# AC Power

As in the case with DC power, the instantaneous electric power in an AC circuit is given by P = VI, but these quantities are continuously varying. Almost always the desired power in an AC circuit is the average power, which is given by

Pavg = VI cosφ

where φ is the phase angle between the current and the voltage and where V and I are understood to be the effective or rms values of the voltage and current. The term cos φ is called the "power factor" for the circuit. RLC series circuit details
 For C = μF
 and L = mH
 and resistance R = ohms
 at angular frequency ω = x10^ rad/s,
 frequency f = x10^ Hz = kHz = MHz

the impedance is

 Z = x10^ ohms at phase φ = degrees.

For an applied rms voltage V = volts,

the rms current will be I = x 10^ amperes.

 and the AC power is given by Pavg = VI cosφ = watts

 The power factor is cos φ =
so the power is reduced to that fraction of what it would be in a DC circuit with the same voltage and current.

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# Instantaneous Power

As in DC circuits, the instantaneous electric power in an AC circuit is given by P=VI where V and I are the instantaneous voltage and current.
 Since then the instantaneous power at any time t can be expressed as  and using the trig identity the power becomes: Averaging this power over a complete cycle gives the average power.

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# Average Power

Normally the average power is the power of interest in AC circuits. Since the expression for the instantaneous power is a continuously varying one with time, the average must be obtained by integration. Averaging over one period T of the sinusoidal function will give the average power. The second term in the power expression above averages to zero since it is an odd function of t. The average of the first term is given by Show
 Since the rms voltage and current are given by and ,
the average power can be expressed as
Pavg = VI cosφ
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# Average Power Integral

Finding the value of the average power for sinusoidal voltages involves the integral The period T of the sinusoid is related to the angular frequency ω and angle θ by Using these relationships, the integral above can be recast in the form: Which can be shown using the trig identity: which reduces the integral to the value 1/2 since the second term on the right has an integral of zero over the full period.
 More detail on integrating trig functions
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