电路分析 基础 电容、电感元件的串联与并联
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串联电容的等效电容u1u_1u1=1C1\frac{1}{C1}C11∫abidx\int^b_a{i}{\rm d}x∫abidxu2u_2u2=1C2\frac{1}{C2}C21∫abidx\int^b_a{i}{\rm d}x∫abidxuuu = u1u_1u1+u2u_2u2
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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
u
1
u_1
u1=
1
C
1
\frac{1}{C1}
C11
∫
a
b
i
d
x
\int^b_a{i}{\rm d}x
∫abidx
u
2
u_2
u2=
1
C
2
\frac{1}{C2}
C21
∫
a
b
i
d
x
\int^b_a{i}{\rm d}x
∫abidx
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
∫abidx =
1
C
\frac{1}{C}
C1
∫
a
b
i
d
x
\int^b_a{i}{\rm d}x
∫abidx
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
1.1.2串联电容的分压
u
1
u_1
u1 =
C
C
1
\frac{C}{C1}
C1Cu =
C
2
C
1
+
C
2
\frac{C2}{C1+C2}
C1+C2C2u
u
1
u_1
u1 =
C
C
2
\frac{C}{C2}
C2Cu =
C
1
C
1
+
C
2
\frac{C1}{C1+C2}
C1+C2C1u
1.2并联电容
1.2.1等效电容
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)
1.2.2并联电容的分流
i
1
i_1
i1=
C
1
C
\frac{C1}{C}
CC1i
i
2
i_2
i2=
C
2
C
\frac{C2}{C}
CC2i
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
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
2.1.2串联电感的分压
u
1
u_1
u1 =
L
1
L_1
L1
d
i
d
t
\frac {di}{dt}
dtdi =
L
1
L
\frac {L1}{L}
LL1u =
L
1
L
1
+
L
2
\frac {L1}{L1+L2}
L1+L2L1u
u
2
u_2
u2 =
L
2
L_2
L2
d
i
d
t
\frac {di}{dt}
dtdi =
L
2
L
\frac {L2}{L}
LL2u =
L
2
L
1
+
L
2
\frac {L2}{L1+L2}
L1+L2L2u
2.2电感的并联
2.2.1等效电感
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
∫abu(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
∫abu(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
∫abu(x)dx =
1
L
\frac{1}{L}
L1
∫
a
b
u
(
x
)
d
x
\int^b_a{u(x)}{\rm d}x
∫abu(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
2.2.2并联电感的分流
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
∫abu(x)dx =
L
L
1
\frac{L}{L1}
L1Li =
L
2
L
1
+
L
2
\frac{L2}{L1+L2}
L1+L2L2i
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
∫abu(x)dx =
L
L
2
\frac{L}{L2}
L2Li =
L
1
L
1
+
L
2
\frac{L1}{L1+L2}
L1+L2L1i
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。
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