Question

A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuit’s current, and the power factor.

Answer 1

Answer:

Impedance = 19.44ohms

Current = 5.14A

Power factor = 0.62

Explanation:

Impedance in an RLC AC circuit is defined as the total opposition to the flow of current in the resistor, inductor and capacitor.

Impedance Z = √R²+(Xl-Xc)²

Where R is the resistance = 12Ω

Inductance L = 0.15H

Capacitance C = 100uF = 100×10^-6F

Since Xl = 2πfL and Xc = 1/2πfC where f is the frequency.

Xl = 2π×50×0.15

Xl = 15πΩ

Xl = 47.12Ω

Xc = 1/2π×50×100×10^-6

Xc = 100/π Ω

Xc = 31.83Ω

Z =√12²+(47.12-31.83)²

Z = √144+233.78

Z = 19.44Ω

Impedance = 19.44ohms

To calculate the circuit current, we will use the expression V=IZ where V is the supply voltage = 100V

I = V/Z = 100/19.44

I = 5.14Amperes

To calculate the power factor,

Power factor = cos(theta) where;

theta = arctan(Xl-Xc)/R

theta = arctan(47.12-31.83)/12

theta = arctan(15.29/12)

theta = arctan1.27

theta = 51.78°

Power factor = cos51.78°

Power factor = 0.62