An AC voltage of the form Av = loo sin 1 000t, where delta v is in volts and t i

Question

An AC voltage of the form Av = loo sin 1 000t, where delta v is in volts and t is in seconds, is applied to a series RLC circuit. Assume the resistance is 480 fi, the capacitance is 5.25 pF, and the inductance is 0.500 H. Find the average power delivered to the circuit. Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with two components of current: one that is 900 out of phase with the voltage and another that is in phase with the voltage. The electric company charges you an extra fee for ?reactive volt-amps'' in addition to the amount you pay for the energy you use. You can avoid the extra fee by installing a capacitor between the power line and your factory. The following problem models this solution. In an RL circuit, a 120-V (rms), 60.0-Hz source is in series with a 20.0-mH inductor and a 25.0-fl resistor (a) What is the rms current? ________ A (b) What is the power factor? (c) What capacitor must be added in series to make the power factor equal to 1? (d) To what value can the supply voltage be reduced if the power supplied is to be the same as before the capacitor was installed?

Explanation / Answer

1)

V = 100*sin1000t

So, angular frequency , W = 1000 rad/s

R = 480 ohm

C = 5.25*10^-6 F

L = 0.5 H

So, inductive reactance, Xl = W*L = 1000*0.5 = 500 ohm

Capacitive reactance, Xc = 1/(WC) = 1/(1000*5.25*10^-6) = 190.5 ohm'

So, impedance of circuit , Z = sqrt(R^2 + (Xl - Xc)^2)

So, Z = sqrt(480^2+(500-190.5)^2) = 571.1 ohm

So, peak current in the cicuit, Ipeak = 100/(571.1) = 0.175 A

So, average power delivered, Pavg = Ipeak*Vpeak/2

= 0.175*100/2 = 8.75 W <----------answer

2)

L = 20 mH = 0.02 H

R = 25 ohm

f = 60 Hz

a)

So, Xl = 2*pi*f*L = 2*pi*60*0.02 = 7.54 ohm

So, impedance, Z = sqrt(R^2+Xl^2) = sqrt(25^2+7.54^2) = 26.1 ohm

So, rms current, Irms = Vrms/Z = 120/26.1 = 4.6 A <------------answer

b)

Power factor = R/Z = 25/26.1 = 0.958

c)

LEt the capacitance added be C

So, Power factor, P = R/sqrt(R^2+(Xl-1/(2*pi*f*C))^2) = 1

So, Xl = 1/(2*pi*f*C)

So, 7.54 = 1/(2*pi*60*C)

So, C = 3.5*10^-4 F

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