AC Circuit in Resonance 1 2 3 4 5 6 A circuit is constructed with an AC generato

Question

AC Circuit in Resonance

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A circuit is constructed with an AC generator, a resistor, capacitor and inductor as shown. The generator voltage varies in time as = Va - Vb = msint, where m = 24 V and = 155 radians/second. At this frequency,the circuit is in resonance with the maximum value of the current Imax = 0.69 A. The capacitance C = 104F. The values for the resistance R and the inductance L are unknown.

1)

What is L, the value of the inductance of the circuit?

mH

2)

What is Umax,C, the value of the maximum energy stored in the capacitor during one cycle?

J

3)

What is U, the total energy dissipated in the circuit in one cycle?

J

4)

What is Q, the quality factor of this ciruit?

5)

What is R, the value of the resistance of the circuit?

6)

Suppose now the value of the capacitance in the circuit is doubled (C’ = 2C) and the inductance is changed appropriately to keep the circuit in resonance at angular frequency = 155 radians/s while the generator voltage and resistance are kept constant. How does Q, the quality factor of the circuit, change, if at all?

Q increases

Q decreases

Q stays the same

Explanation / Answer

1) w = 2 pi f

155 = 2 x pi x f

f = 24.7 Hz

at resonance frequency,

wL = 1/wC

w = 1 / sqrt(LC)

155 = 1 /sqrt(L x 104 x 10^-6)

L = 0.4 H

2) Umax = C e^2 /2 = 104 x 10^-6 x 24^2 2 = 0.03 J

3) rms Power = VI/2 = 24 x 0.69 / 2 = 8.28 W

time period, T = 1/f = 1/ 24.7

average energy = Power x time = 8.28 / 24.7 = 0.335 J

4) Q = 1/R sqrt(L/C)

R= e /I = 24 / 0.69 = 34.8 ohm

Q = (1/34.8) sqrt(0.4 / (104 x 10^-6)) = 1.78

5) R = 34.8 ohm

6) sqrt(LC) = constant

LC= constant

if C is doubled then L will get halved.

then Q will decreases.

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