The same end grounded on picture 1 is =0V, and the reverse and same end are virtual short circuit, so they are 0V, too. Reverse the input resistance is very high and virtual circuit. There are almost no current injection and outflow. Then R1 and R2 are equivalent series. Electric current flowing each component of the series circuit is the same. Namely, the current through the R1 and the current flowing through the R2 is the same.
The current flowing through the R1: I1= (Vi-V-) /R1... A
The current flowing through the R2: I2= (V--Vout) /R2... B
Vout= (-R2/R1) *Vi for solving the above algebraic equation of junior high school.This is the input-output relation of the legendary reverse amplifier.
2) same direction amplifier
In Figure 2, Vi and V- virtual are of short circuit, Vi=V-...... A
Because of the virtual circuit, There is no current inputting or outputting in reverse input. The the current of R1 and R2 is the same. Set this current to be I, it is obtained by Ohm's Law: I=Vout/ (R1+R2)... B
Vi is equal to the partial pressure on R2, that is, Vi=I*R2... C
From ABC to Vout=Vi* (R1+R2) /R2,. This is the legendary formula of the same direction amplifier.
3) adder 1:
Figure 3: According to virtual short circuit: V-=V+=0 ...... A
It is known from the virtual circuit and Kirchhoff's law that the sum of the current of R2 and R1 is equal to the current of the R3, so (V1 – V-) /R1+ (V2 V-) /R2= (V- – Vout) /R3... B
Replace the A, B to V1/R1+V2/R2=Vout/R3 If you take R1=R2=R3, then the upper form becomes -Vout=V1+V2, which is the legendary adder.
4) adder 2:
See Figure four, please. Because of the virtual circuit , there is no current flowing through the OP amp
end. The of the R1 and R2 are equal. Similarly, the current flowing through the R4 and the R3 are equal.
So (V1 - V+) /R1= (V+-V2) /R2... A
(Vout - V-) /R3=V-/R4... B
Known from virtual short circuit: V+=V-...... C if R1=R2, R3=R4, by the above formula you can derive V+= (V1+V2) /2V-=Vout/2. so Vout=V1+V2 is also an adder.