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 ? = 157 radians/second. At this frequency,the circuit is in resonance with the maximum value of the current Imax = 0.7 A. The capacitance C = 110?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’ = 2C) and the inductance is changed appropriately to keep the circuit in resonance at angular frequency ? = 157 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

a)

W=1/sqrt(LC)

=>L =/W2*C =1/1572*(110*10-6)

L=368.8 mH

b)

Xc=1/WC =1/157*110*10-6

Xc=57.9 ohms

Vc =Imax*Xc =0.7*57.9

Vc =40.53 V

Umax,C =(1/2)CVc2 =(1/2)*(110*10-6)*40.532

Umax,C=0.09036 J

c)

dU =Emax*Imax*(pi/W) =24*0.7*(pi/157)

dU=0.336 J

d)

Z=Em/Im =24/0.7 =34.286 ohms

at resonance Z=R

R=34.286 ohms

Q=(1/R)sqrt(L/C) =(1/34.286)sqrt(0.3688/110*10-6)

Q=1.6889

e)

R=34.286 ohms

f)

Q decreases

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