0:04

The topic of this problem is Ohm's Law and

Â also using the well-known equation for Power, as well.

Â The problem is determine the resistance R and the voltage source V sub s.

Â So we have a current source,

Â on the left-hand side of the circuit that also has a voltage associated with it.

Â And that's the V sub S that we're looking for.

Â 0:24

To solve this problem, you have to use the passive sign convention in order

Â to be able to assign the polarities across a voltage drop of the elements.

Â So in this case, we have a voltage drop across the resistor that is

Â unknown as far as the polarity goes.

Â So we first assign a direction for the current.

Â And then, knowing the direction for the current that we've assigned,

Â we can then assign the polarity across the resistor.

Â In the passive sign convention, the current is assumed to enter the positive

Â side of the voltage polarity across the element.

Â So if we assign the current to go clockwise around this single loop circuit,

Â then the voltage drop across the resistor,

Â P sub R is going to be, as shown in the circuit now, and

Â ultimately we're going to decide what the resistance is and

Â what the voltage source of this V sub S is?

Â So we know that using the equation for

Â power, that power = I squared R.

Â And we're given the power in this problem.

Â The power in the resistor is equal to 80 milliwatts.

Â We're also given the current, the current is equal to 4 milliamps.

Â And the problem is on the left-hand side as a current source.

Â So we have everything we need in order to solve for the resistance.

Â So the resistance is going to be equal to P/I squared,

Â and P=80mW, it's positive,

Â I=4mA I'm going to square that and ultimately we end up with a resistance,

Â 2:25

So that solves for our resistance.

Â The next thing we need to find,

Â is we need to find the voltage associated with the current source.

Â So it's the source voltage across the current source.

Â 2:37

We can do this because we know that all that voltage

Â that's across the voltage source is also the voltage across the resistor.

Â It's the only other element in the circuit.

Â So if we could solve for the voltage across the resistor, then we know that

Â that is going to be equal to the voltage across the current source.

Â So the voltage across the resistor,

Â which is equal to V sub S, is going to be equal

Â to the current times the resistance,

Â and so the current is 4 mA.

Â We calculated the resistance from a previous part of the problem, and so

Â we end up with a voltage which is equal to 20 volts.

Â