19. 如右圖所示,有一發電機(內阻為300 Ω)供給負載ZL(阻抗為30 + j40 Ω),現以一 XL 及 XC 匹配電路使負載 ZL 獲得最大功率,請問此匹配電路 XL 及 XC分別為多少歐姆(Ω)?
【用戶】LAI
【年級】小二上
【評論內容】左右切開,左邊取戴維寧,求ZthZth = 300//-jXc +jXL當Zth = ZL*= 30-j40 有最大功率轉移,300//-jXc + jXL= 30 - j40(1/300+1/-jXc)-1 = 30 -j(40+XL) = 30 - jXA (令代數XA = 40+XL 方便運算)1/300+1/-jXc = 1/(30-jXA)(30-jXA)*(-jXc) + 300*(30-jXA) =300*(-jXc)-j30Xc - XA*Xc + 9000 - j300XA = -j300Xc實部:-XA*Xc + 9000 = 0XA * Xc = 9000虛部:-j30Xc - j300XA = -j300Xc270*Xc =300*XAXc = 10/9 * XA 代入 XA * Xc = 900010/9 * XA2 = 9000XA2 = 8100XA = 90XL = XA - 40 = 90 - 40 = 50XA = 9/10 * Xc 代入 XA * Xc = 90009/10 * Xc2 = 9000Xc2 = 10000Xc = 100
【用戶】LAI
【年級】小二上
【評論內容】左右切開,左邊取戴維寧,求ZthZth = 300//-jXc +jXL當Zth = ZL*= 30-j40 有最大功率轉移,300//-jXc + jXL= 30 - j40(1/300+1/-jXc)-1 = 30 -j(40+XL) = 30 - jXA (令代數XA = 40+XL 方便運算)1/300+1/-jXc = 1/(30-jXA)(30-jXA)*(-jXc) + 300*(30-jXA) =300*(-jXc)-j30Xc - XA*Xc + 9000 - j300XA = -j300Xc實部:-XA*Xc + 9000 = 0XA * Xc = 9000虛部:-j30Xc - j300XA = -j300Xc270*Xc =300*XAXc = 10/9 * XA 代入 XA * Xc = 900010/9 * XA2 = 9000XA2 = 8100XA = 90XL = XA - 40 = 90 - 40 = 50XA = 9/10 * Xc 代入 XA * Xc = 90009/10 * Xc2 = 9000Xc2 = 10000Xc = 100