An inductor of 247 mH with a resistance of 37 Ω is connected to a power supply with a maximum voltage of 336 V and a frequency of 71 Hz. Find the current in the circuit. Answer in units of A. 022 (part 2 of 3) 10.0 points Find the phase angle between the current and applied voltage. Answer in units of ◦ . 023 (part 3 of 3) 10.0 points Find the power loss in the inductor. Answer in units of W.

Respuesta :

Answer:

(i) The current in the circuit is 2.044 A

(ii) the phase angle is 71.441⁰

(iii) The power loss in the inductor is 154.58 W

Explanation:

Given;

inductance, L = 247 mH

resistance, R = 37 Ω

maximum voltage, V₀ = 336 V

frequency, f = 71 Hz

The rms voltage is given as;

[tex]V_{rms} = 0.7071V_o\\\\V_{rms} = 0.7071 \ \times \ 336\\\\V_{rms} = 237.586 \ V\\[/tex]

The inductive reactance is given as;

[tex]X_l = \omega \ L\\\\X_l = 2\pi f L\\\\X_l = 2\pi (71)(247 \times 10^{-3})\\\\X_l = 110.202 \ ohms[/tex]

The impedance of the A.C circuit is given as;

[tex]Z = \sqrt{X_l^2 + R^2} \\\\Z = \sqrt{(110.202)^2 + (37)^2}\\\\Z = 116.248 \ ohms[/tex]

(i) The current in the circuit is given as;

[tex]I_{rms} = \frac{V_{rms}}{Z}\\\\ I_{rms} =\frac{237.586}{116.248} \\\\I_{rms} =2.044 \ A[/tex]

(ii) the phase angle is given as;

[tex]tan \phi = \frac{X_l}{R}\\\\tan \phi =\frac{110.202}{37} \\\\ tan \phi =2.9784\\\\\phi = tan^{-1} (2.9784)\\\\\phi = 71.441 ^0 \\\\[/tex]

(iii) The power loss in the inductor is given as;

P = IVcosΦ

P = (2.044)(237.586)Cos(71.441⁰)

P = 154.58 W