Answer:
The voltage input to the inverting terminal is 60μV
Step-by-step explanation:
Given;
open-loop gain, A = 150,000
output voltage, V₀ = 15 V
voltage at the inverting input,
= −40 μV = 40 x 10⁻⁶ V
The relationship between output voltage and voltage at the inverting input is given as;
![V_o = A(V_p -V_n)\\\\15 = 150,000(V_p -(-40*10^(-6)))\\\\15 = 150,000 (V_p +40*10^(-6)) \\\\V_p +40*10^(-6) = (15)/(150,000) \\\\V_p + 40*10^(-6) = 1 *10^(-4)\\\\V_p + 40*10^(-6) = 100 *10^(-6)\\\\V_p = 100 *10^(-6) - 40*10^(-6)\\\\V_p = 60 *10^(-6)\\\\V_p = 60 \ \mu V](https://img.qammunity.org/2021/formulas/engineering/college/61igk9sp67p024uxjc097zct4i04fz8ach.png)
Therefore, the voltage input to the inverting terminal is 60μV