DR_CAN基尔霍夫电路题解法【自留用】
无目录
如图所示电路，输入端电压 e i e_i ei，输出端电压 e o e_o eo，求二者之间关系。
对其中元件进行标号，并将电流环路标号，指出各元件的压降方向：
v值得注意的是：
1）电阻
R
2
R_2
R2同时占据环路I与环路II，按照图中规定的压降方向，其上电流应该为
i
1
−
i
2
i_1-i_2
i1−i2。
2）同理，电阻
R
3
R_3
R3上的电流应该为
i
2
−
i
3
i_2-i_3
i2−i3。
对3条环路，列出3个KVL方程：
环路1：
i
1
R
1
+
(
i
1
−
i
2
)
R
2
=
e
i
(1)
i_1 R_1 + \left( i_1 - i_2 \right) R_2 = e_i \tag{1}
i1R1+(i1−i2)R2=ei(1)
(
i
2
−
i
3
)
R
3
+
e
o
=
(
i
1
−
i
2
)
R
2
(2)
\left( i_2 - i_3 \right) R_3 + e_o = \left( i_1 - i_2 \right) R_2 \tag{2}
(i2−i3)R3+eo=(i1−i2)R2(2)
1
C
∫
0
t
i
3
d
t
=
(
i
2
−
i
3
)
R
3
(3)
\frac{1}{C} \int_0^t {i_3} dt = \left( i_2 - i_3 \right) R_3 \tag{3}
C1∫0ti3dt=(i2−i3)R3(3)同时输出端还有
e
o
=
i
2
R
4
(4)
e_o = i_2 R_4 \tag{4}
eo=i2R4(4)将(1)(2)(3)联立有
1
C
∫
0
t
i
3
d
t
+
i
2
R
4
=
e
i
−
i
1
R
1
(5)
\frac{1}{C} \int_0^t {i_3} dt + i_2 R_4 = e_i - i_1 R_1 \tag{5}
C1∫0ti3dt+i2R4=ei−i1R1(5)同时，由(1)可以推出：
i
1
=
e
i
R
1
+
R
2
+
R
2
R
1
+
R
2
i
2
(6)
i_1 = \frac{e_i}{R_1 + R_2} + \frac{R_2}{R_1 + R_2} i_2 \tag{6}
i1=R1+R2ei+R1+R2R2i2(6)将
i
1
i_1
i1表达式代入(4)：
1
C
∫
0
t
i
3
d
t
=
R
2
R
1
+
R
2
e
i
−
R
1
R
2
R
1
+
R
2
i
2
−
e
o
(7)
\frac{1}{C} \int_0^t i_3 dt = \frac{R_2}{R_1 + R_2} e_i - \frac{R_1 R_2}{R_1 + R_2} i_2 - e_o \tag{7}
C1∫0ti3dt=R1+R2R2ei−R1+R2R1R2i2−eo(7)对其求导
1
C
i
3
=
R
2
R
1
+
R
2
d
e
i
d
t
−
R
1
R
2
R
1
+
R
2
d
i
2
d
t
−
d
e
o
d
t
(8)
\frac{1}{C} i_3 = \frac{R_2}{R_1 + R_2} \frac{de_i}{dt} - \frac{R_1 R_2}{R_1 + R_2} \frac{di_2}{dt} - \frac{de_o}{dt} \tag{8}
C1i3=R1+R2R2dtdei−R1+R2R1R2dtdi2−dtdeo(8)另一方面，由(3)又有
R
2
R
1
+
R
2
e
i
−
R
1
R
2
R
1
+
R
2
i
2
−
e
o
=
(
i
2
−
i
3
)
R
3
⟹
\frac{R_2}{R_1 + R_2} e_i - \frac{R_1 R_2}{R_1 + R_2} i_2 - e_o = \left( i_2 - i_3 \right) R_3 \Longrightarrow
R1+R2R2ei−R1+R2R1R2i2−eo=(i2−i3)R3⟹
i
3
R
3
=
(
R
1
R
2
R
1
+
R
2
+
R
3
)
i
2
+
e
o
−
R
2
R
1
+
R
2
e
i
⟹
i_3 R_3 = \left( \frac{R_1 R_2}{R_1 + R_2} + R_3 \right) i_2 + e_o - \frac{R_2}{R_1 + R_2} e_i \Longrightarrow
i3R3=(R1+R2R1R2+R3)i2+eo−R1+R2R2ei⟹
i
3
=
[
R
1
R
2
R
3
(
R
1
+
R
2
)
+
1
]
i
2
+
1
R
3
e
o
−
R
2
R
3
(
R
1
+
R
2
)
e
i
(9)
i_3 = \left[ \frac{R_1 R_2}{R_3 \left( R_1 + R_2 \right)} + 1 \right] i_2 + \frac{1}{R_3} e_o - \frac{R_2}{R_3 \left( R_1 + R_2 \right)} e_i \tag{9}
i3=[R3(R1+R2)R1R2+1]i2+R31eo−R3(R1+R2)R2ei(9)将(8)(9)联立
C
R
2
R
1
+
R
2
d
e
i
d
t
−
C
R
1
R
2
R
1
+
R
2
d
i
2
d
t
−
C
d
e
o
d
t
=
[
R
1
R
2
R
3
(
R
1
+
R
2
)
+
1
]
i
2
+
1
R
3
e
o
−
R
2
R
3
(
R
1
+
R
2
)
e
i
\frac{C R_2}{R_1 + R_2} \frac{de_i}{dt} - \frac{C R_1 R_2}{R_1 + R_2} \frac{di_2}{dt} - C \frac{de_o}{dt} = \left[ \frac{R_1 R_2}{R_3 \left( R_1 + R_2 \right)} +1 \right] i_2 + \frac{1}{R_3} e_o - \frac{R_2}{R_3 \left( R_1 + R_2 \right)} e_i
R1+R2CR2dtdei−R1+R2CR1R2dtdi2−Cdtdeo=[R3(R1+R2)R1R2+1]i2+R31eo−R3(R1+R2)R2ei又因为
e
o
=
i
2
R
4
e_o = i_2 R_4
eo=i2R4，因此上式化为
C
R
2
R
4
R
1
+
R
2
d
e
i
d
t
+
R
2
R
4
R
3
(
R
1
+
R
2
)
e
i
=
C
(
R
1
R
2
R
1
+
R
2
+
R
4
)
d
e
o
d
t
+
[
R
1
R
2
R
3
(
R
1
+
R
2
)
+
R
4
R
3
+
1
]
e
o
(10)
\frac{C R_2 R_4}{R_1 + R_2} \frac{de_i}{dt} + \frac{R_2 R_4}{R_3 \left( R_1 + R_2 \right)} e_i = C \left( \frac{R_1 R_2}{R_1 + R_2} + R_4 \right) \frac{de_o}{dt} + \left[ \frac{R_1 R_2}{R_3 \left( R_1 + R_2 \right)} + \frac{R_4}{R_3} +1 \right] e_o \tag{10}
R1+R2CR2R4dtdei+R3(R1+R2)R2R4ei=C(R1+R2R1R2+R4)dtdeo+[R3(R1+R2)R1R2+R3R4+1]eo(10)
(10)式即为最终得到的
e
i
e_i
ei与
e
o
e_o
eo关系式。
带入值：
R
1
=
4
3
Ω
,
R
2
=
4
Ω
,
R
3
=
3
Ω
,
R
4
=
2
Ω
R_1 = \frac{4}{3} \Omega, R_2 = 4 \Omega, R_3 = 3 \Omega, R_4 = 2 \Omega
R1=34Ω,R2=4Ω,R3=3Ω,R4=2Ω有
3
2
C
d
e
o
d
t
+
e
o
=
3
4
C
d
e
i
d
t
+
1
4
e
i
.
\frac{3}{2} C \frac{de_o}{dt} + e_o = \frac{3}{4} C \frac{de_i}{dt} + \frac{1}{4} e_i.
23Cdtdeo+eo=43Cdtdei+41ei.
关键点难点解析
注意跨环路的电阻
R
2
R_2
R2与
R
3
R_3
R3上的电流的表达式。