Question

In in in out Q5e Given: Vin-25 cos (4,000t) Volts Rin 40 kΩ (kilo Ohm) Cin-0.1 μF (micro F) Rf-10 Ω (kilo Ohm) Rout-140 (kilo Ohm) R2-10 Ω (Ohm) The opamp has power input rails at +15V and -15V Find the steady-state output voltage Vout which is the voltage across the resistor R out (Degrees) Volts cos (4,000t+ out State the voltage magnitude as positive and state the phase angle as a positive angle (counterclockwise from the origin) in the correct quadrant.

Answer 1

V_in = 25 cos (4000t) R_in = 40 k ohm, C-in = 0.1 mu F, Rf = 10 k ohm X_L = -3/wc = -3/400 times 0.1 times 10^-6 X_L = -j 2500 V_1 = 0v [By virtual ground] apply nodal at v_1 v_in - v_1/R_in - I X_L = v_1 - v_0/Rf v_0/v_in = -Rf/R_in - j C_l = -10k/40 k - j 2500 [v_0/v_in = -10/40 - j 2.5] \r\n Vertical-bar V_0/V_in Vertical-bar = 10/squareroot 40^2 + 2.5^2 = 0.25 Vertical-bar V_0.V_in Vertical-bar = 0.25 angle V_0/V_in = 180 + tan^-1 2.5/40 = 180 + tan^-1 0.0625 angle V_0/V_in = 183.576 degrees V_0 = 25 times Vertical-bar V_0/V_in Vertical-bar cos(4000 t + angle V_0/V_in) V_0 = 25 times 0.25 cos(4000 t + 183.576 degrees) V_0 = 6.25 cos(4000 t + 183.576)

Answer 2

V_in = 25 cos (4000t) R_in = 40 k ohm, C-in = 0.1 mu F, Rf = 10 k ohm X_L = -3/wc = -3/400 times 0.1 times 10^-6 X_L = -j 2500 V_1 = 0v [By virtual ground] apply nodal at v_1 v_in - v_1/R_in - I X_L = v_1 - v_0/Rf v_0/v_in = -Rf/R_in - j C_l = -10k/40 k - j 2500 [v_0/v_in = -10/40 - j 2.5] \r\n Vertical-bar V_0/V_in Vertical-bar = 10/squareroot 40^2 + 2.5^2 = 0.25 Vertical-bar V_0.V_in Vertical-bar = 0.25 angle V_0/V_in = 180 + tan^-1 2.5/40 = 180 + tan^-1 0.0625 angle V_0/V_in = 183.576 degrees V_0 = 25 times Vertical-bar V_0/V_in Vertical-bar cos(4000 t + angle V_0/V_in) V_0 = 25 times 0.25 cos(4000 t + 183.576 degrees) V_0 = 6.25 cos(4000 t + 183.576)

Answer 3

V_in = 25 cos (4000t) R_in = 40 k ohm, C-in = 0.1 mu F, Rf = 10 k ohm X_L = -3/wc = -3/400 times 0.1 times 10^-6 X_L = -j 2500 V_1 = 0v [By virtual ground] apply nodal at v_1 v_in - v_1/R_in - I X_L = v_1 - v_0/Rf v_0/v_in = -Rf/R_in - j C_l = -10k/40 k - j 2500 [v_0/v_in = -10/40 - j 2.5] \r\n Vertical-bar V_0/V_in Vertical-bar = 10/squareroot 40^2 + 2.5^2 = 0.25 Vertical-bar V_0.V_in Vertical-bar = 0.25 angle V_0/V_in = 180 + tan^-1 2.5/40 = 180 + tan^-1 0.0625 angle V_0/V_in = 183.576 degrees V_0 = 25 times Vertical-bar V_0/V_in Vertical-bar cos(4000 t + angle V_0/V_in) V_0 = 25 times 0.25 cos(4000 t + 183.576 degrees) V_0 = 6.25 cos(4000 t + 183.576)