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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。