. A series RLC circuit containing a resistance of 12Ω, an inductive reactance of 47.13 Ω and a capacitive reactance of 31.83 Ω are connected in series across a 100V, 50Hz supply. Calculate the

A series RLC circuit containing a resistance of 12Ω, an inductive reactance of 47.13 Ω and a capacitive reactance of 31.83 Ω are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuits current, power factor

Given that:

R = 12 Ω, [tex]X_L=47.13\ ohm,\ X_C=31.83 \ ohm[/tex], f = 50 Hz,

A) Total circuit impedance (Z) is given by:

[tex]Z=\sqrt{R^2+(X_L-X_C)^2} =\sqrt{12^2+(47.13^2-31.83)^2} =\sqrt{378.09} =19.44\ ohm[/tex]

B) the circuits current (I) is given by:

[tex]I=\frac{V}{Z}=\frac{100}{19.44}=5.14\ A[/tex]

The voltage across the resistor ([tex]V_R[/tex]) = IR= 5.14 × 12 = 61.68 V

The voltage across the inductor ([tex]V_L[/tex]) = [tex]IX_L[/tex] = 5.14 × 47.13 = 242.25 V

The voltage across the capacitor ([tex]V_c[/tex]) = [tex]IX_C[/tex] = 5.14 × 31.83 = 163.5 V

C) The power factor (Θ) is calculated as:

[tex]cos(\theta)=\frac{R}{Z}\\cos(\theta)=\frac{12}{19.44} =0.619\\\theta=cos^{-1}(0.6172)\\\theta=51.9^o\ laggin[/tex]