Question

Your little sister finds some electrical components and decides to assemble an AC circuit using a 6.480x 10^-5 Henry inductor, a 477 micro Farad capacitor and a 104.3 ohm resistor all connected in series using the wall current (120.00 V (rms) at 60.00000 Hertz). She is wondering what results she could expect and asks you to find: A reactance of each compnoent. B) the total impedance. C) the current (rms is fine). D) the phase constant. E) the power loss in each part.

Answer 1

impedence Z^2 = R^2 + (XL-Xc)^2

XL = wL = 2pif L

XL = 2*3.14 * 60 * 6.48*10^-5

XL = 0.0244 ohms inductive reactance

Xc = 1/wC = 1/2pifC

Xc = 1/(2*3.14 * 60 * 477*10^-6)

Xc = 5.56 ohms --- capacitive reactance

part A:

Z^2 = 104.3^2 + (5.56 - 0.0244)^2

Z = 104.44 ohms total impedence

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part C:

Irms = Vrms/Z

Irms = 120/104.44

Irms = 1.148 Amps

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phase tan theta = (XL-Xc)/R

tan theta = ( 0.0244-5.56)/(104.3)

theta = phase = -3.03 deg

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Power loss across L = PL = i^2 XL

PL = 1.148^2 * 0.0244

PL =0.032 Watts

Power loss across Capacitor Pc = i^2 XC

Pc = 1.148^2 * 5.56

Pc = 7.32 Watts

power loss R = i^2 R

Pr = 1.148^2 * 104.32

Pr = 137.48 Watts