# Instantaneous rate of change (two capacitors)

1. Sep 20, 2010

### energychaser

Hello,

I have been pondering how to plot the instantaneous rate of change of one capacitor into another through a resistive impedance R.

So for example you would have

R1 = 100ohms

So one would have Capacitor 1 at 100 volts, discharge through R of 100 ohms into capacitor 2 of zero volts, and 10 farads.

When the process has settled to equilibrium, there will be 9.09 volts on each capacitor, after X time.

I would like to know how long it takes to reach equilibrium, and would think that normal time constants will not quite apply, because the capacitor is not discharging into a resistance only, but into another capacitor through a resistance, therefore you have a static impedance (R1) and a changing impedance (C2).

So what would this discharge graph look like? What would be the function? How long would it take to reach equilibrium at 9.09 volts on each capacitor?

If someone would be willing to explain this to me I would be very grateful!

2. Sep 20, 2010

### Staff: Mentor

Are you familiar with the differential equation that relates capacitor current to the change in the capacitor voltage? Have you been exposed to the "Kirchoff Current Law" equations for solving for circuit voltages and currents?

3. Sep 20, 2010

### energychaser

in answer...Once upon a time yes, I was familiar, now not so. This is an attempt on my part to break the rust loose as it were with an example I have been contemplating.

Point the direction?

4. Sep 20, 2010

### Staff: Mentor

I like to use the KCL equations to work with circuits like the one you described:

http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

And the differential equation relating the current and voltage of a capacitor is:

I = C dv/dt

That is, the current is equal to the capacitance value multiplied by the derivitave of the voltage with respect to time (the time variation of the voltage). So a larger capacitance, or a quicker change in voltage will result in a larger capacitor current.

To solve the circuit you described, write the KCL equations for the two sides of the connecting resistor, solve the simultaneous equations for the output voltage (voltage across the 2nd cap versus time), and use your initial conditions to get the final equation for C2(t). It will most likely involve an exponential e^(something), where the something involves the R and C values, and time.

Can you show us an attempt at doing that? We can talk you through parts if you show us a try...

5. Sep 24, 2010

### energychaser

Okay, I know that by the KCL concept what is leaving one capacitor 1 must be equal to what is entering capacitor 2.

I know that the current at any point is equal to the potential difference between the two capacitors, divided by the total impedance (resistive impedance of the resistor, and impedance of the second capacitor being filled)....so perhaps

(V1-V2)/(R+Z) = I

But I also have this equation

Which is applicable to a constant voltage source which this is not! but it does integrate with respect to time, so this is more like what I want.

So perhaps I would do something like

(V1-V2)/(R+Z) = (V/R)e^(-t/RC)

any pointers?

6. Sep 24, 2010

### energychaser

If i were to quickly restate my objective,

I want to know how integrate or understand how time acts with respect to reaching equilibrium when one capacitor discharges into another. So if I want to know how long it takes the current to reach 1% of its strength at switch close how would I do so?

7. Sep 24, 2010

### Staff: Mentor

You are close, but you need to write the differential equations in the KCL, and then solve them to get the exponential form that you are showing. You also then use the initial conditions V1(0) = V, V2(0) = 0, to figure out what the constants are in the final solution.

(Also, I'm not sure what your Z term is meant to represent.)

So, since this is not homework or schoolwork, I'll help you set up the equations as a learning exercise. This assumes C1 is the left capacitor, which is initially charged up to V, and C2 is the right capacitor, which is initially discharged. There is a resistor and series switch connecting the tops of the two capacitors, and the bottom of each cap is grounded.

For time t = 0+ right after the switch is closed, we can write the two node KCL equations, one for each node at the top of the caps:

$$C1 \frac{dV_1(t)}{dt} + \frac{V_1(t) - V_2(t)}{R} = 0$$

$$C2 \frac{dV_2(t)}{dt} + \frac{V_2(t) - V_1(t)}{R} = 0$$

Now you would assume solutions for V1(t) and V2(t), differentiate them, and plug them back into the equations to solve for the voltages. The V1(t) solution should be an exponential that starts at V and decays to some lower voltage Vf. The V2(t) solution should be an exponential that starts at 0 and rises exponentially to Vf. The time constants will have to be the same (I believe), and my initial guess is that the time constant will be R multiplied by the series combination of the two capacitances C1 and C2 (remember that caps in series do not add; they combine in a different way, right?).

Is that enough to get you going?

8. Sep 24, 2010

### Staff: Mentor

Ah, after re-reading your posts, I see that you are listing Z as the impedance of the 2nd capacitor. Don't worry about that for your main question.

9. Sep 24, 2010

### energychaser

Could you clarify what you mean by "Now you would assume solutions for V1(t) and V2(t)"

10. Sep 24, 2010

### Staff: Mentor

It's one technique for solving differential equations. In cases where you have a pretty good idea of the form of the final solution (like if it's going to be an exponential decay, or a sinusoidal solution), you guess the general form of the solution, with placeholder constants for amplitudes and time constants and phase shifts, etc., and then plug that "guess" back into the differential equation that you are trying to solve.

So as an example, say we take your circuit and short C2, so that the overall circuit is just C1, the switch and the resistor (which now goes to ground on its right side). For time t =0+ when the switch is closed, we can write the single KCL equation on top of C1:

$$C_1 \frac{dV_1(t)}{dt} + \frac{V_1(t)}{R} = 0$$

Guess a solution for V1 as follows:

$$V_1(t) = A e^{Bt}$$

Where A and B are constants that we will figure out with the "boundary conditions" of the circuit in a moment. We also need the first derivative of V1 with respect to time, so just differentiate the equation:

$$\frac{dV_1(t)}{dt} = AB e^{Bt}$$

Now plug these back into the KCL equation that we wrote:

$$C_1 AB e^{Bt} + \frac{A e^{Bt}}{R} = 0$$

Now you can solve for B, which shows you the time constant for the exponential decay.

Next use the fact that V1(0) = V (the starting voltage), which gives you the constant A.

Now you can write the full equation for V1(t).

For the problem with the two capacitors, you do basically the same thing, but the left cap will discharge to some lower Vf < V, and the right cap will charge up from 0 to Vf.