【用戶】許同學
【年級】高一上
【評論內容】此電路可看做一不平衡的電橋,將上半部分40Ω、10Ω、50Ω電阻以 Δ-Y 公式變換﹕假設R1 = 50Ω‚ R2 = 10Ω, R3 = 40Ω Ra = (R3 x R2) / (R1+R2+R3) = (40x10)/(50+10+40) = 4Ω Rb = (R3 x R1) / (R1+R2+R3) = (40x50)/(50+10+40) = 20Ω Rc = (R2 x R1) / (R1+R2+R3) = (10x50)/(50+10+40) = 5Ω總電阻RT = 4 + (20+10) // (5+10) = 14Ω電流I = 28 / 14 = 2A
【用戶】lexmvrk
【年級】小六下
【評論內容】以 Δ-Y 變換,Ra = (R3 x R2) / (R1+R2+R3) = (40x10)/(50+10+40) = 4Ω Rb = 20Ω Rc = 5Ω這裡看不懂