In the following circuit an AC input V_i(t)=(20 mV ) (10^5 t ) is applied at the left end. The amplitude of the output voltage V₀ at the right end across the capacitor will be

Step-by-step solution

Given that, input AC voltage, Comparing Eqs. (i) and (ii), we get =10^5 rad / s , V_ =20 mV =20 10⁻³ ~V The V_i is distributed across R and C . by V_R and V_C , as phasor diagram. Reactance of capacitor, X_C= 1 C = 1 10^5 10⁻⁸ =10^3 =1000 Now, impedance of R - C series circuit. Z= R^2+X_C^2 = (1000)^2+(1000)^2 =1000 2 Now, voltage across capacitor, aligned V_C & =V₀= X_C Z V_ = 1000 1000 2 20 mV & = 2 10 mV & =14.14 mV aligned