11. AC Circuit Power Analysis


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1 . AC Circuit Power Analysis Often an integral part of circuit analysis is the determination of either power delivered or power absorbed (or both). In this chapter First, we begin by considering instantaneous power, the product of the timedomain voltage and timedomain current associated with the element or network of interest Then, we consider the average power in the sinusoidal steady state We end this chapter by introducing the concepts of reactive power, complex power, and the power factor
2 . Instantaneous Power The instantaneous power delivered to any device is i t p t =v t i t For resistive elements, we have v t p t =v t i t =i t R= R For inductive elements d i t t p t =v t i t =L i t =v t v x dx dt L In the case of a capacitor, d v t t p t =v t i t =v t C =i t i x dx dt C v t
3 Example consider a shown series RL circuit excited by a stepvoltage source The familiar current response is V i t = 0 e R t/l u t R The total power delivered by the source V 0 p t =v t i t = e R t/l u t R The power delivered to the resistor V 0 R t /L p R t =i t R= e u t R The power absorbed by the inductor d i t p L t =vl t i t =L i t d t V 0 R t/l R t/l = e e u t R vl t =L d i t =V 0 e R t/l u t dt 3
4 4 Power from Sinusoidal Sources For a sinusoidal source V m cos t The familiar current response is i t =I m cos t where Im= Vm R L = tan L R ThThe instantaneous power delivered to the entire circuit p t =V m I m cos t cos t Using trigonometric identities, we obtain V I p t = m m [cos t cos ] V I V I = m m cos m m cos t Constant Term Double Frequency Term
5 . Average Power The average power over an arbitrary interval from t to t is P= t p t dt t t t When the power is periodic with period T, the average power is calculated over any one period: t T P = t p t dt T x x Average Power in the Sinusoidal Steady State We assume the general sinusoidal voltage and current v t =V m cos t i t =I m cos t The instantaneous power is p t =V m I m cos t cos t = V m I m cos V m I m cos t 5
6 6 The instantaneous power has two terms p t = V m I m cos V m I m cos t constant term periodic term with period T/ Thus, the average power is P = V m I m cos For an ideal resistor ( = =0 ), the average power is P R= V m I m cos 0 V m = V m I m= I m R= R For a purely reactive element ( =±90 ), the average power is P X= V m I m cos 90 The average power delivered to any network =0 composed entirely of ideal inductors and capacitors is zero (no power lost)
7 Find the average power being delivered to an impedance ZL= 8 j Ω by a current I= 5ej0 A Only the 8Ω resistance enters the averagepower calculation, since the j Ω component will not absorb any average power Thus, P = (/)(5)8 = 00 W 7
8 8 Example: Find the average power absorbed by each element The values of I and I are found by any of several methods, such as mesh analysis, nodal analysis, or superposition. They are I =5 j 0= o A I =5 j 5= o A The current through the ohm resistor is I I = j 5=5 90o A So that Im=5 A, the average power absorbed by the resistor P R= I m R= 5 =5 W The average power delivered by the left source P left = 0.8 cos[ o ]=50 W
9 The average power absorbed by the right source P right = cos[0 45 o ]=5 W Maximum Power Transfer An independent voltage source in series with an impedance Zth delivers a maximum average power to that load impedance ZL which is the conjugate of Zth: ZL = Zth* Example: Find the load impedance for max. power transfer 9
10 0.3 Effective Values of Current and Voltage The effective value (root mean square RMS value) of AC sources is equal to the value of the DC sources which delivers the same average power to the resistor The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values T I eff = i t dt T 0 T V eff = v t dt T 0
11 Effective (RMS) Value of a Sinusoidal Waveform Assume the general sinusoidal voltage and current v t =V m cos t which has a period T= The RMS value is V eff = T V cos t dt m T 0 =V m Thus, i t =I m cos t T [ cos t ]dt T 0 Vm T = [ cos t ]dt 0 T V eff = Similarly, we can show that Vm Im I eff =
12 .4 Apparent Power & Power Factor for the sinusoidal voltage v t =V m cos t and current i t =im cos t The average power can be rewritten as P = V m I m cos P =V eff I eff cos The apparent power is defined as VeffIeff and is given the units voltampere VA The power factor (PF) is defined as PF = average power P = apparent power V eff I eff =cos for a purely reactive load, PF=0 generally, 0 PF for a resistive load, PF= 0 An inductive load has a lagging PF 0 A capacitive load has a leading PF
13 3 Example: Find the average power delivered to each of the two loads, the apparent power supplied by the source, and the power factor of the combined loads. The value of I is obtained as I= 60 = 53.3 o A rms 3 j4 o so I eff = A rms, and = 53.3 The average power delivered to the top load P = I m R=I eff R= =88 W The average power delivered to the lower load P =I eff R= =44 W The apparent power of the source V eff I eff = 60 =70 VA The power factor o PF =cos =0.6 lagging
14 4.5 Complex Power The average power P absorbed by the two terminal network is P =V eff I eff cos Using Euler s formula or hence P =Re{V eff I eff ej } P =Re{V eff e j I eff e j } P =Re{Veff Ieff } The complex power S is defined as eff T he Pow e r Tria ngle S=V eff I =P jq where P is the average power and Q is termed as the reactive power The unit of Q is defined as the voltamperereactive (abbreviated VAR) Q=V eff I eff sin The reactive power represents the flow of energy back and forth from the source (utility company) to the inductors and capacitors of the load (customer)
15 Splitting the current phasor Ieff into inphase and outofphase components is another way of visualizing the complex power 5
16 6 Complex powers to loads add: rms values are assumed The complex power can be calculated as S=V I Since I=I+I, we have S=V I I =S S
17 7 Example: Power factor correction An industrial consumer is operating a 50 kw induction motor at a lagging PF of 0.8. The source voltage is 30 V rms. In order to obtain lower electrical rates, the customer wishes to raise the PF to 0.95 lagging. Specify a suitable solution. The average power (the real part of the complex power) is P =V eff I eff cos =50 kw The complex power S is given by S =V eff I eff e j P jcos = e PF PF 50 j 36.8 = e =50 j 37.5 kva 0.8 o =36.8 o The power factor: PF =cos Then =cos PF The apparent power P V eff I eff = PF
18 8 Example: Power factor correction In order to achieve a PF of 0.95, the total complex power must become P jcos PF S=S S = e PF 50 j cos 0.95 = e =50 j 6.43 kva Z= j Thus, the complex power drawn by the corrective load is S =S S = j.07 kva The load impedance of the corrective load can be found as S =V I I =S /V= j 070/ 30= j 9.6 A Therefore, the load impedance is Z = V 30 = = j.5 I j 9.6 is a capacitive load
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