Question

The inverting op-amp shown in the figure has an open-loop gain of 100. The closed-loop gain $\frac{V_0}{V_S}$ is

• A. -8
• B. -9
• C. -10
• D. -11

Answer 1

The correct option is B -9

Given,

$R_1=1K\Omega$

$R_2=10K\Omega$

Open loop gain $A_{OL}=100$

As, it is invering ap-amo.

So,

Closed loop gain $A_r=\frac{V_o}{V_s}=\frac{-\frac{R_2}{R_1}}{1+\left(\frac{1+\frac{R_2}{R_1}}{A_{OL}}\right)}$

$\Rightarrow \frac{V_o}{V_s}=\frac{-\frac{10}{1}}{1+\left(\frac{1+\frac{10}{1}}{100}\right)}$

$=\frac{-10}{1+\left(\frac{1+10}{100}\right)}==\frac{-10}{1+\left(\frac{11}{100}\right)}$

$\Rightarrow \frac{V_o}{V_s}=\frac{-10}{\frac{100+11}{100}}=\frac{-1000}{111}=--9.009$