Resistance with capacitors
With a compressed air system, if there is a restriction in the pipe
leading to the storage tank, the airflow to the tank will be impeded
and the tank will take longer to fill. Likewise, if a resistor is
placed between a capacitor and the voltage source it will take longer
for the capacitor to charge than if the resistor weren't there.
Therefore, the time it takes a capacitor to charge is a product not
only of the capacitance but also of any resistance in the circuit.
RC time constants
When a capacitor is charging it acts like an air tank that is being
filled with compressed air. When you first connect an air compressor to
an air tank, the tank starts to fill quickly. Over time the tank fills
more slowly. This is because at first there is no pressure in the tank
to resist the inflow of air. Over time the tank builds pressure which
pushes back on the incoming air. As the pressure in the tank approaches
the pressure of the compressor the tank fills very slowly. Finally,
there is so much pressure in the tank that the compressor cannot force
any more air into it. This happens when the pressure in the tank equals
the maximum pressure of the compressor. This is quite noticeable when
you fill your tires with an air compressor that has barely more
pressure than the tires. Toward the end the air goes into the tire very
slowly.
A charging capacitor acts the same way. When you first connect a
battery to a capacitor the capacitor charges quickly as the electricity
rushes into the empty capacitor. Over time the capacitor charges more
slowly as the voltage in the capacitor increases and pushes back on the
incoming electricity. When the voltage in the capacitor approaches the
voltage of the battery the capacitor charges very slowly. Finally the
battery cannot force any more electricity into the capacitor. This
happens when the voltage in the capacitor equals the voltage of the
battery.
If you restrict the flow of air into a storage tank, it will take
longer to fill. Likewise, if you restrict the flow of electricity into
a capacitor, it will take longer to charge. Also, a larger tank will
take longer to fill than a smaller tank, just as a capacitor with more
capacitance will take longer to charge than a capacitor with less
capacitance.
Let's make an imaginary demonstration circuit to watch a capacitor as it charges.



Demonstration circuit 

Graph to show voltage and current over time 
The above circuit consists of a 1 ohm resistor, a 1 farad capacitor and
a 10 volt battery connected in series. The SPDT switch can be placed in
two positions. In the up position you have a series circuit going from
the positive terminal of the battery to the resistor, then to the
capacitor and back to the negative terminal of the battery. This
circuit will charge the capacitor. The down position takes the battery
out of the circuit leaving a series circuit with just a resistor and a
capacitor. The current meter and volt meter show that the voltage is
being measured across the capacitor and the current is the total
current in the series circuit. This circuit will discharge the
capacitor.
First lets flip the switch to the discharge position to make sure the
capacitor is completely discharged. We don't want any residual voltage
skewing the demonstration.
First, discharge the capacitor 
Now flip the switch to the charge position. Let's freeze time and look at what happens at that moment.



The moment the switch is moved to the charge position the capacitor looks like a short circuit. 

The voltage is zero (green dot in lowerleft corner) and the current is
at maximum (red dot in upperleft corner,10 amps in this case). 
At the moment you close the switch the capacitor is empty so it offers
no resistance to the inflow of electricity. The capacitor looks like a
short circuit. How can this be? Electrons are streaming onto one plate
of the capacitor. Since electrons repel each other these electrons push
free electrons off the opposite plate. For every electron that enters
the negative side of the capacitor an electron leaves the positive
side. Electrons go in one side and out the other. Even though the
electrons leaving the capacitor are not the same electrons that are
entering the capacitor it looks like electrons are flowing through the
capacitor.



Discharged capacitor.
The dashes represent free electrons on the plates. The two plates have an equal number of free electrons.


Charging capacitor.
Electrons enter the negative electrode and leave the positive
electrode. Current appears to flow through the capacitor (showing
electron flow here because we are following the electrons).^{[1]} 
Again, the capacitor looks like a short circuit. How much voltage can
you have across a short circuit? None. The voltage across the capacitor
is zero. How much current is flowing. With the capacitor acting like a
short circuit all that is left is a 10 volt battery and a 1 ohm
resistor in series. The current is 10 amperes.
Now let's move time forward. Electrons will stream onto the negative
plate and stream off the positive plate. This causes a backup of
voltage across the capacitor and it will begin to impede the inflow of
electrical current. As a filling air tank will begin to push back on
the air flowing into it, the capacitor begins to push back on the
electricity flowing into it. Let's freeze time again after the
capacitor has charged for one second.



After
one second the capacitor has charged to 6.32 volts (63.2% of the
battery voltage) and the current has decreased to 3.68 amps (36.8% of
the maximum current or 10 amps minus 6.32 amps). The voltage has
increased by 63.2% and the current has decreased by 63.2%

A capacitor time constant (RC time
constant), by definition is the time it takes for a capacitor to charge
to 63.2% of the source voltage (battery voltage). It is
calculated by multiplying the resistance by the capacitance. In the
following formula the lowercase Greek letter tau (t) represents the
time constant.
Where:

τ 
=

Time constant in seconds (the time it takes to reach 63.2% of the source voltage). 

R 
=

Resistance in ohms 

C 
=

Capacitance in farads 
With a one ohm resistor and a one farad capacitor we have a time
constant of 1 second (1 X 1 = 1). Therefore it will take one second for
the capacitor to reach 63.2% of the battery voltage. That is 6.32 volts
here. The increased voltage across the capacitor is resisting the
current flow and reducing it proportionally. While the voltage across
the capacitor has increased by 63.2% the current through the circuit
has decreased by 63.2%.
If we double the resistance it will take twice as long for the voltage
across the capacitor to reach 63.2% of the source voltage. The same
thing will happen if we double the capacitance. If you double them both
it will take four times as long for the capacitor to reach 63.2% or the
source voltage. The above circuit has a time constant of 1 second.
Double the resistance or the capacitance and the time constant will
become 2 seconds. Double both the resistance and capacitance and the
time constant will become 4 seconds. We'll come back to that
later. For now let's get back to the demonstration circuit.
We start with 10 volts at the battery and have 6.32 volts across the
capacitor. Where are the rest of the volts? Remember Kirchhoff's
Voltage Law. The voltages in a series circuit must add up to the
battery voltage. With 6.32 volts across the capacitor the remaining
3.68 volts must be across the resistor.
After one time constant the capacitor has 6.32 volts and the resistor
has 3.68 volts. This adds up to the battery voltage of 10 volts,
satisfying Kirchhoff's Voltage Law. 
Since this is a series circuit, whatever current is flowing through the
resistor is flowing through the rest of the circuit. What is that
current? We have 3.68 volts across 1 ohm. We know the voltage so we
divide into it. 3.68 1 = 3.68 for 3.68 amperes. This is the current
we measure anywhere in this series circuit. If time is frozen this
circuit obeys all the rules we expect for DC circuits.
Now let's move forward to two time constants.



The situation after two time constants (two seconds with this circuit).

Now the voltage across the capacitor is 8.65 volts. This is 63.2% of
the difference between 6.32 volts and 10 volts. In the time from the
first time constant to the second time constant the capacitor charged
63.2% of the remaining voltage. This happens for every time constant,
the voltage across the capacitor increases by 63.2% of the remaining
source voltage. The current follows the opposite course. For each time
constant the current decreases by 63.2% of the remaining value.
Now let's see what happens after five time constants.



After five time constants the voltage has essentially reached the source voltage and the current has reached zero.

Now, at 99.3% of the source voltage the capacitor is essentially fully
charged. In reality the voltage will never reach the source voltage. It
will just get closer and closer. However, the voltage is now only 0.7%
away from the battery voltage. It's fully charged. Let's move on.
Capacitors as DC blockers
After five time constants the current has essentially reached zero
amps. When the capacitor first started charging there was a rush of
current but after five time constants it has stopped (or at least is
very low). When you apply a DC voltage to a capacitor you will get a
surge of current then the capacitor quickly blocks current flow. The
capacitor is now essentially what it says on the box, two conductors
separated by an insulator. It's an open circuit. All the source voltage
appears across the open circuit and there is no current flow (see Open
circuits above). This is why
capacitors are said to block DC but pass AC. As long as the voltage remains constant, they do.
However, the capacitor has something that any plain old open circuit
does not have. It has an energy store. This energy can be released to
do work. A charged capacitor is a lot like a battery but doesn't have
anywhere near the energy density of a battery. It's like comparing a
compressed air tank to an air compressor. The compressor will supply
compressed air as long as you supply power to it. A compressed air tank
will quickly run out of air. We will see how this energy store is used
in practical circuits when we study practical circuits in another
volume.
Demonstration:
You can watch a capacitor as an energy store in action using an analog
multimeter. Put the meter in the ohms range and place the probes across
a large capacitor (1,000 microfarads or so). When in the ohms range the
red probe may or may not be positive. You may have to try it both ways.
You should see a very low resistance at first but as the capacitor
charges the resistance will increase eventually reaching infinity. Now
put the meter in a low voltage range and place it across the capacitor.
You should see some voltage. How much depends on the battery used by
the meter to measure resistance. This voltage should go down slowly as
the capacitor discharges.
Now let's change the resistance to 2 ohms and see what happens to the
time constant. Start by discharging the capacitor. Now flip the switch
to the charge position and freeze time when the voltage reaches 6.32
volts.



Same circuit with a 2 ohm resistor. Notice that it takes twice as long for the capacitor to reach 63.2%


This
is the same circuit as to the left. Here the horizontal scale has been
shrunk in half. Notice that the shape of the curves is identical to the
original 5second scale.

This time it took two seconds to reach 63.2% of the battery voltage.
What happened. It's like pinching the hose from the air compressor to
the air tank. The higher resistance caused it to take longer to fill
the tank. The higher resistance does the same thing to the circuit.
Twice the resistance leads to twice the time. Now put the 1 ohm
resistor back in the circuit and change the capacitor to a 2 farad
capacitor. We get exactly the same result. The time constant has
increased to 2 seconds. Using the original horizontal scale on the
graph the curve becomes stretched out. However, if we shrink the
horizontal scale proportionally the curve looks identical to the
original.
Let's change the scale so that it always shows 5 time constants, regardless of how long a time constant is.



Here we have 2 ohms and 1 farad. The time constant is 2 seconds.


Here
we have 2 ohms and 100,000 microfarads. The smaller capacitor charges
10 times as fast so the time constant is only 200 mS (0.2 seconds).
However, no matter how long a time constant is, the curves look
identical if the horizontal scale is in time constants.
