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valentinak56 [21]

3 years ago

14

Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the

phase angle ϕ of the circuit in degrees.

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

<h2>137.69°</h2>

Explanation:

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

ϕ = $tan^{-1} \dfrac{X_l-X_c}{R}$

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 = $tan^{-1} \frac{41.32-64.08}{25} \\\\$

ϕ = $tan^{-1} \frac{-22.76}{25} \\\\\\\\$

$\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0$

Since tan is negative in the 2nd quadrant;

$\phi = 180-42.31^0\\\\\phi = 137.69^0$

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

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