Whereas:
Freq. 5Khz @ 10 Vdc
a. Calculate the Current in the series RL circuit
b. Calculate the Phase Angle
c. Calculate the admittance
Solution:
Xl= 2 π fl
XL= 2 π* 5Khz * 5.1milli Henry
XL= 51π = 160.22
Whereas Z= impedance
Z=( R^2 + XL^2)^ 0.5
Z= 366.83 ohms
Therefore Current in Series RL Circuit
I= Vdc/Z
I= 10V/ 366.83 ohms
a. I= 27.25mA * minimum Current as seen on the sinusoidal waveform by approximation
b. Theta= Inverse Tangent (XL/R)
Theta= Tan-1(160.22/330)
= 25.89°
c. Z=1/Y
366.83 ohms=1/Y
Y= 2.72 milli Siemens
