# 电感

（重定向自電感

${\displaystyle {\mathcal {E}}=-L{\mathrm {d} i \over \mathrm {d} t}}$

## 概述

### 自感

${\displaystyle {\mathcal {E}}=-N{{\mathrm {d} \Phi } \over \mathrm {d} t}=-N{{\mathrm {d} \Phi } \over \mathrm {d} i}\ {\mathrm {d} i \over \mathrm {d} t}}$

${\displaystyle L=N{\frac {\mathrm {d} \Phi }{\mathrm {d} i}}}$

${\displaystyle {\mathcal {E}}=-L{\mathrm {d} i \over \mathrm {d} t}}$

${\displaystyle v=L{{\mathrm {d} i} \over \mathrm {d} t}}$

### 互感

${\displaystyle \Phi _{2}=M_{21}i_{1}}$

• ${\displaystyle M_{21}={\frac {\mu _{0}}{4\pi }}\oint _{\mathbb {C} _{1}}\oint _{\mathbb {C} _{2}}{\frac {\mathrm {d} {\boldsymbol {\ell }}_{1}\cdot \mathrm {d} {\boldsymbol {\ell }}_{2}}{|\mathbf {X} _{2}-\mathbf {X} _{1}|}}}$

### 推導

${\displaystyle \Phi _{2}(t)=\int _{\mathbb {S} _{2}}\mathbf {B} _{1}(\mathbf {X} _{2},t)\cdot \mathrm {d} \mathbf {a} _{2}}$

${\displaystyle \mathbf {B} _{1}(\mathbf {X} _{2},t)=\nabla _{2}\times \mathbf {A} _{1}(\mathbf {X} _{2},t)}$

${\displaystyle \Phi _{2}(t)=\int _{\mathbb {S} _{2}}[\nabla _{2}\times \mathbf {A} _{1}(\mathbf {X} _{2},t)]\cdot \mathrm {d} \mathbf {a} _{2}=\oint _{\mathbb {C} _{2}}\mathbf {A} _{1}(\mathbf {X} _{2},t)\cdot \mathrm {d} {\boldsymbol {\ell }}_{2}}$

${\displaystyle \mathbf {A} _{1}(\mathbf {X} _{2},t)\ {\stackrel {def}{=}}\ {\frac {\mu _{0}i_{1}}{4\pi }}\oint _{\mathbb {C} _{1}}{\frac {\mathrm {d} {\boldsymbol {\ell }}_{1}}{|\mathbf {X} _{2}-\mathbf {X} _{1}|}}}$

${\displaystyle \Phi _{2}(t)={\frac {\mu _{0}i_{1}}{4\pi }}\oint _{\mathbb {C} _{1}}\oint _{\mathbb {C} _{2}}{\frac {\mathrm {d} {\boldsymbol {\ell }}_{1}\cdot \mathrm {d} {\boldsymbol {\ell }}_{2}}{|\mathbf {X} _{2}-\mathbf {X} _{1}|}}}$

${\displaystyle M_{21}={\frac {\mathrm {d} \Phi _{2}}{\mathrm {d} i_{1}}}={\frac {\mu _{0}}{4\pi }}\oint _{\mathbb {C} _{1}}\oint _{\mathbb {C} _{2}}{\frac {\mathrm {d} {\boldsymbol {\ell }}_{1}\cdot \mathrm {d} {\boldsymbol {\ell }}_{2}}{|\mathbf {X} _{2}-\mathbf {X} _{1}|}}}$

${\displaystyle \Phi _{1}(t)={\frac {\mu _{0}i_{1}}{4\pi }}\oint _{\mathbb {C} _{1}}\oint _{\mathbb {C} '_{1}}{\frac {\mathrm {d} {\boldsymbol {\ell }}_{1}\cdot \mathrm {d} {\boldsymbol {\ell }}'_{1}}{|\mathbf {X} _{1}-\mathbf {X} '_{1}|}}}$

${\displaystyle L={\frac {\mathrm {d} \Phi }{\mathrm {d} i}}={\frac {\mu _{0}}{4\pi }}\oint _{\mathbb {C} }\oint _{\mathbb {C} '}{\frac {\mathrm {d} {\boldsymbol {\ell }}\cdot \mathrm {d} {\boldsymbol {\ell }}'}{|\mathbf {X} -\mathbf {X} '|}}}$

${\displaystyle \mathbf {X} _{1}=\mathbf {X} '_{1}}$時，這積分可能會發散，需要特別加以處理。另外，若假設閉合迴路為無窮細小，則在閉合迴路附近，磁場會變得無窮大，磁通量也會變得無窮大，所以，必須給予閉合迴路有限尺寸，設定其截面半徑${\displaystyle r_{0}}$超小於徑長${\displaystyle \ell _{0}}$

### 電感與磁場能量

${\displaystyle N_{k}\Phi _{k}=\sum _{n=1}^{K}L_{k,n}i_{n}}$

${\displaystyle v_{k}=-{\mathcal {E}}_{k}=N_{k}{\frac {\mathrm {d} \Phi _{k}}{\mathrm {d} t}}=\sum _{n=1}^{K}L_{k,n}{\frac {\mathrm {d} i_{n}}{\mathrm {d} t}}=L_{k}{\frac {\mathrm {d} i_{k}}{\mathrm {d} t}}+\sum _{n=1,\ n\neq k}^{K}M_{k,n}{\frac {\mathrm {d} i_{n}}{\mathrm {d} t}}}$

${\displaystyle k}$條閉合迴路的電功率${\displaystyle p_{k}}$

${\displaystyle p_{k}=i_{k}v_{k}}$

${\displaystyle W_{1}=\int i_{1}v_{1}\mathrm {d} t=\int _{0}^{I_{1}}i_{1}L_{1}\mathrm {d} i_{1}={\frac {1}{2}}L_{1}I_{1}^{2}}$

${\displaystyle W_{2}=\int i_{2}v_{2}\mathrm {d} t=\int _{0}^{I_{2}}i_{2}L_{2}\mathrm {d} i_{2}+\int _{0}^{I_{2}}I_{1}M_{1,2}\mathrm {d} i_{2}={\frac {1}{2}}L_{2}I_{2}^{2}+M_{1,2}I_{1}I_{2}}$

${\displaystyle W_{k}=\int i_{k}v_{k}\mathrm {d} t=\int _{0}^{I_{k}}i_{k}L_{k}\mathrm {d} i_{k}+\sum _{n=1}^{k-1}\int _{0}^{I_{k}}I_{n}M_{n,k}\mathrm {d} i_{k}={\frac {1}{2}}L_{k}I_{k}^{2}+\sum _{n=1}^{k-1}M_{n,k}I_{n}I_{k}}$

${\displaystyle W={\frac {1}{2}}\sum _{k=1}^{K}L_{k}I_{k}^{2}+\sum _{k=1}^{K}\sum _{n=1}^{k-1}M_{n,k}I_{n}I_{k}={\frac {1}{2}}\sum _{k=1}^{K}L_{k}I_{k}^{2}+{\frac {1}{2}}\sum _{k=1}^{K}\sum _{n=1,n\neq k}^{K}M_{n,k}I_{n}I_{k}}$

## 串聯與並聯電路

### 串聯電路

${\displaystyle v_{k}=L_{k}{\frac {\mathrm {d} i_{k}}{\mathrm {d} t}}}$

${\displaystyle i=i_{1}=i_{2}=\cdots =i_{n}}$

${\displaystyle v=v_{1}+v_{2}+\cdots +v_{n}=L_{1}{\frac {\mathrm {d} i_{1}}{\mathrm {d} t}}+L_{2}{\frac {\mathrm {d} i_{2}}{\mathrm {d} t}}+\cdots +L_{n}{\frac {\mathrm {d} i_{n}}{\mathrm {d} t}}=(L_{1}+L_{2}+\cdots +L_{n}){\frac {\mathrm {d} i}{\mathrm {d} t}}}$

${\displaystyle L_{eq}=L_{1}+L_{2}+\cdots +L_{n}}$

• 假設兩個電感器分別產生的磁場或磁通量，其方向相同，則稱為「串聯互助」，以方程式表示，
${\displaystyle L_{eq}=L_{1}+L_{2}+2M}$
• 假設兩個電感器分別產生的磁場或磁通量，其方向相反，則稱為「串聯互消」，以方程式表示，
${\displaystyle L_{eq}=L_{1}+L_{2}-2M}$

${\displaystyle L_{eq}=(M_{11}+M_{22}+M_{33})+(M_{12}+M_{13}+M_{23})+(M_{21}+M_{31}+M_{32})}$

${\displaystyle L_{eq}=(M_{11}+M_{22}+M_{33})+2(M_{12}+M_{13}+M_{23})}$

#### 推導

${\displaystyle -v+L_{1}{\frac {\mathrm {d} i}{\mathrm {d} t}}+M{\frac {\mathrm {d} i}{\mathrm {d} t}}+L_{2}{\frac {\mathrm {d} i}{\mathrm {d} t}}+M{\frac {\mathrm {d} i}{\mathrm {d} t}}=0}$

${\displaystyle v=(L_{1}+L_{2}+2M){\frac {\mathrm {d} i}{\mathrm {d} t}}}$

${\displaystyle L_{eq}=L_{1}+L_{2}+2M}$

### 並聯電路

${\displaystyle {\frac {1}{L_{eq}}}={\frac {1}{L_{1}}}+{\frac {1}{L_{2}}}+\cdots +{\frac {1}{L_{n}}}}$

• 假設兩個電感器分別產生的磁場或磁通量，其方向相同，則稱為「並聯互助」，以方程式表示，
${\displaystyle L_{eq}={\frac {L_{1}L_{2}-M^{2}}{L_{1}+L_{2}-2M}}}$
• 假設兩個電感器分別產生的磁場或磁通量，其方向相反，則稱為「並聯互消」，以方程式表示，
${\displaystyle L_{eq}={\frac {L_{1}L_{2}-M^{2}}{L_{1}+L_{2}+2M}}}$

#### 推導

${\displaystyle -v+L_{1}{\frac {\mathrm {d} i_{1}}{\mathrm {d} t}}+M{\frac {\mathrm {d} i_{2}}{\mathrm {d} t}}=0}$
${\displaystyle -v+L_{2}{\frac {\mathrm {d} i_{2}}{\mathrm {d} t}}+M{\frac {\mathrm {d} i_{1}}{\mathrm {d} t}}=0}$

${\displaystyle {\frac {\mathrm {d} i_{2}}{\mathrm {d} t}}={\frac {L_{1}-M}{L_{2}-M}}\ {\frac {\mathrm {d} i_{1}}{\mathrm {d} t}}}$

${\displaystyle i=i_{1}+i_{2}}$

${\displaystyle {\frac {\mathrm {d} i_{1}}{\mathrm {d} t}}={\frac {L_{2}-M}{L_{1}+L_{2}-2M}}\ {\frac {\mathrm {d} i}{\mathrm {d} t}}}$

${\displaystyle v={\frac {L_{1}L_{2}-M^{2}}{L_{1}+L_{2}-2M}}\ {\frac {\mathrm {d} i}{\mathrm {d} t}}}$

${\displaystyle L_{eq}={\frac {L_{1}L_{2}-M^{2}}{L_{1}+L_{2}-2M}}}$

## 鏡像法

• 一條筆直的載流導線與導體牆之間的距離為${\displaystyle d/2}$
• 兩條互相平行、載有異向電流的導線，彼此之間的距離為${\displaystyle d}$

## 非線性電感

「大信號電感」是用來計算磁通量，以方程式定義為

${\displaystyle L_{s}(i)\ {\stackrel {\mathrm {def} }{=}}\ {\frac {N\Phi }{i}}={\frac {\Lambda }{i}}}$

「小信號電感」是用來計算電壓，以方程式定義為

${\displaystyle L_{d}(i)\ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathrm {d} (N\Phi )}{\mathrm {d} i}}={\frac {\mathrm {d} \Lambda }{\mathrm {d} i}}}$

${\displaystyle v(t)={\frac {\mathrm {d} \Lambda }{\mathrm {d} t}}={\frac {\mathrm {d} \Lambda }{\mathrm {d} i}}{\frac {\mathrm {d} i}{\mathrm {d} t}}=L_{d}(i){\frac {\mathrm {d} i}{\mathrm {d} t}}}$

## 簡單電路的自感

${\displaystyle \quad {\frac {r^{2}N^{2}}{3\ell }}\left\{-8w+4{\frac {\sqrt {1+m}}{m}}\left(K\left({\sqrt {\frac {m}{1+m}}}\right)-\left(1-m\right)E\left({\sqrt {\frac {m}{1+m}}}\right)\right)\right\}}$

${\displaystyle ={\frac {r^{2}N^{2}\pi }{\ell }}\left\{1-{\frac {8w}{3\pi }}+\sum _{n=1}^{\infty }{\frac {\left(2n\right)!^{2}}{n!^{4}\left(n+1\right)\left(2n-1\right)2^{2n}}}\left(-1\right)^{n+1}w^{2n}\right\}}$
${\displaystyle ={\begin{cases}{\frac {r^{2}N^{2}\pi }{\ell }}\left(1-{\frac {8w}{3\pi }}+{\frac {w^{2}}{2}}-{\frac {w^{4}}{4}}+{\frac {5w^{6}}{16}}-{\frac {35w^{8}}{64}}+...\right)\ ,&w\ll 1\\rN^{2}\left\{\left(1+{\frac {1}{32w^{2}}}+O({\frac {1}{w^{4}}})\right)\ln \left(8w\right)-{\frac {1}{2}}+{\frac {1}{128w^{2}}}+O({\frac {1}{w^{4}}})\right\}\ ,&w\gg 1\end{cases}}}$

${\displaystyle N}$：捲繞匝數
${\displaystyle r}$：半徑
${\displaystyle \ell }$：長度
${\displaystyle w=r/\ell }$
${\displaystyle m=4w^{2}}$
${\displaystyle E,K}$橢圓積分

（高頻率）
${\displaystyle {\frac {\mu _{0}\ell }{2\pi }}\,\ln {\frac {\;r_{\text{o}}}{\;r_{i}}}}$ ${\displaystyle r_{\text{o}}}$：外半徑
${\displaystyle r_{i}}$：內半徑
${\displaystyle \ell }$：長度

${\displaystyle a}$：導線半徑

${\displaystyle {\frac {\mu _{0}}{\pi }}\left(b\ln {\frac {2b}{a}}+d\ln {\frac {2d}{a}}-\left(b+d\right)\left(2-{\frac {Y}{2}}\right)+2{\sqrt {b^{2}+d^{2}}}\right.}$

${\displaystyle \left.-b\cdot \operatorname {arsinh} {\frac {b}{d}}-d\cdot \operatorname {arsinh} {\frac {d}{b}}+O\left(a\right)\right)}$

${\displaystyle a}$：導線半徑
${\displaystyle b}$：邊長
${\displaystyle d}$：邊寬
${\displaystyle b,d\gg a}$

${\displaystyle {\frac {\mu _{0}\ell }{\pi }}\left(\ln {\frac {d}{a}}+Y/2\right)}$ ${\displaystyle a}$：導線半徑
${\displaystyle d}$：距離
${\displaystyle d\geq 2a}$
${\displaystyle \ell }$：長度

（高頻率）
${\displaystyle {\frac {\mu _{0}\ell }{\pi }}\operatorname {arcosh} \left({\frac {d}{2a}}\right)={\frac {\mu _{0}\ell }{\pi }}\ln \left({\frac {d}{2a}}+{\sqrt {{\frac {d^{2}}{4a^{2}}}-1}}\right)}$ ${\displaystyle a}$：導線半徑
${\displaystyle d}$：距離
${\displaystyle d\geq 2a}$
${\displaystyle \ell }$：長度

${\displaystyle {\frac {\mu _{0}\ell }{2\pi }}\left(\ln {\frac {2d}{a}}+Y/2\right)}$ ${\displaystyle a}$：導線半徑
${\displaystyle d}$：距離
${\displaystyle d\geq a}$
${\displaystyle \ell }$：長度

（高頻率）
${\displaystyle {\frac {\mu _{0}\ell }{2\pi }}\operatorname {arcosh} \left({\frac {d}{a}}\right)={\frac {\mu _{0}\ell }{2\pi }}\ln \left({\frac {d}{a}}+{\sqrt {{\frac {d^{2}}{a^{2}}}-1}}\right)}$ ${\displaystyle a}$：導線半徑
${\displaystyle d}$：距離
${\displaystyle d\geq a}$
${\displaystyle \ell }$：長度

• ${\displaystyle Y=1/2}$：電流均勻地分佈於整個導體截面。
• ${\displaystyle Y=0}$：集膚效應，電流均勻地分佈於導體表面。
• 對於高頻率案例，假若導體彼此移向對方，另外會有屏蔽電流流動於導體表面，含有參數${\displaystyle Y}$的表達式不成立。

## 參考資料

1. ^ Heaviside, O. Electrician. Feb. 12, 1886, p. 271.見該文集的再版
2. ^ Glenn Elert. The Physics Hypertextbook: Inductance. 1998–2008.
3. ^ Michael W. Davidson. Molecular Expressions: Electricity and Magnetism Introduction: Inductance. 1995–2008.
4. ^ Bansal, Rajeev, Fundamentals of engineering electromagnetics illustrated, CRC Press: pp. 154, 2006, ISBN 9780849373602
5. ^ Alexander, Charles; Sadiku, Matthew, Fundamentals of Electric Circuits 3, revised, McGraw-Hill: pp. 564–565, 2006, ISBN 9780073301150
6. ^ Ghosh, Smarajit, Fundamentals of Electrical and Electronics Engineering, PHI Learning Pvt. Ltd.: pp. 113–117, 2004, ISBN 9788120323162
7. ^ Lorenz, L. Über die Fortpflanzung der Elektrizität. Annalen der Physik. 1879, VII: 161–193.（這表達式給出面電流流動於圓柱體表面的電感）.
8. ^ Elliott, R. S. Electromagnetics. New York: IEEE Press. 1993.對於均勻電流分佈，答案裏不應該有常數 -3/2。
• Frederick W. Grover. Inductance Calculations. Dover Publications, New York. 1952.
• Griffiths, David J. Introduction to Electrodynamics (3rd ed.). Prentice Hall. 1998. ISBN 0-13-805326-X.