Capacitor charging/discharging equation, maths

Hello,
Can anybody explain me the capacitor charge/discharge equation derivation in detail?
Places I check, they start at either complicated point or in other easier version, some of the steps are omitted.

Any pointers/ links will be helpful too.

Thanks.

Places I check, they start at either complicated point or in other easier version, some of the steps are omitted.

Click to expand...

Better you give the link reference and pinpoint what exactly you did not understand, instead of raising an open ended question.

Places I check, they start at either complicated point or in other easier version, some of the steps are omitted

Click to expand...

Please cite a reference so that we can start with a starting point; otherwise I will have to draw some diagrams to begin...

-Begin at the beginning and go on till you come to the end; then stop- Lewis Caroll

The mathematical constant e is in the formula. I experimented with a graphing program until I found that the equation below creates a plot which resembles the classic discharge graph. It assumes time constant=1.
V starts at 1V.
0.368 = 1/e.

V = 0.368 ^ t

After 5 seconds (5 time constants) the capacitor is nearly discharged, as according to convention.

After 5 seconds (5 time constants) the capacitor is nearly discharged, as according to convention.

Click to expand...

Another interesting concept is the half-time; this is the time needed for the voltage to drop by 50%. In the example above, you will have voltage reduce by 50% after 0.6931 secs. This number is ln(2) and also crops up when you expect it the least.

If the OP had shown few more interest in this issue since the opening of the thread, we could also explore the similarity among mathermatical models of different physical phenomens, for example understanding the discharge of the capacitor as the discharge of a fluid, which flows with more intensity as more charged is the deposit of liquid. The fact is that certain concepts have to be acquired intuitively before one proceed to learn something few palpable to human senses, such as the electric current.

Can anybody explain me the capacitor charge/discharge equation derivation in detail...

Click to expand...

I do not know how much detail you want, but I can try.

I am not drawing a diagram and you need to have a diagram to get most out of this description...

1. Consider a power supply (battery) of const voltage V and zero internal resistance; it can supply infinite current.

2. We have resistor of resistance R connected in series with a capacitor of capacitance C. The positive end of the battery is connected to the resistor via a switch and the free end of the capacitor is connected to the negative end of the battery.

3. Set up initial conditions: at time t=0, the capacitor has no potential across it and the switch is closed.

4. Now consider the case at time t; the current is i(t) and the voltage is v(t). The total voltage across the circuit is V=i(t).R + v(t); v(t) is the voltage across the capacitor.

5. Capacitor formula is C=dQ/dV; or CdV=dQ; also dQ=i(t)dt; therefore CdV=i(t)dt; The voltage across the capacitor will be then integral (1/C)(i(t)dt) with limit from 0 to t. Put this in the previous equation.

6. Hence is final equation is V= V(t) + (1/RC)(dV(t)/dt)

7. This equation is a simple differential equation and can be solved easily.

Hello,
Thanks you all for your responses.
here is more on my question.

I am going through this(attachment)

I dont get how it goes to vN(t) = K e -t/RC step.

Thank you

If you are clear about the previous step, then it is simple. Rearrange the equation to get:

RC dV/dt=-V; more rearragement and you get : (dV/dt)/V=-(1/RC); or dV/V=-(1/RC)dt; then integrate to obtain: ln(V)=-t/(RC); or V=V(0) exp(-t/RC)

Thank you very much. It is clear now. I made mistakes while doing this on my own.