Hi, I'm Jacob Block, and I'm an EECE student here at Georgia Tech.
Today I'll be going over and extra problem on RC Circuits.
So, in this extra problem, were going to be looking at an RC circuit.
And in this, so, for here, we have a a 5 volt source for a 300k resistor, a 10
microfarad capacitor, a switch. and the switch closes at t equals 0, and
a 150k ohm resistor. So, in this problem, we want to know what
is the voltage across the capacitor for all time t.
So, let's take this setup the circuit in two stages.
So, the first stage we're going to look at before the switch is flipped.
Before the switch is flipped, we have an open circuit here.
And that means that we have simply a resistor and a capacitor.
And this capacitor, since this circuit has been running for a really long time,
for t less than than 0, the capacitor is filled up.
And for a filled capacitors, we can replace those with an open circuit.
So, really, quickly we know there is no current because of the open circuit.
So, this top node has to have a voltage of 5 relative to this bottom node.
So, the voltage across here, Vc or, t less than 0, is 5 volts.
Now, we have our switch flipped. There's kind of 2 so now that our switch
is flipped, we have our voltage source, resistor, capacitor, and our parallel
resistor. 5 volts, 300k, 10 microfarads, and 150k.
So, how do we analyze this circuit now? Well, we can set up a KCL at this node.
And then, use the currents calculate the current in this top branch, the capacitor
branch, and this resistor branch. So, let's go ahead and do that.
So, we kind of blow up this node, here. We know that the the voltage at this node
is Vc. And the voltage at this top node is 5.
And then, we have our resistor in the middle.
So, the current through this resistor is 5 minus Vc over 300k, and similarly for
this branch. We know that the voltage is Vc here.
And the voltage is 0 here. And so, we know the current over across
the resistor is Vc minus 0 over 150k. And finally, for the bottom branch, the
current through a capacitor is the capacitance times the change in voltage.
So, we can write our, our KCL, so the sum of the currents in have to be equal to
the sum of the currents out. That tells us that we have 5 minus Vc.
on the left side, over 300k. And on the right side, we have Vc over
150k, plus, this 10 micro microfarad capacitor times dVc, dt.
So, lets and using, so now that we have our, kind of our KCL equation, we can and
it's a first order differential equation with a constant forcing function.
Or, a constant source on it. We can rewrite it in terms that we've
seen in the lecture slides. So, let's, let's go ahead and do that.
So, by moving our, our equation around, we get dVc, dVc of t, dt plus 1 times Vc
of t. equal to 10 E to the negative oops, sorry
5 3rds equal to five thirds. So, so this is our differential equation
for the for that node. And it describes the currents entering
and leaving that node. And we also know that Vc at t equal 0,
right when the switch flips, is 5 volts. So, using the formula that we have in the
notes, Vc of t is equal to 5 3rds, the k over a times 1 minus E to the negative a,
which is 1 times t. plus the initial condition, 5, times E to
the negative at, which is just t. And this is valid for all time t after 0.
And we can kind of simplify that in one, in one more step.
So, Vc of t is out of 5 or t less than 0. And 5 3rds plus 10 3rds, e to the
negative t, or t greater than 0, after the switch is flipped.
Let's go ahead and, and graph that solution.
So, the voltage starts at 5. And this extends out to however, I mean,
the circuit's been running for a really long time, so this gets extended out to
the left. And we know that the circuit ends up at 5
3rds because as t goes to infinity, this term drops out and we're just left with 5
3rds. So, it kind of approaches this level
here. And in between, we have this transient
response of 5 3rds plus 10 10 3rds e to the negative t.
So, this exp, there's an exponential curve that connects these two points.
And we know that this so this, this, this switch flips at 0.
And 1 second later, if we plug in 1 here, we drop significantly in our value, and
this, kind of, value here, after one, one time step, is 2.89 volts.
And if we actually look at our circuit, we can kind of see where this negative 1
comes from. For our C circuits, we can calculate a
tao constant, which is equal to RC. And this RC, the R for this C is the
parallel combination of this 300k and 150k.
So, when you combine those two parallel resistors of 1 over 300k plus 1 over
150k, and multiply it by our 10 microfarad capacitor, you get a time
constant of 1 second. And that's exactly what shows up in our
differential equation here. So in this problem, we looked at an RC
circuit with a switch. And we analyzed the circuit before and
after the switch. And then looked at, and then calculated
the transient response, and then plotted the steady state before the switch was
flipped, the transient. And then, the steady state after the
switch was flipped.