We have the resistance $R = 50 \mathrm{k} \Omega = 50 \times 10^3 \Omega$, the capacitance $C = 0.04 \mu \mathrm{F} = 0.04 \times 10^{-6} \mathrm{F}$, and the input voltage $v_{i} = 10 \sin 50t \mathrm{mV} = 10 \times 10^{-3} \sin 50t \mathrm{V}$.
Show more…