以下均为个人理解，如有错误，请大佬指出，小生立马就改正。

以下均为理想化的模型。

电路分析 基础 电容、电感元件的串联与并联

1.电容的串并联（与电阻相反）1.1串联电容1.1.1等效电容1.1.2串联电容的分压1.2并联电容1.2.1等效电容1.2.2并联电容的分流2.电感的串并联（与电阻一致）2.1串联电感2.1.1等效电感2.1.2串联电感的分压2.2电感的并联2.2.1等效电感2.2.2并联电感的分流

1.电容的串并联（与电阻相反）

1.1串联电容

1.1.1等效电容

u1u_1u1​=1C1\frac{1}{C1}C11​∫abidx\int^b_a{i}{\rm d}x∫ab​idx

u2u_2u2​=1C2\frac{1}{C2}C21​∫abidx\int^b_a{i}{\rm d}x∫ab​idx

uuu = u1u_1u1​+u2u_2u2​ = (1C1\frac{1}{C1}C11​+1C2\frac{1}{C2}C21​)∫abidx\int^b_a{i}{\rm d}x∫ab​idx = 1C\frac{1}{C}C1​∫abidx\int^b_a{i}{\rm d}x∫ab​idx

C = (1C1\frac{1}{C1}C11​+1C2\frac{1}{C2}C21​) = C1C2C1+C2\frac{C1C2}{C1+C2}C1+C2C1C2​

1.1.2串联电容的分压

u1u_1u1​ = CC1\frac{C}{C1}C1C​u = C2C1+C2\frac{C2}{C1+C2}C1+C2C2​u

u1u_1u1​ = CC2\frac{C}{C2}C2C​u = C1C1+C2\frac{C1}{C1+C2}C1+C2C1​u

1.2并联电容

1.2.1等效电容

i1i_1i1​=C1C_1C1​dudt\frac {du}{dt}dtdu​

i2i_2i2​=C2C_2C2​dudt\frac {du}{dt}dtdu​

iii = i1i_1i1​+i2i_2i2​ = (C1C_1C1​+C2C_2C2​)dudt\frac {du}{dt}dtdu​ = CCCdudt\frac {du}{dt}dtdu​

C = (C1C_1C1​+C2C_2C2​)

1.2.2并联电容的分流

i1i_1i1​= C1C\frac{C1}{C}CC1​i

i2i_2i2​= C2C\frac{C2}{C}CC2​i

2.电感的串并联（与电阻一致）

2.1串联电感

2.1.1等效电感

u1u_1u1​ = L1L_1L1​didt\frac {di}{dt}dtdi​

u2u_2u2​ = L2L_2L2​didt\frac {di}{dt}dtdi​

uuu = u1u_1u1​+u2u_2u2​ = (L1L_1L1​+L2L_2L2​)didt\frac {di}{dt}dtdi​ = LLLdidt\frac {di}{dt}dtdi​

L = L1L_1L1​ + L2L_2L2​

2.1.2串联电感的分压

u1u_1u1​ = L1L_1L1​didt\frac {di}{dt}dtdi​ = L1L\frac {L1}{L}LL1​u =L1L1+L2\frac {L1}{L1+L2}L1+L2L1​u

u2u_2u2​ = L2L_2L2​didt\frac {di}{dt}dtdi​ = L2L\frac {L2}{L}LL2​u =L2L1+L2\frac {L2}{L1+L2}L1+L2L2​u

2.2电感的并联

2.2.1等效电感

i1i_1i1​=1L1\frac{1}{L1}L11​∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx

i2i_2i2​=1L2\frac{1}{L2}L21​∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx

iii = i1i_1i1​+i2i_2i2​ = (1L1\frac{1}{L1}L11​+1L2\frac{1}{L2}L21​)∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx =1L\frac{1}{L}L1​∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx

L = (1L1\frac{1}{L1}L11​+1L2\frac{1}{L2}L21​) = L1L2L1+L2\frac{L1L2}{L1+L2}L1+L2L1L2​

2.2.2并联电感的分流

i1i_1i1​ = 1L1\frac{1}{L1}L11​∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx = LL1\frac{L}{L1}L1L​i =L2L1+L2\frac{L2}{L1+L2}L1+L2L2​i

i2i_2i2​ = 1L2\frac{1}{L2}L21​∫abu(x)dx\int^b_a{u(x)}{\rm d}x∫ab​u(x)dx = LL2\frac{L}{L2}L2L​i =L1L1+L2\frac{L1}{L1+L2}L1+L2L1​i

注意：

以上虽然是关于两个电容或两个电感的串联和并联等效，但其结论可以推广到n 个电容或 n 个电感的串联和并联等效。