Answer:
The voltage input to the inverting terminal is 60μV
Explanation:
Given;
open-loop gain, A = 150,000
output voltage, V₀ = 15 V
voltage at the inverting input, [tex]V_n[/tex] = −40 μV = 40 x 10⁻⁶ V
The relationship between output voltage and voltage at the inverting input is given as;
[tex]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} = \frac{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[/tex]
Therefore, the voltage input to the inverting terminal is 60μV
