Question

The values of L, C and R for a circuit are 1H, 9F and $3\Omega $. What is the quality factor for the circuit at resonance?
A. 1
B. 9
C. $\dfrac{1}{9}$
D. $\dfrac{1}{3}$

Answer 1

Hint: Remember that we use quality factor to define the sharpness of resonance. Also recall that the circuit is said to be in resonance when inductive reactance and capacitive reactance are equal to each other. From this relation, you could get the expression for resonant frequency and then you could substitute the same in the expression of the quality factor that represents the sharpness of resonance and hence we get the required relation to solve the question by simply substituting the given values.

Formula used:
Expression for inductive reactance,
${{X}_{L}}=\omega L$
Expression for capacitive reactance,
${{X}_{C}}=\dfrac{1}{\omega C}$
Expression for resonant frequency,
${{\omega }_{0}}=\dfrac{1}{\sqrt{LC}}$
Expression for quality factor of a resonance circuit,
$Q=\dfrac{{{\omega }_{0}}L}{R}$
$Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

Complete step by step answer:

Let us first understand what resonance is.
The amplitude of current for a RLC circuit is given by,
${{i}_{m}}=\dfrac{{{v}_{m}}}{Z}$ ……………… (1)
Where ${{v}_{m}}$ = amplitude of voltage, Z = impedance.
We know that, the impedance of a series LCR circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{C}}-{{X}_{L}} \right)}^{2}}}$
Where, R=resistance,
${{X}_{C}}=\dfrac{1}{\omega C}$ =capacitive reactance,
${{X}_{L}}=\omega L$ = inductive reactance
Substituting this value in equation (1), we have,
${{i}_{m}}=\dfrac{{{v}_{m}}}{\sqrt{{{R}^{2}}+{{\left( {{X}_{C}}-{{X}_{L}} \right)}^{2}}}}$
Resonance is known to be the condition at which we have minimum impedance. In other words, when the capacitive reactance is equal to the inductive reactance, that condition is resonance. That is,
${{X}_{C}}={{X}_{L}}$ …………….. (2)
At such a condition impedance Z becomes,
$Z=\sqrt{{{R}^{2}}+0}=R$
Let us call the particular frequency at which we achieve this condition as resonant frequency${{\omega }_{0}}$.
 Equation (2) now becomes,
$\dfrac{1}{{{\omega }_{0}}C}={{\omega }_{0}}L$
${{\omega }_{0}}=\dfrac{1}{\sqrt{LC}}$ ………………. (3)
Quality factor comes into picture when we talk about the sharpness of resonance.
Sharpness of resonance is given by,
$\dfrac{{{\omega }_{0}}}{2\Delta \omega }=\dfrac{{{\omega }_{0}}L}{R}$
Here, the ratio $\dfrac{{{\omega }_{0}}L}{R}$ gives the quality factor. It is represented by ‘Q’. Therefore,
$Q=\dfrac{{{\omega }_{0}}L}{R}$ …………………….. (4)
Substituting (3) in (4), we get,
$Q=\dfrac{L}{\sqrt{LC}\times R}$
$Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ ……………….. (5)
Now, we could directly substitute the given values in the question in equation (5) and then find the quality factor for the given LCR circuit.
In the question, we are given,
$L=1H$
$C=9F$
$R=3\Omega $
Substituting these values in equation (5), we get,
$Q=\dfrac{1}{3}\sqrt{\dfrac{1}{9}}$
$Q=\dfrac{1}{3}\times \dfrac{1}{3}$
$Q=\dfrac{1}{9}$
Hence, the quality factor for the circuit at resonance with the given values of L, C and R is $\dfrac{1}{9}$ .
So, the answer to the question is option C.

Note:
From equation (4), we realize that, larger the value of Q, smaller is the bandwidth $(2\Delta \omega )$ and sharper is the resonance. We could also define the selectivity of the circuit using Q. Sharper the resonance is, more selective the circuit is. That is, the circuits with large values of Q are said to be more selective.