Question

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of begin quantity omega times t end quantity, where V is the maximum voltage, omega is the angular frequency, and t is the time. The current supplied by this source that flows through this resistor is described with the function i of t is equal to I times cosine of begin quantity omega times t end quantity, where I is the maximum current. What is the average power supplied by this AC source?

Answer 1

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

`p(t)=v(t)i(t)`

So the average power is:

`P=(1)/(T)\intop_(0)^(T)p(t)dt`

But:

`v(t)=v_(m)cos(\omega t) \\ \\ i(t)=i_(m)cos(\omega t)`

So:

`P=(1)/(T)\intop_(0)^(T)v_(m)cos(\omega t)i_(m)cos(\omega t)dt \\ \\ P=(v_(m)i_(m))/(T)\intop_(0)^(T)cos^(2)(\omega t)dt \\ \\ But: cos^(2)(\omega t)=(1+cos(2\omega t))/(2)`

`P=(v_(m)i_(m))/(T)\intop_(0)^(T)((1+cos(2\omega t))/(2) )dt \\\\P=(v_(m)i_(m))/(T)\intop_(0)^(T)[(1)/(2)+(cos(2\omega t))/(2)]dt \\\\P=(v_(m)i_(m))/(T)[(1)/(2)(t)\right|_0^T +(sin(2\omega t))/(4\omega) \right|_0^T] \\ \\ P=(v_(m)i_(m))/(2T)[(t)\right|_0^T +(sin(2\omega t))/(2\omega) \right|_0^T] \\ \\ P=(v_(m)i_(m))/(2)`

In terms of RMS values:

`V_(RMS)=V=(v_(m))/(√(2)) \\ \\ I_(RMS)=I=(i_(m))/(√(2)) \\ \\ Then: \\ \\ P=VI`