What is the phase angle of an AC series circuit that is constructed of a 14.5-Ω resistor along with 16.5-Ω inductive reactance and 9.41-Ω capacitive reactance?

26.06°

Given an RLC circuit [a circuit containing a capacitor, inductor and resistor], the phase angle (Φ), which is the difference in phase between the voltage and the current in the circuit, is given by;

Φ = tan⁻¹ [ ]            --------------------------(i)

Where;

= inductive reactance of the circuit

= capacitive reactance of the circuit

R = resistance of the circuit

From the question;

= 16.5 Ω

= 9.41 Ω

R = 14.5 Ω

Substitute these values into equation (i) as follows;

Φ = tan⁻¹ [ ]    

Φ = tan⁻¹ [ ]

Φ = tan⁻¹ [ 0.4890]

Φ =  26.06°

Therefore the phase angle of the AC series circuit is 26.06°

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