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1. A series RLC current has a resistance of 25 ohms, the capacitance of 0.80 micro-F,...

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littlecomet1992 | Student, Undergraduate | eNoter

Posted June 21, 2013 at 2:20 PM via web

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1. A series RLC current has a resistance of 25 ohms, the capacitance of 0.80 micro-F, and in an inductance of 250 mH. The current is connected to a variable frequency source with a fix rms output of 12 V.

a) If the source frequency is 60 Hz, calculate Z.

b) Find the rms current

c) What is the phase angle

d) Determine the power factor

e) Calculate the average power.

Tagged with ac, circuits, physics, science

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 12, 2013 at 11:30 AM (Answer #1)

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The circuit is series. It means the current I is the same though all circuit components (R, L and C). The diagram of voltage drops on R, L and C and also the total voltage drop is attached below.

1.

`X_L = 2*pi*F *L =2*pi*60*0.250 = 94.25 Omega`

`X_C = 1/(2*pi*F*C) =1/(2*pi*60*0.8*10^-6) =3315.72 Omega`

`Z = sqrt(R^2 + (X_L -X_C)^2)= sqrt(25^2 +(3315.72-94.25)^2) =3221.57 Omega`

2.

`I = U/Z = 12/3221.56 =3.72*10^-3 A= 3.72 mA`

3.

`tan(theta) = (X_L -X_C)/R = -3221.47/25 = 128.86`

`theta = 89.55 degree`

4.

Power factor is

`f = "ResistivePower"/"ApparentPower" =P/Pa =(I^2*R)/(I^2*Z) =R/Z = 25/3221.56 =0.0078`` `

5.

Resistive power is

`P = I^2*R = (3.72*10^-3)^2*25 =3.46*10^-4 W`

Apparent power is

`Pa = I^2*Z =(3.72*10^-3)^2 *3221.56 =0.0446 VA`

Reactive power is

`P_R = I^2*(X_C-X_L) = (3.72*10^-3)^2 *3221.47 =0.0446 VAR`

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