Answer:
(a) 4.04 ohm
(b) 6.93 A
(c) 8.53°
Step-by-step explanation:
f = 12 kHz = 12000 Hz
Vo = 28 V
R = 4 ohm
L = 30 micro Henry = 30 x 10^-6 H
C = 8 micro Farad = 8 x 10^-6 F
(a) Let Z be the impedance



Z = 4.04 Ohm
(b) Let Io be the amplitude of current


Io = 6.93 A
(c) Let the phase difference is Ф


tan Ф =0.15
Ф = 8.53°