## 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 ca

Question

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?

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3 years 2021-09-02T23:38:07+00:00 2 Answers 7 views 0

26.06°

## Explanation:

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°

Explanation: the phase angle cosθ is given as

cosθ = R/Z

Where R = resistive reactance = 14.5 ohms

Z = impeadance = √R^2 +(Xl – Xc)^2

Where Xl = inductive reactance = 16.5 ohms and Xc= capacitive reactance = 9.41 ohms

By substituting the parameters, we have that

Z = √14.5^2 + (16.5^2 – 9.41^2)

Z = √210.25 + (272.25 – 88.5481)

Z = √210.25 + 183.7019

Z = √393.9519

Z = 19.85 ohms

Z = 19.85 ohms, R = 14.5 ohms

cosθ = R/Z = 14.5/19.85

cosθ = 0.7531