5-1-2PerformanceParameters（性能参数）评价电能(电压、电流)质量的几个指标As a base case, wefirst consider linearload.Example1:RwireSuppose weplugaresistorload+Fuseinto the wall socket.vR/2V, sin otP= Vms ms = I ms RThe fuse is rated for a specified rms current, above that, itwill blow so that dissipation in Rwire does not start a fireWitha 220V,15Afuse,weget 3.3kWmax.fromwallsocket (neglecting Rwire)
Example 1: Suppose we plug a resistor load into the wall socket. The fuse is rated for a specified rms current, above that, it will blow so that dissipation in Rwire does not start a fire. With a 220V, 15A fuse, we get 3.3kW max. from wall socket (neglecting Rwire ). = I 2 P = Vrms Irms rms R Rwire Fuse R + V - 2Vs sint 5-1-2 Performance Parameters (性能参数) 评价电能(电压、电流)质量的几个指标 As a base case, we first consider linear load
Example 2: Suppose instead we plug a inductor load into the wall socketRwireneglecting Rwire+Fusei(o1) =- 2y, co(α)vsin OtOLPv(ot)i(ot)d(ot) = 0While we draw no real power, we still draw current.V2i?(ot)d(wt) = OL2元220V, 50Hz, L=40mH-→lrms=17.5A>15A. So, the fuse will stillblow (to protect the wall wire). In this case, we are not utilizingE.hesaurtoeweallm
neglecting Rwire P = 0 v(t)i(t)d(t) = 0 s L 2V cos(t) i(t) = − 2 While we draw no real power, we still draw current. rms I = 0 2 V i 2 (t)d(wt) = s 2 L 1 220V, 50Hz, L=40mH→Irms=17.5A >15A. So, the fuse will still blow (to protect the wall wire). In this case, we are not utilizing E t -m ha eil: sxjohuua r n cg@ebjwtu.eedlu . .cn Example 2: Suppose instead we plug a inductor load into the wall socket. Rwire Fuse L + V - s 2V sint
1 Defining certain indices (指标)(1) Defining the Power Factor(功率因数 ： a measureof the utilization of the sourcePP---realaveragepower（平均功率）PFV,I,Vsls---apparentpower（视在功率For the resistor PF=1,best utilization.For the inductor, PF=O, worst utilization
Vs Is P PF = (1) Defining the Power Factor (功率因数) ： a measure of the utilization of the source. P-real average power (平均功率) Vs Is -apparent power (视在功率) For the resistor PF=1, best utilization. For the inductor, PF=0, worst utilization. 1 Defining certain indices (指标)
LinearLoads(线性负载)：The sinusoidal current drawnby linear load has zero distortion.LinearLoadP=V,I, cosdVPPFcosdV,I,(b)(a)P1PFV..PF>l, and PF are inversely proportional, unity power factorLoads（单位功率因数负载thatdrawpowerattheminimumrms current.☆
V PF PF I = P PF = s s Vs Is P 1 = cos P =Vs Is cos ➢Is and PF are inversely proportional, unity power factor loads (单位功率因数负载) that draw power at the minimum rms current. ☆ Linear Loads (线性负载): The sinusoidal current drawn by linear load has zero distortion
idistortion(-i,-is))Rg/o(b)Ti(a)n(t)=i(t)-isi(t)distortionNonlinear Loads（非线性负载）[i2(t) -dt = Is1+I disortionrms:1 TP=V,Isi cos@P(t).dtSPFCOSV,Is1= Ecosd(t)·dtistortior
idistortion (t) = is (t) −is1 (t) = 1 1 1 2 2 2 1 T1 distortion 1 T1 s1 distortion 1 T1 i 2 distortion (t)dt T I i 2 s1(t)dt T I = I s1 +I T is rms：Is = (t) dt = Nonlinear Loads （非线性负载） 1 cos1 = cos s s s Is1 V I I P PF = = P =Vs Is1 cos