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?

Respuesta :

Answer:

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

[tex]p(t)=v(t)i(t)[/tex]

So the average power is:

[tex]P=\frac{1}{T}\intop_{0}^{T}p(t)dt[/tex]

But:

[tex]v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)[/tex]

So:

[tex]P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}[/tex]

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

In terms of RMS values:

[tex]V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI[/tex]

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