Α b sin 2 π ft γ then the average power 05 x a2 b2

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α ) + B sin (2 π f’t + γ ) ; then the average power = 0.5 x [ A2 + B2 ] /R d) If v(t) = A sin (2 π ft + α ) + B ; then the average power = [0.5A2 + B2] /R e) If v(t) = A sin (2 π ft + α ) + B sin (2 π f’t + γ ) + C sin (2 π f’’t + δ ) + D; then the average power = [0.5 x (A2+B2+C2) + D2] /R Power in AC Circuits
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The instantaneous power delivered to a load (watts) = v(t) x i(t) = p(t) The average power delivered to a load = 1/T x the integral from 0 to T of: p(t) dt ; where T= 1 period of the waveform p(t) Example: v(t) = A sin (2 π ft + α ) + B sin (2 π f’t + γ ) ; and a capacitive load i(t) = C d[v(t)]/dt = C x [A x 2 π f x cos (2 π ft + α ) + B x 2 π f’ x cos (2 π f’t + γ )] p(t) = v(t) x i(t ) = [A sin (2 π ft + α ) + B sin (2 π f’t + γ )] x C x [A x 2 π f x cos (2 π ft + α ) + B x 2 π f’ x cos (2 π f’t + γ )] But, using a trigonometric identity: sin (2 π ft + α ) cos (2 π f’t + γ ) = { sin [ 2 π( f-f’)t + α - γ ] + sin [ 2 π( f+f’)t + α + γ ] } /2 We can then calculate, separately, the average value of each of the six (6) non- zero parts of v(t) x i(t), and then add those results. The associated six (6) periods are: T1= 1/(2f), T2= 1/(2f’), T3= 1/(f-f’), T4=1/(f+f’). The result is that the average value of v(t) x i(t) = 0 + 0 + 0 + 0 + 0 + 0 = 0 Power in AC Circuits
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The instantaneous power delivered to a load (watts) = v(t) x i(t) = p(t) . For a resistive load , the instantaneous power is {v(t)}2/R= {i(t)}2 R The average power delivered to a load = 1/T x the integral from 0 to T of: p(t) dt ; where T= 1 period of the waveform p(t) If the voltage waveform, below, is across a resistor having value R (Ohms), then: (1/T) x the integral from 0 to T of: v2(t)/R dt = (1/T) x [1/R] x [T/2+ 9T/2] = 5/R Watts 1 0 -3 v(t) T/2 T/2 Power in AC Circuits
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