[理工] [電磁] 互感,磁能

看板Grad-ProbAsk作者CGMS (~海~)時間13年前 (2010/09/18 13:57)推噓5(5推 0噓 6→)

留言11則, 6人參與討論串1/1

有關台大考古題：94年台大電磁學(B) 第4題： Consider two coupled circuit, having self-inductances L1 and L2, that carry currents I1 and I2, respectively. The mutual inductance between the circuits is M. Find the ratio I1/I2 that makes the stored magnetic energy W2 a minimum. 意思大概是有兩個線圈,自感分別為L1及L2,且分別載有電流I1、I2, 兩線圈的互感是M,求I1和I2的比值(I1/I2)能使磁能W2為最小值以下是孫超群的解法：I2固定 W2(I1,I2) = 1/2(L1*I1^2) + 1/2(L2*I2^2) + M*I1*I2 將I2^2提出 = I2^2 [ 1/2 L1(I1/I2)^2 + 1/2(L2) + M(I1/I2) ] 令x = I1/I2 W2(x) = I2^2 [ 1/2 L1(x^2) + Mx + 1/2 L2] 然後將W(x)對x微分 dW2(x)/dx = 0 = I2^2 (L1*x + M) 可求得x = -M/L1 答案即是 I1/I2 = -M/L1 想問為什麼知道W2微分後等於零,有最小值？或是有其它的解法，希望有人可以回答，非常感謝！ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 220.142.176.70

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youmehim

09/18 14:56, , 1^F

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youmehim

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ntust661

09/18 15:03, , 3^F

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