Simple RC Circuits
RC time constant
Consider a capacitor initially charged to V_{init}, and connected
to a resistor:

Here we will apply the Kirchoff Voltage Law (described in the next
chapter) for this loop (this is similar to the Loop Law we used in Hydraulic
systems): V_{r}+V_{c}=0 or V_{r}=V_{c}
Because the same current runs through each element (current is a through
variable): i_{r} = i_{c}. 
One of the homework problems which may be assigned will let you explore
this discharge numerically. Here, we'll do it analytically.
Using the constitutive law of the capacitor and resistor
V_{c} = 1/C q_{c}
V_{r} = R i_{r}

(5)

and the relation between charge and current, q_{c}' =
i_{c} , together with the voltage and current relations next
to the figure above, we can obtain a differential equation of motion
q_{c}' =  (1/RC) q_{c}

(6)

If you would rather see it in terms of current, take its time derivative:
i_{c}' =  (1/RC) i_{c}

(7)

Or in terms of voltage, using (5)
V_{c}' =  (1/RC) V_{c}

(8)

All of these equations of motion have an easy solution, e.g.
V_{c} = A e ^{ t /RC}

(9)

which you should check. You might wonder how we guessed this solution 
as you will learn as you study differential equations further in the next
course, a differential equation can often be recognized as belonging to
a particular class of equations which has a particular type of solution.
If you look at this equation, you see that it states that the derivative
of something is equal to a constant times itself. And what is the derivative
of itself??? A simple exponential. The exact form of the exponential can
be obtained by trial and error, or by using constants and matching terms
such that the original diff eq and initial/boundary conditions are satisfied.
In (9), we have already obtained the constant in the exponential itself
such that the differential equation is satisfied, but A can be any constant
and (9) holds true. However for this solution to be valid for the problem
at hand, we must choose A such that we match the stated initial condition
that V_{c} = V_{init} at t=0. For that to be so,
A must be V_{init}, so the specific solution is
V_{c} = V_{init} e ^{ t /RC}

(10)

You can see that the voltage (and the current, and the charge) all decay
away with time. The decay time constant is RC. This is a result worth remembering.
Whenever there's a capacitor and a resistor around, there is a time period
t=RC which is characteristic of the action.
Charging up a capacitor
That was discharing a capacitor, but what happens if the capacitor is initially
discharged, then placed in a circuit with a resistor and a battery? Here
is such a circuit. Note that there is a switch in the circuit above the
battery which will be closed at time t=0. At t=0, V_{C}=0.

To analyze this circuit, we will apply the Kirchoff Voltage Law that
we will see in more detail in the next section. The polarities of my meters
has been assigned consistently with our convention:
V_{R}+V_{C}  V_{B} = 0
Because the same current runs through each element (current is a through
variable): i_{R }= i_{C }= i 
Our constitutive laws of the two elements are:
V_{C} = 1/C q_{C}
V_{R} = R i_{R}
Again, starting with our current/charge relation for the capacitor,
q_{C}' = i_{C} = V_{R}/R = ( V_{B
} V_{C })/R
q_{C}' = V_{B }/R  (1/RC) q_{C}
Since we are not so interested in charge per se, but more so in the voltage
changing across the capacitor, we can rewrite this diffeq using the constitutive
law of the capacitor again:

(11)

Again, we have a differential equation in which the derivative of something
is equal to a constant times itself, plus another constant. We anticipate
an exponential solution for this and the following is the solution for
(11):

(12)

Again, you should check this solution to convince yourself that it indeed
satisfies the diffeq (11) and the initial condition, V_{C}(t=0)=0.
So, what does this equation mean? Let's plot it, using V_{B}=9
volts, R=2 ohms, C=10F:

This is the voltage in the capacitor as a function of time. It charges
up exponentially until the capacitor reaches the voltage of the battery,
9 volts. After this point in time, nothing happens! 
So, if after a certain time, V_{C}=constant, what does that mean?
This is the steadystate condition, meaning no more changes with
time. Thus in steady state, V'_{C}= 0. Since the constitutive law
of the capacitor states that V_{C}= 1/C q_{C}, which we
can also write as V'_{C}= 1/C i_{C}, therefore in steady
state i_{C}=0! In fact, if we plot the current in the system as
a function of time, we see that it starts off at some initial value, then
decays to zero at the same time as the voltage in the capacitor reaches
9 volts:
From these two examples, we see that RC circuits have a time constant
associated with them which is R*C (convince yourself that the units of
ohms*farads really equals seconds!) and that such circuits could effectively
be used as timers. We also see that the capacitor allows a current through
only until it has reached an equilibrium voltage, at which point the current
stops.