Asked by chan
An AC voltage of the form Δv = (100 V) sin(1000t) is applied to a series RLC circuit. Assume the resistance is 400 , the capacitance is 4.60 µF, and the inductance is 0.500 H. Find the average power delivered to the circuit.
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Answers
Answered by
Henry
v = Vmax*sinWt
v = 100*sin(1000t)
Vrms = 0.707 * 100 = 70.7 Volts.
W = 2pi*F = 1000
6.28*F = 1000
F = 159 Hz.
Xl = W*L = 1000 * 0.5 = 500 Ohms.
Xc=1/W*C=1/(1000*4.6*10^-6)=-217.4 Ohms
tan A = (Xl+Xc)/R
tanA = (500+(-217.4))/400 = 0.7065
A = 35.2o
Z = R/cos A = 400/cos35.2 = 489.5 Ohms
[35.2o].
Irms = Vrms/Z =
70.7/489.5[35.2]= 0.1444A[-35.2o]
P=Irms^2 * R = (0.1444)2 * 400= 8.34 W.
NOTE: All of the power is consumed in the resistor.
v = 100*sin(1000t)
Vrms = 0.707 * 100 = 70.7 Volts.
W = 2pi*F = 1000
6.28*F = 1000
F = 159 Hz.
Xl = W*L = 1000 * 0.5 = 500 Ohms.
Xc=1/W*C=1/(1000*4.6*10^-6)=-217.4 Ohms
tan A = (Xl+Xc)/R
tanA = (500+(-217.4))/400 = 0.7065
A = 35.2o
Z = R/cos A = 400/cos35.2 = 489.5 Ohms
[35.2o].
Irms = Vrms/Z =
70.7/489.5[35.2]= 0.1444A[-35.2o]
P=Irms^2 * R = (0.1444)2 * 400= 8.34 W.
NOTE: All of the power is consumed in the resistor.
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