A circuit is constructed with an AC generator, a resistor, capacitor and inducto

Question

A circuit is constructed with an AC generator, a resistor, capacitor and inductor as shown. The generator voltage varies in time as ? = Va - Vb = ?msin?t, where ?m = 24 V and ? = 147 radians/second. At this frequency,the circuit is in resonance with the maximum value of the current Imax = 0.61 A. The capacitance C = 172?F. 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

Explanation / Answer

1.

angular resonant frequency

W=1/sqrt(LC)

=>L=1/W^2*C =1/147^2*(172*10^-6)

L=269.05 mH

b)

Capacitive reactance

Xc=1/WC =1/147*(172*10^-6)

Xc=39.55 ohms

Voltage across capacitor

Vc=Imax*C =0.61*39.55

Vc=24.126 Volts

so maximum energy stored in capacitor is

Umax,c=(1/2)*C*Vc^2=(1/2)*(172*10^-6)*24.126^2

Umax,c=0.05 J

c)

Total energy dissipated in 1 cycle is

U=Emax*Imax*(pi/W) =24*0.61*(pi/147)

U=0.313 J

d)

Impedance

Z=Emax/Imax =24/0.61

Z=39.34 ohms

at resonance

XL=Xc

so impedance

R=Z=39.34 ohms

so Q-factor is

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

Q=(1/39.34)*sqrt(0.269/172*10^-6)

Q=1.005

e)

R=39.34 ohms

f)

Q decreases

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