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?

Answer 1

Answer: cosθ = 0.7531

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

Answer 2

26.06°

Step-by-step 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 [
`(X_(L) - X_(C))/(R)`] --------------------------(i)

Where;

`X_(L)` = inductive reactance of the circuit

`X_(C)` = capacitive reactance of the circuit

R = resistance of the circuit

From the question;

`X_(L)` = 16.5 Ω

`X_(C)` = 9.41 Ω

R = 14.5 Ω

Substitute these values into equation (i) as follows;

Φ = tan⁻¹ [
`(16.5 - 9.41)/(14.5)`]

Φ = tan⁻¹ [
`(7.09)/(14.5)`]

Φ = tan⁻¹ [ 0.4890]

Φ = 26.06°

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