Charging a capacitor with another capacitor

10

Hi everyone, when charging an uncharged capacitor with a charged capacitor, with a resistor in series, why do we treat the capacitors as being in series and not in parallel when calculating the total capacitance of the circuit?
Thanks

Reply 1

As6dghn48

2

Hi

Reply 2

snowblob

OP

10

Original post

by Nitrotoluene

A resistor (R) will move positive charge from capacitor C1 to capacitor C2 when these capacitors are linked. The charge moves between the capacitors until both acquire the same voltage value Vf. A series equivalent model helps to understand how the charges and voltages distribute across different capacitors. During the steady-state condition both capacitors receive equal charge quantities. The total voltage of all capacitors in the equivalent circuit provides the final voltage of each capacitor.
Here is my 2 cents!

Thanks for your reply - what do you mean by the last line about the total voltage providing the final voltage of each capacitor?

Reply 3

Eimmanuel

Study Forum Helper

15

Original post

by snowblob

Hi everyone, when charging an uncharged capacitor with a charged capacitor, with a resistor in series, why do we treat the capacitors as being in series and not in parallel when calculating the total capacitance of the circuit?
Thanks

Short answer: the 2 capacitors are considered to be in parallel.

When the charged capacitor C1 is connected in series with a resistor and an uncharged capacitor C2,

•

the charges on the charged capacitor will distribute among C1 and C2 to reach same p.d. NOT same charges,

•

the total charge on C1 initially is the sum of the respective charges on C1 and C2 at equilibrium.

Recall from what you have learnt when the capacitors are connected in series and parallel.
When the capacitors are connected in series, the charges on capacitors connected in series are the same.

When the capacitors are connected in parallel, the total charge on capacitors connected in parallel is the sum of the charges on the individual capacitors.

Reply 4

Eimmanuel

Study Forum Helper

15

Original post

by Nitrotoluene

A resistor (R) will move positive charge from capacitor C1 to capacitor C2 when these capacitors are linked. The charge moves between the capacitors until both acquire the same voltage value Vf. A series equivalent model helps to understand how the charges and voltages distribute across different capacitors. During the steady-state condition both capacitors receive equal charge quantities. The total voltage of all capacitors in the equivalent circuit provides the final voltage of each capacitor.
Here is my 2 cents!

Both statements (in red) are wrong.

They are not corrected "fully" in this post.

Both capacitors will have same charges at steady state unless the capacitances are the same.

The voltage across each capacitor is the same at steady state.

It is WRONG to say the "total voltage" at steady state is 2*Vf in this post. This is like saying when we connect 2 cells of same emf of 1.5 V in parallel, the total emf becomes 3.0 V. This is wrong physics!

Reply 5

snowblob

OP

10

Original post

by Eimmanuel

Short answer: the 2 capacitors are considered to be in parallel.
When the charged capacitor C1 is connected in series with a resistor and an uncharged capacitor C2,

•

the charges on the charged capacitor will distribute among C1 and C2 to reach same p.d. NOT same charges,

•

the total charge on C1 initially is the sum of the respective charges on C1 and C2 at equilibrium.

Recall from what you have learnt when the capacitors are connected in series and parallel.
When the capacitors are connected in series, the charges on capacitors connected in series are the same.
When the capacitors are connected in parallel, the total charge on capacitors connected in parallel is the sum of the charges on the individual capacitors.

Thanks, this makes sense, but I wanted to know which formula you would use when working out the time constant of your circuit, to estimate when both capacitors would reach the same voltage (after about 5 time constants) - would you not treat the capacitors as being in series to work out the effective capacitance?

Reply 6

Eimmanuel

Study Forum Helper

15

Original post

by snowblob

Thanks, this makes sense, but I wanted to know which formula you would use when working out the time constant of your circuit, to estimate when both capacitors would reach the same voltage (after about 5 time constants) - would you not treat the capacitors as being in series to work out the effective capacitance?

I don’t know if there is easy way to get the time constant.
Fundamentally, I would write KVL and solve the differential equation.

Is this whole thread of questions originated from an exam question or your school question?

Charging a capacitor with another capacitor

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