0:03

The topic of this problem is Energy Storage Elements.

The problem is to find R, the resistance in our circuit,

if v1(t) = 1e to the -200t V for t greater than or equal to 0.

So if we look at our circuit below, it has two sources, a voltage source on the left

hand side and a current source on the right hand side of the circuit.

Our voltage v1 is the voltage that's dropped across the 50 ohm resistor at

the top of the circuit.

And the resistance that we're looking for

is the resistor that connects the top and the bottom of our circuit as shown.

So how do we find that resistance?

We can use a number of different methods to find it.

Perhaps the easiest way to do this is to use Kirchhoff's Voltage Laws and

sum the voltages around this loop.

So if we sum the voltages around this loop,

the sum has to be equal to 0 at any instant in time.

We alternative could have looked at another circuit, or another mesh, and

perhaps looked at the right side of the mesh and try to do something with it, but

what we would find is we don't know the voltage across this current source.

So if we try to use Kirchhoff's Voltage Law on this

right hand side of the circuit, we'd have to introduce an additional variable for

the voltage drop across this current source.

And ultimately that would give us too many unknowns for

the number of equations that we can generate.

So if we look at this, again, left most loop and sum the voltage around it,

then that should give us at least one equation that we could look at to see if

we can resolve R from our knowledge.

So starting in the lower left hand corner of this loop, and

again, summing the voltages around this loop, we would have,

1:56

Let's first assign the loop and

a loop current, call I1.

We would have starting in the lower left hand corner and

proceeding right into the negative polarity of the voltage source, so

it's -3e to the -200t for the voltage drop across the source.

And then we have the voltage drop V1 in addition to that.

So I'm going to add it in there as V1, and we know what that voltage is,

it's given to us in the problem statement as well.

And we have the voltage drop across 100 millihenry inductor.

And we know the voltage drop across the inductor is the inductance,

which is 100 millihenries or 0.1, and

then the derivative, Of the current.

And we know the current is

equal to V1/50 ohms.

That's our current through the inductor.

And so that is the first three of our voltage drops around our loop, and

we have one more left, which is associated with the resistance, R.

And so the resistance R has two currents flowing through it.

It has the current that we just identified, and so

the voltage drop is going to be the resistance, which is R,

times the current that's flowing through it.

And what currents are flowing through it?

We have this current, which is flowing through this resistor and

inductor at the top of the circuit that flows into here,

as well as the current from the source on the right hand side of our circuit.

3:48

So we've already said that the voltage V1 is dropped

across a 50 ohm resistor, and we can get the current

flow through the top part of our circuit by using V1/50.

And then we also have, again, this other current,

which is 10e to the -200t microamps.

4:35

We get an equation which has a series of terms which have this

exponential e to the -200t as part of the expression.

And so hopefully, when we finish this,

what we'll find is that we can sum the coefficients,

and the sum of those coefficients must be equal to 0.

And by doing that, we'll be able to resolve what R is.

So there's our first two terms.

Now if we take this derivative,

we will end up with a -0.4e to the -200t here,

plugging in for V1(t) and solving.

5:30

And so it's going to be R times,

plug in for V1 and pulling out

the exponential part of that,

[1/50 + 0.01]e to the -200t.

5:59

And this is in milliamps, so the 0.01 that we

have in our expression comes from this 10.

And so we end up with these terms, which,

after they're summed, must be equal to 0.

So we can take the coefficients, the- 3,

+1,- 0.4, + R/50 + 0.01R,

and we can set those equal to 0.

And if we do that,

6:55

We can see clearly that we can solve for R, and if we do that,

we end up with a resistance equal to 80 ohms.

So, if we want a V1(t), which is 1e to the -200t, and we're looking for

the resistance value, then we can use this Kirchhoff's Voltage Law and

sum voltages around our leftmost loop and ultimately get on the expression that

allows us to solve for the resistance, and we end up with 80 ohms.