Question

Obtain the resonant frequency ${{\omega }_{r}}$ of a series LCR circuit with$L=2.0H$, $C=32\mu F$ and$R=10\Omega $ . What is the Q-value of this circuit?

Answer 1

Hint: Firstly recall the condition for which a circuit is said to be in resonance and thus derive the expression for resonant frequency of an LCR circuit. You could then substitute the given values to get the resonant frequency here. Now you could recall the expression for quality factor in terms of the given quantities and then substitute to get the answer.

Formula used:
Resonant frequency,
${{\omega }_{r}}=\dfrac{1}{\sqrt{LC}}$
Quality factor,
$Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

Complete step by step solution:
We are given the values of inductance of an inductor, capacitance of a capacitor and resistance of resistor in the given question. From these values we are asked to find the resonant frequency${{\omega }_{r}}$ and also the quality factor or the Q-value of this circuit.
Given values:
$L=2.0H$
$C=32\mu F$
$R=10\Omega $
We know that the resonant frequency is the natural vibrating frequency of an object. It is denoted by${{\omega }_{r}}$An LCR circuit is said to be in resonance when the inductive reactance$\left( {{X}_{L}} \right)$ is equal to that of capacitive reactance$\left( {{X}_{C}} \right)$and thus only opposition to current flow in such a circuit will be the resistance R. So, at resonance,
${{X}_{L}}={{X}_{C}}$
$\Rightarrow {{\omega }_{r}}L=\dfrac{1}{{{\omega }_{r}}C}$
$\Rightarrow {{\omega }_{r}}^{2}=\dfrac{1}{LC}$
$\therefore {{\omega }_{r}}=\dfrac{1}{\sqrt{LC}}$
Now substituting the given values we get,
${{\omega }_{r}}=\dfrac{1}{\sqrt{2.0\times 32\times {{10}^{-6}}}}$
$\therefore {{\omega }_{r}}=125rad/s$
Basically the Q factor of a resonant circuit is the measure of quality or goodness of a resonant circuit. It can also be defined as the ratio of power stored to the power dissipated in a given circuit. It is given by the expression,
$Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$
Substituting the given values we get,
$Q=\dfrac{1}{10}\sqrt{\dfrac{2.0}{32}}$
$\Rightarrow Q=\dfrac{0.25}{10}$
$\therefore Q=2.5$
Therefore, we found the value of resonant frequency to be 125rad/s and the value of quality factor to be 2.5.

Note: We could also define the selectivity of the circuit using the quality factor Q. Sharper the resonance is, more selective the circuit is, that is, the circuit with large values of Q is said to be more selective. We also have a relation connecting the resonant frequency with the quality factor which given by,
$Q=\dfrac{L}{R}{{\omega }_{r}}$