Generally, if a network of resistors is connected between the inverting input and the output, negative feedback exists.

First, we verify that negative feedback is present. Assume a positive value for v_i, which results in a negative output voltage of very large magnitude. Part of this negative voltage is returned through the resistor network and opposes the original input voltage. Thus, we conclude that negative feedback is present.
Next, we assume the conditions of the summing point constraint:

v_i \text{ and }i_i=0

Then, we apply Kirchhoff’s current law, Kirchhoff’s voltage law, and Ohm’s law to analyze the circuit. To begin, we notice that v_{\text{in}} appears across R_1 (\text{ because } v_i = 0).
Hence, we can write

i_1=\frac{v_{\text{in}}}{R_1} \quad \quad \quad (13.8)

Next, we apply Kirchhoff’s current law to the node at the right hand end of R_1 , to obtain

i_2=i_1 \quad \quad \quad (13.9)

(Here, we have used the fact that i_i = 0.)
Writing a voltage equation around the loop through v_i, R_2 , \text{ and } R_3, we obtain

R_2i_2=R_3i_3 \quad \quad \quad (13.10)

(In writing this equation, we have used the fact that v_i = 0.) Applying Kirchhoff’s current law at the top end of R_3 yields

i_4=i_2+i_3 \quad \quad \quad (13.11)

Writing a voltage equation for the loop containing v_o, R_4 , \text{ and } R_3 gives

v_o=-R_4i_4-R_3i_3 \quad \quad \quad (13.12)

Next, we use substitution to eliminate the current variables (i_1 , i_2 , i_3 , \text{ and } i_4 ) and obtain an equation relating the output voltage to the input voltage. First, from Equations 13.8 and 13.9, we obtain

i_2=\frac{v_{\text{in}}}{R_1} \quad \quad \quad (13.13)

Then, we use Equation 13.13 to substitute for i_2 in Equation 13.10 and rearrange terms to obtain

i_3=v_{\text{in}}\frac{R_2}{R_1R_3} \quad \quad \quad (13.14)

Using Equations 13.13 and 13.14 to substitute for i_2 \text{ and } i_3 in Equation 13.11, we find that

i_4=v_{\text{in}}\left(\frac{1}{R_1}+\frac{R_2}{R_1R_3} \right) \quad \quad \quad (13.15)

Finally, using Equations 13.14 and 13.15 to substitute into 13.12, we obtain

v_o=-v_{\text{in}}\left(\frac{R_2}{R_1}+\frac{R_4}{R_1}+\frac{R_2R_4}{R_1R_3} \right) \quad \quad \quad (13.16)

Therefore, the voltage gain of the circuit is

A_v=\frac{v_o}{v_{\text{in}}}=-\left(\frac{R_2}{R_1}+\frac{R_4}{R_1}+\frac{R_2R_4}{R_1R_3} \right) \quad \quad \quad (13.17)

The input impedance is obtained from Equation 13.8:

R_{\text{in}}=\frac{v_{\text{in}}}{i_1}=R_1 \quad \quad \quad (13.18)

Inspection of Equation 13.16 shows that the output voltage is independent of the load resistance. Thus, the output appears as an ideal voltage source to the load. In other words, the output impedance of the amplifier is zero.
Evaluating the voltage gain for the resistor values given (R_1 = R_3 = 1 \text{ k} \Omega \text{ and } R_2 = R_4 = 10 \text{ k}\Omega) yields

A_v=-120

In the basic inverter circuit of Figure 13.5, the voltage gain is given by Equation 13.5, which states that

A_v=-\frac{R_2}{R_1}

Therefore, to achieve a voltage gain of –120 with R_1 = 1 \text{ k}\Omega, \text{ we need } R_2 = 120 \text{ k}\Omega.
Notice that a resistance ratio of 120:1 is required for the basic inverter, whereas the circuit of Figure 13.6 has a ratio of only 10:1. Sometimes, there are significant practical advantages in keeping the ratio of resistances in a circuit as close to unity as possible. Then, the circuit of Figure 13.6 is preferable to the basic inverter shown in Figure 13.5.