Capacitor Charge Calculation: C1 and C2 in Series

In summary, two parallel plate capacitors C1 and C2 are connected in series across a battery of emf V with a switch S. C1 has initial charge q0 and C2 is uncharged. The final charges on the two capacitors are given by q0 + (VC1 - q0)C2/(C1 + C2) and (VC1 - q0)C2/(C1 + C2), respectively. However, there may be some issues with the expressions and further clarification is needed to determine their accuracy.
  • #1
i_island0
123
0
there are two parallel plate capacitors C1 and C2 connected in series across a battery of emf V with a switch S. The capacitor C1 has some initial charge q0 while C2 is uncharged.
I have connected the positive terminal of battery (of emf V) to the positive plate of capacitor C1.
Final charge on the two capacitors that i got are:
Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)
is my answer correct?
 
Physics news on Phys.org
  • #2
In the series combination of the capacitors charge on each plate must be the same irrespective of the initial charge.
 
  • #3
sorry sir, but i m not able to understand your point. can you then tell me the charge on both capacitors as a function of time.
 
Last edited:
  • #4
i_island0 said:
Final charge on the two capacitors that i got are:
Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)
is my answer correct?

I think these expressions need some clarification and examination. Is VC1 intended to represent the initial "voltage drop" across C1, that is,

VC1 = q0/C1 ?

If so, there is a problem: VC1 and q0 have different units (voltage vs. charge), so the difference (VC1 - q0) in your expressions would not be meaningful.

Perhaps it would be helpful to show how you arrived at these results. You will not be able to construct functions of time describing the charges, since this idealized problem has no resistances in the circuit: you will not be able to calculate current. In any case, I don't think your expression can be correct.
 
Last edited:

FAQ: Capacitor Charge Calculation: C1 and C2 in Series

What is the formula for calculating the total capacitance of two capacitors in series?

The formula for calculating the total capacitance of two capacitors in series is C = (C1*C2)/(C1+C2).

How do I determine the individual capacitance values for capacitors in series?

To determine the individual capacitance values, you can rearrange the formula C = (C1*C2)/(C1+C2) to solve for either C1 or C2. You can then use this value to find the remaining capacitance value.

Can I add more than two capacitors in series and still use the same formula?

Yes, the formula for calculating the total capacitance of capacitors in series can be extended to more than two capacitors. The formula becomes C = (C1*C2*C3*...)/(C1+C2+C3+...).

What happens to the total capacitance when capacitors in series have different capacitance values?

When capacitors in series have different capacitance values, the total capacitance will always be less than the smallest individual capacitance value. This is because the smaller capacitor will charge faster and limit the overall charge in the circuit.

Why is it important to calculate the total capacitance in a circuit?

Calculating the total capacitance in a circuit is important because it determines the amount of charge that can be stored in the circuit. This information is necessary for properly designing and analyzing electronic circuits.

Similar threads

Replies
3
Views
1K
Replies
9
Views
2K
Replies
3
Views
2K
Replies
19
Views
2K
Replies
5
Views
1K
Replies
1
Views
3K
Back
Top