电路分析 基础 电容、电感元件的串联与并联

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。

u 1 u_1 u1​= 1 C 1 \frac{1}{C1} C11​ ∫ a b i d x \int^b_a{i}{\rm d}x ∫ab​idx u 2 u_2 u2​= 1 C 2 \frac{1}{C2} C21​ ∫ a b i d x \int^b_a{i}{\rm d}x ∫ab​idx u u u = u 1 u_1 u1​+ u 2 u_2 u2​ = ( 1 C 1 \frac{1}{C1} C11​+ 1 C 2 \frac{1}{C2} C21​) ∫ a b i d x \int^b_a{i}{\rm d}x ∫ab​idx = 1 C \frac{1}{C} C1​ ∫ a b i d x \int^b_a{i}{\rm d}x ∫ab​idx

C = ( 1 C 1 \frac{1}{C1} C11​+ 1 C 2 \frac{1}{C2} C21​) = C 1 C 2 C 1 + C 2 \frac{C1C2}{C1+C2} C1+C2C1C2​

u 1 u_1 u1​ = C C 1 \frac{C}{C1} C1C​u = C 2 C 1 + C 2 \frac{C2}{C1+C2} C1+C2C2​u u 1 u_1 u1​ = C C 2 \frac{C}{C2} C2C​u = C 1 C 1 + C 2 \frac{C1}{C1+C2} C1+C2C1​u

i 1 i_1 i1​= C 1 C_1 C1​ d u d t \frac {du}{dt} dtdu​ i 2 i_2 i2​= C 2 C_2 C2​ d u d t \frac {du}{dt} dtdu​

i i i = i 1 i_1 i1​+ i 2 i_2 i2​ = ( C 1 C_1 C1​+ C 2 C_2 C2​) d u d t \frac {du}{dt} dtdu​ = C C C d u d t \frac {du}{dt} dtdu​

C = ( C 1 C_1 C1​+ C 2 C_2 C2​)

i 1 i_1 i1​= C 1 C \frac{C1}{C} CC1​i i 2 i_2 i2​= C 2 C \frac{C2}{C} CC2​i

u 1 u_1 u1​ = L 1 L_1 L1​ d i d t \frac {di}{dt} dtdi​ u 2 u_2 u2​ = L 2 L_2 L2​ d i d t \frac {di}{dt} dtdi​

u u u = u 1 u_1 u1​+ u 2 u_2 u2​ = ( L 1 L_1 L1​+ L 2 L_2 L2​) d i d t \frac {di}{dt} dtdi​ = L L L d i d t \frac {di}{dt} dtdi​

L = L 1 L_1 L1​ + L 2 L_2 L2​

u 1 u_1 u1​ = L 1 L_1 L1​ d i d t \frac {di}{dt} dtdi​ = L 1 L \frac {L1}{L} LL1​u = L 1 L 1 + L 2 \frac {L1}{L1+L2} L1+L2L1​u u 2 u_2 u2​ = L 2 L_2 L2​ d i d t \frac {di}{dt} dtdi​ = L 2 L \frac {L2}{L} LL2​u = L 2 L 1 + L 2 \frac {L2}{L1+L2} L1+L2L2​u

i 1 i_1 i1​= 1 L 1 \frac{1}{L1} L11​ ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx i 2 i_2 i2​= 1 L 2 \frac{1}{L2} L21​ ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx i i i = i 1 i_1 i1​+ i 2 i_2 i2​ = ( 1 L 1 \frac{1}{L1} L11​+ 1 L 2 \frac{1}{L2} L21​) ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx = 1 L \frac{1}{L} L1​ ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx

L = ( 1 L 1 \frac{1}{L1} L11​+ 1 L 2 \frac{1}{L2} L21​) = L 1 L 2 L 1 + L 2 \frac{L1L2}{L1+L2} L1+L2L1L2​

i 1 i_1 i1​ = 1 L 1 \frac{1}{L1} L11​ ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx = L L 1 \frac{L}{L1} L1L​i = L 2 L 1 + L 2 \frac{L2}{L1+L2} L1+L2L2​i i 2 i_2 i2​ = 1 L 2 \frac{1}{L2} L21​ ∫ a b u ( x ) d x \int^b_a{u(x)}{\rm d}x ∫ab​u(x)dx = L L 2 \frac{L}{L2} L2L​i = L 1 L 1 + L 2 \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。