# Q-factor of an RLC circuit

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I am given the following question:

A circuit is formed by an inductance L in series with a capacitance C. The capacitor leaks current which can be treated as due to a parallel resistance R. The circuit is driven by a supply of angular frequency ω. On the assumption that R ≫ 1/ωC find an expression for the Q of the circuit.

I firstly tried to make use of the definition of Q-factor of the resonant frequency over the FWHM. However, that did not work since it gave an answer of the resonant frequency. I then tried to estimate the energy stored and the energy dissipated over a cycle to find an expression of the Q-factor. However, it was in vain as well since, in my expression, when I limit the frequency to the resonant frequency, it tends to 0 rather than infinity as it should.

How am I supposed to approach this question?

(additionally, may I ask is it appropriate, in this case, to assume that the energy stored is
1/2C$V^2$? I've assumed it in my case)

asked Jan 10, 2017

## 1 Answer

1 vote

The frequency-dependent Q factor is defined as
$Q=\omega \frac{energy.stored}{power. loss}$.

The stored energy oscillates between capacitor and inductor. The total energy stored is equal to the maximum energy stored in the capacitor, which is $\frac12 CV_0^2$ where $V_0$ is the peak voltage across the capacitor. The average power loss in the resistor is $\frac{V_{rms}^2}{R}$ where $V_{rms}=\frac{V_0}{\sqrt2}$ is the rms voltage across the resistor.

$Q$ is usually quoted at the resonant frequency of the circuit. The condition $RC \gg 1/\omega$ means that the resistance is small enough not to affect the resonant frequency of the $LC$ circuit, which is $\omega_0=\frac{1}{\sqrt{LC}}$.

Therefore
$Q=\frac{1}{\sqrt{LC}}\frac{\frac12CV_0^2}{V_0^2/2R}=\frac{RC}{\sqrt{LC}}=R\sqrt{\frac{C}{L}}$

This result is confirmed later in the wikipedia article under RLC Circuits.

See also section 4 of this document from Cambridge University.

answered Jan 11, 2017 by (28,746 points)
edited Jan 12, 2017