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 epsilon = V_a - V_b = epsilon_m sin omega t, where epsilon_m = 120 V and omega = 649 radians/second. At this frequency, the circuit is in resonance with the maximum value of the current I_max = 0.58 A. The capacitance C = 160 mu F. The values for the resistance R and the inductance L are unknown. What is L, the impedance of the circuit? mH What is U_max,C, the value of the maximum energy stored in the capacitor during one cycle? J What is Delta U, the total energy dissipated in the circuit in one cycle? J What is Q, the quality factor of this circuit? What is R, the value of the resistance of the circuit? Ohm 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 omega = 144 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)

angular resonant frequency

W=1 / sqrt(LC)

L=1 / W2C =1 / 1442*(160*10-6)

L = 0.301 H = 301 mH

(b)

Maximum Energy stored in capacitor during one cycle is

Umax,c = Umax,L =(1/2)L*Imax2

Umax,c = (1/2)*0.301*0.582

Umax,c = 0.0507 J

(c)

Total Energy dissipated in one cycle is

dU =PavT =(1/2)EmaxImax*(2pi/W)

dU =pi*Emax*Imax / W

dU = 3.14*24*0.58 / 144

dU = 0.303 J

(d)

Quality factor of the circuit is,

Q = 2pi*(Umax / dU)

Q = 2*3.14*(0.0507 / 0.303)

Q = 1.048

(e)

Resistance is,

R = Emax / Imax

R = 24 / 0.58

R = 41.37 ohms

(f)

Since, there is no change in Umax,c and dU.

so, Q stays the same.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.