In an LCR circuit,R = 16.0 Ohm, C = 31.2 muF, L = 9.20 mH, and epsilon = epsilon

Question

In an LCR circuit,R = 16.0 Ohm, C = 31.2 muF, L = 9.20 mH, and epsilon = epsilonm sin omega t, with epsilon m = 45 V, and omega = 3000 rad/s. For time f = 0.442 ms find (a) the rate at which energy is being supplied by the generator, (b) the rate at which energy is being stored in the capacitor, (c) the rate at which energy is being stored in the inductor, and (d) the rate at which energy is being dissipated in the resistor, (e) What is the meaning of a negative result for any of parts (a), (b), and (c)? (f) Show that the results of parts (b), (c), and (d) sum to the result of part (a).

Explanation / Answer

capacitive reactance is XC = 1/(w*C) = 1/(3000*31.2*10^-6) = 10.68 ohm

inductive reactnace is XL = w*L = 3000*9.2*10^-3 = 27.6 ohm

Resistance is R = 16 ohm

impedence Z = Sqrt(R^2+(XL-XC)^2) = Sqrt(16^2+(27.6^2-10.68^2) = 30.06 ohm

in t = 0.442mS

emf e = 45*sin(3000*0.442*10^-3) =45*0.97 = 43.65 V

current I = e/Z = 43.65/30.06 = 1.45 A

A) Power P = V*I = 43.65*1.45 = 63.4 W

B) PC = 0.5*C*V^2/t = 0.5*31.2*10^-6*15.486*15.486/(0.442*10^-3) = 8.46 W

Vc = i*Xc = 1.45*10.68 = 15.486 V
C) PL = 0.5*L*i^2/t = 0.5*9.2*10^-3*1.45*1.45/(0.442*10^-3) = 21.88 W

D) P = i^2*R = 1.45*1.45*16 = 33.64 W

E) energy being diissipatinmg is represented by the negative sign

F) 63.4 = 8.46+33.64+21.88 whichj is almost all equal

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