Capacitors in sereis charging & discharging time?

I know that when two caps are connected in sereis to a 12 volt battery as shown below, the 12 volts are divided between the two caps in reverse.



and voltage will remain fixed value on each capacitor for a long time

When I connect a 100KΩ resistor in parallrl to the lower (10u) capacitor, I noticed that the voltage on this capacitor decreases, and at the same time the voltage on the upper capacitor is increasing.



When I connect a 100KΩ resistor in parallrl to the upper (22u) capacitor, I noticed that the voltage on the upper capacitor decreases, and at the same time the voltage on the lower capacitor is increasing.



My questions are:
(1) Why is voltage decreasing on the capacitor connected to the 100K resistor?

(2) In fig.2,,, At what time (t =??) will the voltage at the lower capacitor (10u) = 3 volt??
thanks

As you said, if we have the capacitors (C1, C2) only, connected to the generator VG, then the voltage between C1 and C2 will be:

Vx=VG*C1/(C1+C2)

When you connect the resistor to the upper capacitor (C1) it will start to discharge through the resistor, while C2 will be further charged. Then V(C1) will decrease and V(C2) will increase.
If the resistor is, instead, connected in parallel to C2, it will discharge through the resistor while C1 will be further charged. Then V(C2) will decrease and V(C1) will increase.

In any case must be: V(C1)+V(C2)=VG

To calculate the voltage behaviour with respect to time we can write the following equations (let's suppose R in parallel with C1):

VG=V1+V2 (V1 and V2 are voltages across the two capacitors)
V1=R*IR (IR current through the resistor)
IC=C1*V1' (IC current through C1, V1' is the first derivative of V1)
IG=C2*V2'
IG=IC+IR

Developing the calculations:

R*(C1+C2)*V2'+V2=R*C1*VG'+VG

Homogenous (that is VG=0) solution

V2=K*exp[-t/(R*C)] where C=C1+C2

considering steady state (VG'=0) for initial voltage V2(0)=Vo then

V2(t)=(Vo-VG)*exp[-t/(R*C)]+VG obtained substituting the solution in the differential equation for t=0

But when we connect the generator to the circuit a step (from 0 to VG) will be generated, so VG' !=0. Then Vx will quickly reaches the value Vx=VG*C1/(C1+C2) from which will go to VG following the previous exponential formula.
If Vo+VG*C1/(C1+C2) < VG then V2 will increase otherwise will decreas.

Thank you for trying to help
I have reviewed many books and sites the net
All those sources are talking about one capacitor connected with one resistor in series.
the following image is found in most sources:

In my circuit:

My question was: at what time will the voltage on the 10u cap decrease to 3 volts?
I need the formula
thanks

samy555 said:

My question was: at what time will the voltage on the 10u cap decrease to 3 volts?
I need the formula

Click to expand...

It's the half life period: t_1/2 = RC*ln2

69ms for your above circuit.

No
I've built the circuit and noted that it takes about 20-25 seconds to get to 4 volts
I hope to find a mathematical formula to find out in advance

samy555 said:

I've built the circuit and noted that it takes about 20-25 seconds to get to 4 volts

Click to expand...

You probably built the circuit with electrolytic capacitors (elcos), isn't it? Elcos use to have considerable leakage currents, which increase with the size of their capacitance, their voltage, and additionally strongly with temperature. So in this case the upper elco with its larger leakage current may well recharge the lower one for quite a while.

Simulation shows that the time constant (without those leakage currents) is RC = R * (C1+C2), in your case 320ms.

V1/V2 = C2/C1 -> V1 = 2.2*V2 ; V1+V2=12V = 3.2*V2 -> V2 = 12/3.2 = 3.75V ; V1 = 12V - 3.75V = 8.25V

Discharge to half of the original voltage (i.e. from 8.25V -> 4.125V): t_1/2 = RC * ln 2 = 222 ms

Discharge from 8.25V to 3V: t_3V = RC * (ln 8.25 - ln 3) = 1.01*RC = 324 ms , cf. the PDF below!

View attachment cap-discharge.pdf


You'd see these lower time values, if you'd use foil capacitors instead of elcos.

Re: Capacitors in sereis charging &amp; discharging time?

erikl said:

You probably built the circuit with electrolytic capacitors (elcos), isn't it? Elcos use to have considerable leakage currents, which increase with the size of their capacitance, their voltage, and additionally strongly with temperature. So in this case the upper elco with its larger leakage current may well recharge the lower one for quite a while.

Simulation shows that the time constant (without those leakage currents) is RC = R * (C1+C2), in your case 320ms.

V1/V2 = C2/C1 -> V1 = 2.2*V2 ; V1+V2=12V = 3.2*V2 -> V2 = 12/3.2 = 3.75V ; V1 = 12V - 3.75V = 8.25V

Discharge to half of the original voltage (i.e. from 8.25V -> 4.125V): t_1/2 = RC * ln 2 = 222 ms

Discharge from 8.25V to 3V: t_3V = RC * (ln 8.25 - ln 3) = 1.01*RC = 324 ms , cf. the PDF below!

View attachment 96931


You'd see these lower time values, if you'd use foil capacitors instead of elcos.

Click to expand...

First thank you very very much for that excellent answer

Secondly, I apologize for the mistake, the resistor was 1 mega-ohm and not 10 kilo ohms

Third: the mathematical formulas you provided gave me the same practical results

I have a questions:
How are the the 10u cap discharged when it is still connected to a battery?
thank you again

- - - Updated - - -

erikl said:

You probably built the circuit with electrolytic capacitors (elcos), isn't it? Elcos use to have considerable leakage currents, which increase with the size of their capacitance, their voltage, and additionally strongly with temperature. So in this case the upper elco with its larger leakage current may well recharge the lower one for quite a while.

Simulation shows that the time constant (without those leakage currents) is RC = R * (C1+C2), in your case 320ms.

V1/V2 = C2/C1 -> V1 = 2.2*V2 ; V1+V2=12V = 3.2*V2 -> V2 = 12/3.2 = 3.75V ; V1 = 12V - 3.75V = 8.25V

Discharge to half of the original voltage (i.e. from 8.25V -> 4.125V): t_1/2 = RC * ln 2 = 222 ms

Discharge from 8.25V to 3V: t_3V = RC * (ln 8.25 - ln 3) = 1.01*RC = 324 ms , cf. the PDF below!

View attachment 96931


You'd see these lower time values, if you'd use foil capacitors instead of elcos.

Click to expand...

First thank you very very much for that excellent answer

Secondly, I apologize for the mistake, the resistor was 1 mega-ohm and not 10 kilo ohms

Third: the mathematical formulas you provided gave me the same practical results

I have a questions:
How are the the 10u cap discharged when it is still connected to a battery?
thank you again

Re: Capacitors in sereis charging &amp; discharging time?

samy555 said:

How are the 10u cap discharged when it is still connected to a battery?

Click to expand...

With ideal capacitors (no leakage current at all) and no discharge resistance, there's virtually no netto current, i.e. no steady charge flow through the series capacitor connection, the 2 capacitors just share evenly their (once received) total charge: Vtot = V1 + V2 ; Q1 = Q2 ; Qtot = Q1 + Q2 = C1V1 + C2V2 --> V1/V2 = C2/C1 .

The discharge resistor allows for total discharge of the lower cap; the original Qtot - because of the law of charge conservation - has to assemble at the upper cap: Qtot = C2*Vtot = C2*(V1 + V2) .

Because both effects (redistribution of the capacitors' charge, and discharge of C1) occur at the same time, i.e follow the same timing behaviour, the sum of C1+C2 appears in the equation of the time constant. The charge-redistribution of both caps - resp. the discharge of C1 - both follow this time constant R*(C1+C2).

Re-charging from the battery isn't possible (in case of ideal capacitors, i.e. no leakage current), because - apart from the charge re-distribution - a steady current flow through such a cap isn't possible (infinite high - or non-existent - parallel resistance).


Not sure if I was able to present this explanation in an understandable form.