# X ir1t 80 sin2377t 80 x 1 cos 754t 2 80w 40w 0w the

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x iR1(t) = 80 sin2(377t) = 80 x [1- cos (754t)] /2 80W 40W 0W The average power is 40 watts : corresponding to one (1) 40W light bulb

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Power in AC Circuits R1=360 L1 =1H ~ j377 Vutil(t) = 120 x 1.414 sin (2 π 60t) Volts = Vutil(t) = 170 sin (377t) Volts iR1(t) = Vutil(t) /360 = 0.47 sin (377t) Amperes iL1(t)= - [170 Volts/( 377 radians/s x 1H )] cos (377t) = -0.45 cos (377t) Amperes iTotal(t) = 0.65 sin (377t - 0.76) Amperes; where 0.76 radians = 44o iR1 iL1 +
Power in AC Circuits R1=360 L1 =1H (actual~ j650 + 50 Ω29 Vutil(t) = 120 x 1.414 sin (2 π 60t) Volts = Vutil(t) = 170 sin (377t) Volts iR1(t) = Vutil(t) /360 = 0.47 sin (377t) Amperes (measured 0.43 A peak) iL1(t)= - [170 Volts/( 377 radians/s x 1H )] cos (377t) = - 0.45 cos (377t) Amperes (measured 0.26A peak) iTotal(t) = 0.65 sin (377t - 0.76) Amperes; where 0.76 radians = 44o (Measured: 0.51A peak) Note: [(0.43A)2 + (0.26A)2]0.5 = 0.50A iR1 iL1 + Measured: 394

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Voltage and Current Waveforms Vutil(t) = 170 sin (377t) Volts iTotal(t) = 0.65 sin (377t - 0.76); where 0.76 radians = 44o
Power in AC Circuits Vutil(t) = 170 sin (377t) Volts iR1(t) = Vutil(t) / 360 = 0.47 sin(377t) Amperes iL1(t)= - [170/( 377 x 1 )] cos (377t) Amperes = - 0.45 cos (377t) Amperes Instantaneous power (watts) = Vutil(t) x {iR1(t) + iL1(t)} = 80 sin2(377t) - 77 sin(377t) x cos(377t)= 40 - 40cos (754t) - 38 sin(754t) = 40 - β sin(754t + φ ) , where β = [(40)2 + (38)2] 0.5 = 55 95W - 15W 40W 0 W The average power is 40 watts : corresponding to one (1) 40W light bulb

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Power in AC Circuits R1=360 L1=1H Vutil(t) = 120 x 1.414 sin (2 π 60t) Volts = Vutil(t) = 170 sin (377t) Volts iR1(t) = 0.47 sin (377t) Amperes iL1(t) = -0.45 cos (377t) Amperes iC1(t) = 0.45 cos (377t) Amperes iR1 iL1 iC1 C1 If we pick C1 such that (377/s) x L1 = 1/[(377/s) x C1], then the power generating facility will not have to provide the excess peak power associated with the inductor We need: C1= 1/[L1 x (377/s)2]= .000007F = 7 μ F [Based on measured results, we need 4uF] Note: C1 must be a capacitor that is designed to have a 170V (peak) AC voltage across it, with no DC offset. Use two (2) electrolytic capacitors “back-to-back” +
Calculating the average of p(t) 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) T= 1 period Av power = (1/4s) [ the area under p(t) in the interval t=0 to t=4s] = (1/4s) [(6W x 2s)/2 + (6W x 1s) + (3W x 1s)] = (1/4s) [6J + 6J + 3J] = 3.75W p(t) t 6W 3W 4s 3.75 W

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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) Example: v(t) = A sin (2 π ft + α ) + B sin (2 π f’t + γ ) ; and a resistive load p(t) = v2(t)/R = [ A sin (2 π ft + α ) + B sin (2 π f’t + γ ) ]2/R= { [A sin (2 π ft + α )]2 + [B sin (2 π f’t + γ )]2 + 2AB sin (2 π ft + α ) sin (2 π f’t + γ ) } /R But, using a trigonometric identity: sin (2 π ft + α ) sin (2 π f’t + γ ) = { cos [ 2 π( f-f’)t + α - γ ] - cos [ 2 π( f+f’)t + α + γ ] }/2 We can then calculate, separately, the average value of each of the six (6) parts of v2(t)/R, and then add those results. The average value of a constant = that constant. For the sinusoidal terms, the associated four (4) periods are: T1= 1/(2f), T2= 1/(2f’), T3= 1/(f-f’), and T4=1/(f+f’). The result is that the average value of : v2(t)/R = [0.5 x A2/R + 0.5 x B2/R + 0 + 0 + 0 + 0 Power in AC Circuits
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) Special cases for resistive loads: a) If v(t) = V sin (2 π ft + φ ); then the average power = 0.5 x V2/R b) If v(t) = A sin (2 π ft + α ) + B sin (2 π ft + γ ) = V sin (2 π ft + φ ); then the average power = 0.5 x V2/R c) If v(t) = A sin (2 π ft +

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