电路如图所示,若ui=0.2V,输出电压uo为多少?

I1 = If = Ui/R1 = 0.2/100k = 2uA
设Rf,R2,R3相连的结点为A点，则A点电压
Ua = -If * Rf = -2uA * 50k = -0.1V
I3 = Ua/R3 = -0.1/0.5k = -0.2mA
由节点电流定律得
I2+If-I3 = 0
I2 = I3-If = -0.2mA - 2uA = -0.202mA
Uo = U2+Ua = I2*R2+Ua = -0.202mA * 50k -0.1 = -10.2V

不好意思，之前的答案有点错的。

-9.7v