
For an RLC circuit, driven with voltage of amplitude $vm$ and frequency ${\omega _0} = \dfrac{1}{{\sqrt {LC} }}$ the current exhibits resonance. The quality factor, $Q$ is given by
a) $\dfrac{R}{{({\omega _0}C)}}$
b) $\dfrac{{CR}}{{{\omega _0}}}$
c) $\dfrac{{{\omega _0}L}}{R}$
d) $\dfrac{{{\omega _0}R}}{L}$
Answer
208.2k+ views
Hint: The quality factor is defined as the voltage magnification of the circuit at resonance. Electrical resonance occurs in an electric circuit at a particular resonant frequency when the impedances or admittances of circuit elements cancel each other.
Complete step by step answer:
Let’s define all the data given in the question:
Voltage amplitude $ = vm$
Frequency, ${\omega _0} = \dfrac{1}{{\sqrt {LC} }}$
It is given that the current exhibits resonance.
Thus frequency ${\omega _0}$when subjected to resonance:
$ \Rightarrow {\omega _0} = \dfrac{1}{{\sqrt {LC} }}\left\{ {{X_L} = {X_C}} \right\}$
We need to find the quality factor, $Q$
The goodness or quality of a resonant circuit is described by the term quality factor, Q of a resonant circuit. If the figure is a higher value that corresponds to a narrower bandwidth.
The quality factor, Q is given by, $Q = \dfrac{{{\omega _0}}}{{Bw}}$
Where, $Bw = $Band width,
We know the value of bandwidth by the formula,
$Bw = \dfrac{R}{L}$
Applying this value of band width to the formula of the quality factor, we get,
$ \Rightarrow Q = \dfrac{{{\omega _0}}}{{Bw}}$
$ \Rightarrow Q = \dfrac{{{\omega _0}L}}{R}$
From The above equation, we get the value for the quality factor, that is,
The quality factor, $Q = \dfrac{{{\omega _0}L}}{R}$
So the final answer will be Option(C).
Note: A resonant L-C-R circuit’s sharpness of resonance can be determined by the ratio of resonant frequency with the selectivity of the circuit. And this ratio is called the Quality Factor or the Q-factor. The goodness or quality of a resonant circuit is described by the term quality factor, Q of a resonant circuit. If the figure is a higher value that corresponds to a narrower bandwidth, which is desirable in many applications. More formally, Q is the ratio of power stored to power dissipated in the circuit.
Complete step by step answer:
Let’s define all the data given in the question:
Voltage amplitude $ = vm$
Frequency, ${\omega _0} = \dfrac{1}{{\sqrt {LC} }}$
It is given that the current exhibits resonance.
Thus frequency ${\omega _0}$when subjected to resonance:
$ \Rightarrow {\omega _0} = \dfrac{1}{{\sqrt {LC} }}\left\{ {{X_L} = {X_C}} \right\}$
We need to find the quality factor, $Q$
The goodness or quality of a resonant circuit is described by the term quality factor, Q of a resonant circuit. If the figure is a higher value that corresponds to a narrower bandwidth.
The quality factor, Q is given by, $Q = \dfrac{{{\omega _0}}}{{Bw}}$
Where, $Bw = $Band width,
We know the value of bandwidth by the formula,
$Bw = \dfrac{R}{L}$
Applying this value of band width to the formula of the quality factor, we get,
$ \Rightarrow Q = \dfrac{{{\omega _0}}}{{Bw}}$
$ \Rightarrow Q = \dfrac{{{\omega _0}L}}{R}$
From The above equation, we get the value for the quality factor, that is,
The quality factor, $Q = \dfrac{{{\omega _0}L}}{R}$
So the final answer will be Option(C).
Note: A resonant L-C-R circuit’s sharpness of resonance can be determined by the ratio of resonant frequency with the selectivity of the circuit. And this ratio is called the Quality Factor or the Q-factor. The goodness or quality of a resonant circuit is described by the term quality factor, Q of a resonant circuit. If the figure is a higher value that corresponds to a narrower bandwidth, which is desirable in many applications. More formally, Q is the ratio of power stored to power dissipated in the circuit.
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