Pre-Calculus?

Question

Does anyone know how to do this problem and can explain to me? thanks.

(Note: j = imaginary number)
the active power of an ac circuit whose current I=a+bj and voltage V=c+dj is given by P= ac+ bd. Find P when I=5(cos 45 degrees + j sin 45 degrees) and V=110 (cos 30 + j sin30) by first changing the numbers to rectangular form.

B. find the impedance Z in polar form of the circuit. Note: Z=V / I

Answer 1

To get you started, try to multiply out all of the terms to get your polar coordinates into some decimal form: V= 110*cos 30 + (110 sin 30) j then use your calculator to convert polar to rectangular and divide the answer.

When you put an impedance load on an AC circuit the current and voltage aren't always in phase. This equation describes what the two waveforms look like. In this case it looks like you have a domestic US voltage, like at a wall outlet, of 110 Vac, and your load - say a lightbulb - is causing some amount of current to flow through it. By looking at what the lightbulb does to the voltage and current of the AC signal you can tell the impedance of the load.

Answer 2

Change numbers to rectangular form: I = 5 cos(45)+j5sin(45) = 3.54 + 3.54 j, V = 110 cos(30) + j 110 sin(30) = 95.26 + j 55. Follows: P = ac + bd = 3.54*95.26 + 3.54*55 = 532 w.
Since cos(x)+jsin(x) = e^(jx), the polar forms will be I = 5e^(j45) and V = 110e^(j30). Then, Z = V/I = (110 /5)e^(-j15) = 22(cos(15) - j sin(15)) = 21.25 - j 5.69.