Power **new** 2003/06/13 10:29:17.887 GMT-5 2003/06/13 13:33:01.272 GMT-5 Bill Wilson wlw@rice.edu Bill Wilson wlw@rice.edu Elizabeth Gregory lizzardg@rice.edu Jeffrey Silverman jsilv@rice.edu How to calculate the power that flows into and out of a line, as well as average power. You might be tempted to now say that P in V in I in , but that is incorrect for sinusoidal excitation. V in and I in are phasors! So let's digress for a second to see (or review, I hope) how to find power when the voltage and current are phasor quantities. What really matters is not the absolute phase angle of the two quantities, but rather the phase angle between them. Suppose we have a voltage phasor, V which has zero phase angle and a complex impedance Z Z θ z . Obviously, the current is given by I ˜ V ˜ Z ˜ V Z θ z To find power, we can not work just with phasors, we have to go back to the complete function of time as well so we write: V t V ω t I t V Z ω t θ z I t I ω t θ z The power as a function of time is given as P t I t V t V V ω t ω t θ z We remember a useful trig identity: A B A B A B Hence: ω t θ z ω t θ z ω t θ z which makes P t P t ω t 2 θ z ω t ω t θ z We are really interested in finding average power since energy which flows into and then back out of the line does no work for us. Clearly the second term in (going as ω t ω t ) has an average value of zero, and so we can forget about it. Time for one more trig identity: A 2 1 2 1 2 2 A 2 ω t has zero average value as well, so we are left with the following for the average value of the power P t P t V I 2 θ z V 2 2 Z θ z Note that one useful way that people sometimes use to express this is to say P t 1 2 V ˜ V ˜ * Back to our example: V in 4.1838 and I in 0.14421 Thus P in t 1 2 4.18 0.144 59 Watts 0.155 As an alternative way of calculating the power into the line note that we know the magnitude of the current through both the capacitor and the resistor of the apparent Z in . They are just two elements in series, and so they both have the same current flowing through them, namely, I in . No power is dissipated in the capacitor, so we could just as well have said P in t 1 2 I 2 R 1 2 0.144 2 15 0.155 and gotten the answer in an even easier fashion! (Note that we still have to keep the factor of "1/2" to account for the time average of a sinusoidal product.) For reasons I do not understand, students have always had an aversion to finding power. It is not that hard, and in the end, is usually the "bottom line" with regard to how a system will perform. Go back over this section until it makes sense, as you may see power crop up someplace else one of these days!