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(sharon) #1

Alternating Current Analysis 231


where

V


T


RMS vtdt

T


(^12)
0


()



and

I


T


itdt
T
MS pt dt

TT
R PAVG

(^12)
00


() (


∫∫
and

1


)


R.3.20 Let the current through and voltage across an arbitrary load be given by

i(t) = Im sin(t)

and

v(t) = Vm sin(t + ) (an RL equivalent circuit since v(t) leads i(t) by )

Then the instantaneous power is given by

p(t) = i(t) ⋅ v(t) = Im Vm sin(t) ⋅ sin(t + )

Using trigonometric identities

pt

VI VI


t

VI


( )mmcos( ) mmcos( )cos( )mmsin( )sin( t)
22

2


2


  2


where

VI V I
mm m m VIRMS RMS
2 2




2


.


Let VRMS IRMS = A
then, p(t) = A cos(θ) − A cos(θ) cos(2ωt) + A sin(θ) sin(2ωt).
R.3.21 Let us explore the resistive case, where θ = 0°. Then p(t) of R.3.20 becomes

p(t) = A − A cos(2t)

And the average power, often referred as the real power, is given by

PA


VI


AVGmmVIRMS RMS
2

(in watts)

PAVG = real(VRMS IRMS*) (the character * denotes the complex conjugate of IRMS)

The energy dissipated by the resistor R, in the form of heat over one full cycle, is
given by

WR = VRMS IRMS T (in joules)
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