Question

Chapter 31, Problem 046

An alternating emf source with a variable frequency f_d is connected in series with a 57.0 ? resistor and a 27.0 ?F capacitor. The emf amplitude is 11.0 V. Consider a phasor diagram for phasor V_R (the potential across the resistor) and phasor V_C (the potential across the capacitor). (a) At what driving frequency f_d do the two phasors have the same length? At that driving frequency, what are (b) the phase angle in degrees, (c) the angular speed at which the phasors rotate, and (d) the current amplitude?

Answer 1

(a) If they have same length means
V_R = V_C
IR = IX_C
Where X_C = 1/wC , therefore
R = 1/wC
w = 1/CR
2f = 1/RC
therefore the driving frequency
f = 1/2RC = 1/(2*57*27*10^-6)
f = 218.32 Hz
(b) The phase angle
= tan^-1(-V_C/V_R)
since V_C = V_R , therefore
= tan^-1(-1) = -45 ^o
(c) The angular speed
w = 1/RC = 1/(57*27*10^-6) = 649.77 rad/s
(d) Current is given by
I = V/Z
Where Z = (R² +X_C²)^1/2
X_C = 1/wC = 1/(649.77*27*10^-6) = 57
Z = (57² +57²)^1/2 = 80.61
I = 11/80.61
= 0.136 A