A series AC circuit contains a resistor, an inductor of 210 mH, a capacitor of 5

Question

A series AC circuit contains a resistor, an inductor of 210 mH, a capacitor of 5.30 ^F, and a generator with Al/max = 240 V operating at 50.0 Hz. The maximum current in the circuit is 150 mA. (a) Calculate the inductive reactance. |65.94 * -O (b) Calculate the capacitive reactance. |601 -O (c) Calculate the impedance. Impedance is the effective resistance as a function of frequency. If you know the maximum current and the maximum voltage, how do you find the impedance? kohm (d) Calculate the resistance in the circuit.

Explanation / Answer

Here ,

L = 0.510 H

C = 5.3 *10^-6 F

Vmax = 240 V

f = 50 Hz

Imax = 0.150 A

a)

inductive reactance = 2pi *f * L

inductive reactance = 2pi * 50 * 0.510

inductive reactance = 65.94 Ohm

b)

capacitive reactance = 1/(2pi * f * C)

capacitive reactance = 1/(2pi * 50 * 5.3 *10^-6)

capacitive reactance = 601 Ohm

c)

Impedance = Vmax/Imax

Impedance =240/.150

Impedance = 1600 Ohm

d)

let the resistance is R

as impedance = sqrt(R^2 + (XL - XC)^2)

1600 = sqrt(R^2 + (601 - 65.94)^2)

solving for R

R = 1507.9 Ohm

the resistance R is 1507.9 Ohm

e)

for the phase

phase angle = arctan((-601 + 65.95)/1507.9)

phase angle = -19.6 degree

the phase angle between current and generator voltage is -19.6 degree

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