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Consider a series RLC circuit where \( R = 25.0 \, \Omega \), \( C = 35.5 \, \mu F \), and \( L = 0.0940 \, H \), which is driven at a frequency of 70.0 Hz. Determine the phase angle \( \phi \) of the circuit in degrees.

Answer :

The phase angle of the given RLC circuit is approximately -42.0 degrees, which indicates a capacitive nature of the circuit.

To determine the phase angle [tex]\\( \phi )[/tex] of the RLC circuit, we first need to calculate the inductive reactance[tex]( X_L )[/tex] and capacitive reactance [tex]( X_C )[/tex].

  1. Inductive Reactance [tex]( X_L)[/tex]:
    [tex]X_L = 2 \pi f L = 2 \pi (70.0 \text{ Hz}) (0.0940 \text{ H}) = 41.4 \Omega[/tex]
  2. Capacitive Reactance (\\( X_C \\)):
    [tex]X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (70.0 \text{ Hz}) (35.5 \times 10^{-6} \text{ F})} \approx 63.9 \Omega[/tex]
  3. Calculate the Phase Angle [tex](\phi )[/tex]:
    The phase angle [tex]\phi[/tex] is given by
    [tex]\tan \phi = \frac{X_L - X_C}{R} \approx \frac{41.4 \Omega - 63.9 \Omega}{25.0 \Omega} \approx -0.90[/tex]
  4. Therefore, [tex]\phi \approx \tan^{-1}(-0.90) \approx -42.0^\circ[/tex]

The negative sign indicates that the circuit is capacitive in nature.

Answer:

137.69°

Explanation:

The phase angle of an RLC circuit ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

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