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Chapter 28: Problem 73

For \(R L C\) circuits in which the resistance isn't too high, the \(Q\) factor may be defined as the ratio of the resonant frequency to the difference between the two frequencies where the power dissipated in the circuit is half the power dissipated at resonance. Using suitable approximations, show that this definition leads to \(Q=\omega_{0} L / R,\) with \(\omega_{0}\) the resonant frequency.

Short Answer

Expert verified
By analyzing the power dissipation in the RLC circuit, it is possible to show that the quality factor in the case of low resistance satisfies the relation \( Q = \omega_0L/R \), where \( \omega_0 \) is the resonant frequency, \( L \) is the inductance, and \( R \) is the resistance.

Step by step solution

01

Write down the formula for the power

The power dissipated in a circuit with resistance \( R \) is given by \( P = I^2R \), where \( I \) is the current. The current in an RLC circuit is given by \( I = V/R \), where \( V \) is the amplitude of the voltage. thus, the power becomes \( P = V^2 / R \). At resonance, the magnitude of the impedance \( Z \) is minimized. It is also equal to \( R \), which gives us \( P_{res} = V^2 / R \).
02

Find the power at half resonance

Since we are looking for the points where the power is half the resonant power, we equate \( P_{res}/2 = V^2 / 2R \) to the general formula for the power \( P = V^2 / |Z| \), where \( Z = R + j \, (\omega L - 1/\omega C) \), yielding \( 2R = R + j \, (\omega L - 1/\omega C) \). The imaginary part of the impedance accounts for the difference in frequencies \( \Delta \omega \).
03

Solve for the quality factor

The quality factor \( Q \) is defined as \( Q = \omega_0 / \Delta \omega \), where \( \omega_0 \) is the resonant frequency. From the previous step, equating the real parts, we can see that the frequency shift \( \Delta \omega \) equals \( R/L \), thereby we get \( Q = \omega_0L/R \), which is the desired result.

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Most popular questions from this chapter

A 1.2 - \(\mu\) F capacitor is connected across a generator whose output is \(V=V_{\mathrm{p}} \sin 2 \pi f t,\) with \(V_{\mathrm{p}}=22 \mathrm{V}, f=60 \mathrm{Hz},\) and \(t\) in seconds. Find (a) the peak current and the magnitudes of (b) the voltage and (c) the current at \(t=6.5 \mathrm{ms}\)

One-eighth of a cycle after the capacitor in an \(L C\) circuit is fully charged, what are the following as fractions of their peak values: (a) capacitor charge, (b) energy in the capacitor, (c) inductor current, (d) energy in the inductor?

Resonance is a phenomenon which occurs in RLC circuit, during which impedance due to capacitor and inductor become equal and peak current is achieved in the circuit. At resonance the impedance of the circuit is lowest and equal to the resistance in the circuit The value of peak current / is given as follows: \(I=\frac{V}{r}\) Here, \(V\) is voltage in RLC circuit and \(r\) is resistance in the circuit.

A series \(R L C\) circuit has resistance \(100 \Omega\) and impedance \(300 \Omega\) (a) What's the power factor? (b) If the rms current is 200 mA, what's the power dissipation?

When a particular inductor and capacitor are connected across the same AC voltage, the current in the inductor is higher than in the capacitor. Is this true at all frequencies?
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