The topic of this problem is AC steady state power. The problem is to determine the values for the resistance R and the inductance L in the circuits shown below, such circuits has a voltage source is 12cos at 4t volts And has the resistor inductor and in parallel with one another. We also know that the power delivered by the voltage source, the complex power, is 18 plus j9 volt amps. So we're going to use this information to find the appropriate values for R and L based on the power that's delivered by the source. We use our well-known equation for complex power, complex power is equal to the voltage times the Conjugate of the current divided by two. So if we substitute first of all to find I of T. If we substitute end values for the complex power and the voltage then we can find the value for the conjugate of the current in the circuit. And so doing that, we have two times the complex power 18 plus J9 divided by the voltage which is 12 with a phase angle of zero degrees. To solve this we end up with the complex current which is equal to 3 plus j 1.5. And magnitude angle format It's 3.35 in an angle of 26.6 degrees for our complex conjugate of the current. So we know that If we know the current was flows in to the RL combination, and we know the voltage drop across the RL parallel combination, then we can find our values for R and L. And we do that by first of all recognizing that these are impervious values and there in parallel with one another, so. We know that the total impedance for those two elements is going to be 1 over R. This is 1 over the complex impedance plus, 1 over j omega L. And so if we plug in values for that, We plug in our omega value, which is four and we set that equal to V over I, but it's one over Z, so it's I divided by V. Then we can find our values for R and L. So going ahead and doing that, first of all putting in our values for I and V, we have 3.35 the angle of 26.6 degree for our current and our voltage value is 12 at an angle of 0 degrees. Our current I is a complex conjugate of the current value we found here. So I need to put a negative sign in front of the 26.6 degrees if we're looking at the current and not the complex conjugate of the current. And if we continue with that and break it down into rectangular coordinates that we have. 0.25 minus J 0.125. And again, that's equal to one over R plus one over J4L. And so if we equate the real parts of the two expressions and also the imaginary parts of the two expressions we can find our value for R and L. Our R, which is 1 over 0.25 comes out to be 4 and our inductance comes out to be 2 Henries.
