Okay, so now, we're going to be talking about capacitors, the actual physical
devices. So, we'll start by presenting how these
capacitors work in a system, identify the behavior in DC circuits.
And then graphically represent the relationships between current, voltage,
power, and energy within capacitors. To get some feel for how these devices
are actually going to be used and how they work.
From out previous class we talked about capacitance, we talked about finding the
capacitance of things. And got some idea of the relationship
between current and voltage with capacitors and it was similar to what we
did with resistance. And we now have some idea about what
capacitance means, basically, it's how well this device holds holds these
electric fields. So, how much charge is going to be stored
by that capacitor for any amount of voltage that we're putting across it.
After we finish talking about capacitors, we're going to be talking about
inductors, which are different devices. But you'll notice a lot of similar
behavior, a lot of parallelism between the two.
And so, it can be good to go back and forth and look at them together to see
how what things are similar and what things are different.
The objectives for this particular lesson are to allow you to analyze the
capacitors when they are placed in parallel and series configurations.
To analyze DC circuits that have capacitors in them.
DC meaning that there's no changes in the system, everything is a constant value.
And then calculating the energy of a capacitor.
We'll do a calculation and a derivation for that.
And then we'll also sketch these current and voltage and power energy curves to
see the relationships between them. So, here we've placed capacitors into a
parallel configuration. Due to Kirchhoff's current law, we know
that i, the current coming in, is equal to i1 plus i2, the two currents that are
flowing. We also know that because these devices
are in parallel, the voltage across them is equivalent, thanks to Kirchoff's
voltage law. That allows us to do this derivation.
Since i is equal to i1 plus i2, and these currents can then be related by voltages
C1 dv dt plus C2 dv dt. If you factor out the Cs, put the dv dt
together, then you see that C1 plus C2 is our capacitance.
Since capacitance was basically defined to be the C that gives us this behavior C
dv dt so, there's our C. If we place them in series instead, now
we're going to have to use the, the other equation but we're still making use of
Kirchhoff's laws. The current is equivalent between the
two. Now the voltages are different.
And my voltage ab when we're using this notation the first letter is the one with
the plus the second letter is the one with the minus.
That voltage is going to be equal to v1 plus v2.
So let's start form there, Vab is equal to v1 plus v2.
Using our integration properties, finding voltage as a function of current, we get
this. And after moving things around, doing
some manipulation, here we make use of the fundamental theorem of calculus and
take a derivative. We eventually come to this, where i is
equal to C dv dt, where that is our C. What we're doing is we're inverting each
of the Cs, adding them together, and then inverting the result.
So, you might think back to the resistors, capacitors in series behave
very similarly to resistors in parallel. Capacitors in parallel behave a lot like
resistors in series in the sense that to find equivalence, here we're doing this
inverse sum and then inverting. And for the parallel, we just added up
the individual capacitances, just like with the resistors in series adding up
individual resistances. So, its good to notice that, that
combination so you can remember these equations.
When we put these things into a dc circuit what will happen is, since
everything is constant nothings changing in time.
The capacitors going to find it's way into an equilibrium.
Where it finds its way into that equilibrium is because charge is then
pushed onto the plates, to a point where it's kind of at this natural voltage.
And the, essentially, for an analysis point of view, is equivalent to taking
this capacitor, and just removing it. And if I want to know, for example, the
voltage across this capacitor, what I do in this DC system, start off by removing
the capacitor. Now, this gap here, the voltage for this
gap is the voltage across the capacitor. Now, since I don't have any current
flowing through this resistor here, I can't have any voltage across this
resistor because it's just dangling off the end.
So this is really unimportant for our analysis, we can pretend like it doesn't
even exist at this point, for the purposes of what we're doing.
At this point the system reduces to essentially a voltage divider, we have
one volt here one ohm here and three ohms here.
Which means that the drop across this resistor is three fourths of a volt and
since this voltage is measuring the difference from this node to this node.
It's essentially in parallel with this resistor.
So, this voltage here is also going to be equal to three fourths of a volt.
To find the stored energy in a capacitor we're going to start by doing an energy
calculation and we'll start from what we know.
First we know that power is equal to current times voltage.
Secondly we know we can calculate energy, by integrating that power and then adding
our initial energy. Finally we know that the relationship
between current and voltage in the capacitor is i is equal to C dv vt.
So we will plug in for p i times v in our energy calculation and then replace i
with C dv dt to give us this equation. After doing a little bit of manipulation
this is doing a change of variables. If you kind of get lost in this step go
back and review changes of variables from calculus.
That leads us to this equation, and finally we get this result.
The energy stored in a capacitor is equal to one half Cv squared.
So the bigger the voltage, the bigger the energy is stored by our capacitor and
also it's scaled by the capacitance of our capacitor as well.
And we'll take that and apply it to a graphically seeing how all of these
things relate. So, first of all we're going to take a
curve that represents the voltage through the capacitor.
And here notice that our capacitance is going to be one farad, it's a really big
capacitance. But it makes it easy for doing this
example, it makes the math very simple and clear.
If I want to find the current from the voltage, i is C dv dt.
Since C is 1, this basically just amounts to finding the derivative of the voltage,
or the slope of this line. So, the slope of this line it goes up
two, in two seconds. So the slope is one.
So our current, in this graph, is one amp.
So this flattens out here, it has zero slope.
Which means our current goes to zero. And then since it drops down here, at
twice the rate, our slope is 2, our current here is going to be minus 2 amps.
To find our power, which is in this graph, we're going to be combining these
two by multiplication. So, if I multiply these two functions, I
start at zero going up to 2 times 1, so it's going to increase like this.
Then at this point where this flattens out the current goes to zero so, so does
our power. And this goes down but this is now a
negative current, so we take minus 2 times 2 volts which puts us at this
point. And then we're just basically multiplying
a constant value by this line so we get a curve linearly increasing here.
So this is what our power's going to be. Now remember this is not a problem, power
can jump as much as it likes. But something that's interesting to
notice is that because of this behavior, the C dv dt behavior, our voltage in the
capacitor will always be continuous. If it wasn't continuous, that would mean
that we had an unbounded current. And since that's not possible, if you
ever get a result where your voltages across your capacitors are jumping
around, you've made a mistake. But it's fine if current does, current
can jump around as much as it likes in a capacitor.
To calculate our energy here, we're basically going to be integrating this
line, and so we know it starts at zero. We're going to assume that, well, maybe
not, that's going to be zero. And as we integrate over this triangle,
this is going to then increase. And since this is a line and we're
integrating it, basically integrating by x.
So, we're going to get something that's and X square, so it's something that's
parabolic. So it's going to parabolically increase
and we can find this final point by integrating the area under that curve.
So, 2 times 2 times one half, which brings us up to 2 joules.
Now one of the problems that my students often made is that as this power drops
down to zero here, they would say, okay great, that means my energy comes down to
zero. It does not, remember, energy is
something that has to be continuous. You can't instantaneously change energy,
because to do so would require unbounded power.
It's just going to go flat. This power being zero just in just means
that no power's being generated or consumed.
So it goes flat here, and then this is basically backwards.
And this is now going to parabolically decrease until it reaches back at zero.
Since the area under this curve is the same as the area under this one.
In order to check our work, we remember that w is equal to one half, Cd squared.
[INAUDIBLE] C here is 1, and we do one half times all these values squared.
We'll see that this, if you square it and divide it by 2, gives us this as well.
And so we come at it from two different directions, we get the same answer.
It's a great way to check your work. And if they match, you probably did it
correctly. because, so to summarize, we calculated
capacitance for capacitors in both parallel and series configurations, and
did the derivations. Now, the derivations were kind of gone
over very quickly, I put all of the, the steps there.
But if you got lost by any of those derivations, it might be a good thing for
you to go to the forums. And ask questions there, to make sure
that you really understand what we were doing.
We also identified how capacitors in DC circuits behave like open circuits.
We derived an equation for energy stored in a capacitor as an electric field.
And then we showed graphically the relationships between our voltage, our
current, our power and energy in capacitors.
Now that we've talked about capacitors, the next topic of discussion will be
inductors. Where capacitors hold their energies as
electric fields, we're now going to be looking at magnetic fields.
And so we're going to be talking a little bit about how electricity and magnetism
interreact. and that allow us to start talking about
something called inductance, so until then