0:03

The topic of this problem is Energy Storage Elements.

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

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

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

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

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

Â the top of the circuit.

Â And the resistance that we're looking for

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

Â So how do we find that resistance?

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

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

Â sum the voltages around this loop.

Â So if we sum the voltages around this loop,

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

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

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

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

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

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

Â the voltage drop across this current source.

Â And ultimately that would give us too many unknowns for

Â the number of equations that we can generate.

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

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

Â we can resolve R from our knowledge.

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

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

Â 1:56

Let's first assign the loop and

Â a loop current, call I1.

Â We would have starting in the lower left hand corner and

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

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

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

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

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

Â And we have the voltage drop across 100 millihenry inductor.

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

Â which is 100 millihenries or 0.1, and

Â then the derivative, Of the current.

Â And we know the current is

Â equal to V1/50 ohms.

Â That's our current through the inductor.

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

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

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

Â It has the current that we just identified, and so

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

Â times the current that's flowing through it.

Â And what currents are flowing through it?

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

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

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

Â 3:48

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

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

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

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

Â which is 10e to the -200t microamps.

Â 4:35

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

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

Â And so hopefully, when we finish this,

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

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

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

Â So there's our first two terms.

Â Now if we take this derivative,

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

Â plugging in for V1(t) and solving.

Â 5:30

And so it's going to be R times,

Â plug in for V1 and pulling out

Â the exponential part of that,

Â [1/50 + 0.01]e to the -200t.

Â 5:59

And this is in milliamps, so the 0.01 that we

Â have in our expression comes from this 10.

Â And so we end up with these terms, which,

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

Â So we can take the coefficients, the- 3,

Â +1,- 0.4, + R/50 + 0.01R,

Â and we can set those equal to 0.

Â And if we do that,

Â 6:55

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

Â we end up with a resistance equal to 80 ohms.

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

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

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

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

Â