inverting op-amp circuit
Why is -Vout equal to the voltage going through Rf?


(1)  The voltage at the inverting terminal is 0V* so, by KVL,

$$v_{OUT} = 0V - i_{R_F}R_F $$

where the reference direction for \$i_{R_F}\$ is from left to right through \$R_F\$

(2) By KCL,

$$i_{R_i} = i_{R_F} $$

since the inverting input is an open circuit.

(3) By Ohm's Law,

$$i_{R_i} = \dfrac{v_{IN}}{R_i}$$

(remember, the voltage at the inverting terminal is 0V).

Thus:

$$v_{OUT} = 0V - i_{R_F}R_F = -i_{R_i}R_F =  -\dfrac{v_{IN}}{R_i}R_F$$

or

$$\dfrac{v_{OUT}}{v_{IN}} = -\dfrac{R_F}{R_i}$$

*For an ideal op-amp with negative feedback, the inverting and non-inverting input voltages are equal.