charging capasitor formula

[Prototyp_V1.0]

Given a capacitor and a resistor in series
gnd - capacitor - resistor - voltageSource

There is a formula for voltage over the capacitor for a given time. Question is - how du we get there. I know about integrating and the math, but not the path :?:

Given that the capacitor is no charget at all, I know the formula will be:
\[Ucap = Usource \cdot (1 - \exp (\frac{-T}{RC}))\]

But how to get there?

My guess is that there must be set up a equation that is to be integrated ∫
Second, when the integration operation is done, there is a new equation left that consist of ln to some function (i guess this ln contains source voltage and capacitor voltage because the exp function.

 

[FvM]

The differential equation is:

I = C•dUc/dt = (Us-Uc)/R

You could either try integrate the equation or - that's more simple - show that the wellknown exponential function is a solution to the above equation.

 

[halls]

Ok, the voltage drop of the entire circuit lays on the resistor and the capacitor, this is:
\[U_s = U_{R}(t) + U_{c}(t)\]

Where \[U_s\] is the voltage of the source, \[U_{R}(t) = R\cdot{i(t)}\] is the voltage drop in the resistor, and \[U_C(t) = \frac{q(t)}{C}\] is the voltage drop in the capacitor.

The voltage drop in the resistor is based on Ohm's basic law, V = I.R, while the voltage drop in the capacitor can be understood by common sense: the capacity of a capacitor is a constant attribute, and the voltage drop means the potential difference between its plates, which is a charge-dependent value.

*EDIT: The capacity of a capacitor is given by the relation between the charge in its plates and the voltage drop, like: \[C = \frac{Q}{V}\] where Q is the charge and V is the voltage drop. In our case, the charge will be changing as time passes, that's why i used the notation in lower case and t-parameter dependent: \[U_C(t) = \frac{q(t)}{C}\]
*END_EDIT*

In the other hand, we know the current flow is, by definition, the variation in time of the charge of the circuit (which in this case, will be the same current for every element in the circuit), this is:

\[i(t) = \frac{dq(t)}{dt}\]

so the main equation remains:

\[U_s = R\cdot{i(t)} + \frac{1}{C}\cdot{\int_{t_0}^{t}i(t)\cdot{dt}}\]

if we differentiate this equation in time, assuming the source voltage is constant in time, we get:

\[0 = R\cdot{di(t)} + \frac{1}{C}\cdot{i(t)\cdot{dt}}\]

So, regrouping:

\[\frac{-1}{RC}\cdot{dt} = \frac{di(t)}{i(t)}\]

And now, we integrate the whole equation again:

\[\frac{-1}{RC}\cdot{(t-t_0)} = Ln(\frac{i(t)}{i_0})\]

Now we have to take exponentials in all the equation and regroup, so:

\[i(t) = i_0\cdot{e^{\frac{-1}{RC}\cdot{(t-t_0)}}\]

With this you have the expression for the current in the circuit, but still depending on two variables, initial current [text]i_0\] and initial time, \[t_0\]. We can consider the initial time as \[t_0=0\], as a reference point. Now, the initial current will be more tricky.

Imagine there is a switch between the source and the resistor, and we have it off, so the circuit is not working. Current here is null, while voltage drop on the resistor is null and so, voltage drop in the capacitor is null, thus null charge. In the precise moment we switch the circuit on, just in the same exact moment, current will start flowing, but there still will be no charge in the capacitor.

This means all the source voltage will be dropped in the resistor (remember we are still in the exact moment of switching on) and therefore, the value for the initial current, will be \[i_0 = \frac{U_S}{R}\]

Now we can write the exact expression for the current:

\[i(t) = \frac{U_S}{R}\cdot{e^{\frac{-1}{RC}\cdot{t}}}\]

Given this, we can calculate the expression for the voltage drop of the capacitor:

\[U_C = \frac{q(t)}{C}\]

\[U_C = \frac{1}{C}\cdot{\int_{t_0}^{t}i(t)\cdot{dt}}\]

\[U_C = \frac{1}{C}\cdot{\frac{U_S}{R}\cdot{\int_{0}^{t}e^{\frac{-1}{RC}\cdot{t}}\cdot{dt}}}\]

\[U_C = \frac{1}{C}\cdot{\frac{U_S}{R}\cdot{(-RC)}\cdot{\int_{0}^{t}\cdot{\frac{-1}{RC}}e^{\frac{-1}{RC}\cdot{t}}\cdot{dt}}}\]

\[U_C = \frac{1}{C}\cdot{\frac{U_S}{R}\cdot{(RC)}\cdot{\int_{t}^{0}\cdot{\frac{-1}{RC}}e^{\frac{-1}{RC}\cdot{t}}\cdot{dt}}}\]

\[U_C = U_S\cdot\int_{t}^{0}\cdot{\frac{-1}{RC}}e^{\frac{-1}{RC}\cdot{t}}\cdot{dt}\]

\[U_C = U_S\cdot\left( e^{\frac{-1}{RC}\cdot{0}} - e^{\frac{-1}{RC}\cdot{t}} \right)\]

And finally, you have the equation that you started with:

\[U_C = U_S\cdot\left(1-e^{\frac{-1}{RC}\cdot{t}} \right)\]

I hope this is not confusing and is helpful

 

[Prototyp_V1.0]

Wow with big letter W. Thanks a lot.

I'm reading an almost fully understand, but what is \[g(t)\] ?

 

[halls]

Oops sorry, it was meant to be a q(t), not a g(t). It is often used to denote the charge in some element/system/whatever.

It represents the charge in the capacitor, which is variable in time. As long as current flows through the capacitor it gets charged (positive charges in one plate and negative ones in the other) creating a potential difference which is denoted as \[U_S\].

I have edited the explanation for a better understanding, hope it works. Anyway, feel free to ask whatever you want.

 

[Prototyp_V1.0]

Ok, that make sense. Thanks