Question

A power supply that provides a voltage of 39.0 V·sin(163·t), where t is in seconds, is connected across a resistor R = 111 Ω. Calculate the average power dissipated in the resistor.

Calculate the voltage drop across the resistor.

Calculate the voltage drop across the capacitor.

Calculate the phase angle for the circuit.

Answer 1

as given, V(t) = 39.0 V sin(163t)

só, V_max = 39 V ,

só, I_max = V_Max / R

= 39 /111 = 0.35 A

só, I_R = I_max. sin wt

= 0.35 * sin (163t)

Average power dissipated in the resistor , P_avg. = V_o² / 2R

= 39² / 2*111 = 6.85 W

Assuming only resistor is connected in the circuit

só, Voltage drop resistor at any time t , V_R = I_R * R

= 0.35 * sin (163.t) * 111 = 39 sin (163t)

If there is a capacitor is connected to the circuit too, for a capacitor Let C be the capacitance of the capacitor

i_c = C . dV/dt

= C * d ( 39.0 V sin(163t)) / dt

= C * 39*163* cos (163t),

and, capacitive impedence, X_c = 1 / wC = 1/ (163 * C)

so, V_c = I_c * X_c = C * 39*163* cos (163t), *  1/ (163 * C) = 39 .cos (163 t),

phase angle, = arctan (X_c / R) = arctan(V_c / V_R)