正在加载图片...
Resistive Circuits with Steady State Sources Consider again a resistive network,but this time with a sinusoidal source What is the instantaneous power consumed by R? p()=V()i1() V1(t) Vp sin(@t)[Vp sin(@t)]/RI =[V,2/R]sin2(o。t) V3(t) What is the average power consumed by R? PAvG=1/TPi(t)dt v(t)=V。sin(oot) =1/T [Vp2/R]sin2(t)dt PAVG=VP2/R1 Just as before,we can use Kirchoff's Laws: VEFF2 R1,since VEPF=Vp/V2 In general, i(t)R1-V(t)=0 i(t)=V()/R1 PAVG=VEFF IEFF vo(t)=vs(t) Says that Effective Values of a sinusoid produce average power equivalent to comparable DC values Basic Electronics-Special Lecture for TIPP 2011 5 Gary Drake,Argonne National Lab-Session 2Basic Electronics – Special Lecture for TIPP 2011 5 Gary Drake, Argonne National Lab – Session 2 Resistive Circuits with Steady State Sources  Consider again a resistive network, but this time with a sinusoidal source vs(t) = Vp sin(  t) Just as before, we can use Kirchoff’s Laws: i1(t) R1 – vs(t)= 0 i1(t) = vs(t) / R1 vO(t) = vS(t) v1(t) vs(t)   What is the instantaneous power consumed by R1? p1(t) = v1(t) i1(t) = Vp sin(o t) [Vp sin(o t)] / R1 = [Vp2 / R1] sin2(o I1 t)   R1 What is the average power consumed by R1? PAVG = 1/T f p1(t) dt = 1/T f [VP2/R1] sin2(o t) dt PAVG = ½ VP2 / R1 = VEFF2 / R1, since VEFF = VP / S2 In general, PAVG = VEFF IEFF 0 T 0 T  Says that Effective Values of a sinusoid produce average power equivalent to comparable DC values  
<<向上翻页向下翻页>>
©2008-现在 cucdc.com 高等教育资讯网 版权所有