal?

In theory, the potential difference across each capacitor in a series circuit can be equal, but it would require all three capacitors to have the same capacitance value. In most cases, the capacitance values of the capacitors will differ, resulting in a different potential difference across each capacitor.

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