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