Question

Calculate resonant frequency and Q-factor of a series L-C-R circuit containing a pure inductor of inductance $3\,\,H,$ Capacitor of capacitance $27\mu F$ and resistor of resistance $7.4\Omega .$

Answer 1

Hint: In series, LCR circuit, resonant frequency corresponds to the maximum current and Q-factor is related to sharpness of curve. Their values can be calculated from this dependence on the LCR circuit.

Formula used:
1. Resonance frequency,
\[{\omega _r} = \dfrac{1}{{\sqrt {LC} }}\]
Where L represent the inductance
And C represent the capacitance
2. Q-factor,
Q$ = \dfrac{{{\omega _r}L}}{R}$
Where R represents the resistance.

Complete step by step answer:
We know that in a series LCR circuit, for resonance to occur, the condition is that the current through the circuit has to be at its maximum value.
Now, current is given by
${I_0} = \dfrac{{{V_o}}}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}$
Where ${I_0}$is current amplitude
R is resistance, ${V_0}$ is amplitude of voltage
${X_L}$ is inductive reactance
${X_C}$ is capacitive reactance
And, for ${I_0}$ to be maximum,
${X_L} - {X_C} = 0$
$ \Rightarrow {X_L} = {X_C}$
$\omega L = \dfrac{1}{{\omega c}}\,\,\,..........\left( {as\,\,{X_C} = \dfrac{1}{{\omega c}},\,\,{X_L} = \omega L} \right)$
Now, at resonance $\omega = {\omega _r}$
$ \Rightarrow {\omega _r}^2 = \dfrac{1}{{LC}}$
$ \Rightarrow {\omega _r} = \dfrac{1}{{\sqrt {LC} }}\,\,............\left( 1 \right)$
Where ${\omega _r}$ represent resonance frequency
Now, here inductance, L$ = 3H$
Capacitance, C$ = 27\mu F = 27 \times {10^{ - 6}}F$ and Resistance, $R$$ = 7.4\Omega $
So, substituting these values in equation (1), we get
${\omega _r} = \dfrac{1}{{\sqrt {3 \times 27 \times {{10}^{ - 6}}} }}$
$ \Rightarrow {\omega _r} = 111.11\,\,Hz$
Hence, resonance frequency is $111.11\,\,Hz$
Now, Quality factor is given as
Q$ = \dfrac{{{\omega _r}L}}{R}$
Substituting the values, we get Quality factor,
Q$ = \dfrac{{111.11 \times 3}}{{7.4}}$
Q$ = 45.04$
Hence, quality factor of this series LCR circuit is $45.04$ and resonance frequency is $111.11\,\,Hz$

Note:
Remember that Quality factor has no units as it is the ratio of two similar quantities i.e.
Q$ = \dfrac{{{\omega _r}}}{{{\omega _2} - {\omega _1}}} = \dfrac{{{\omega _r}}}{{2\Delta \omega }}$
Where ${\omega _2} - {\omega _1},$ is band width and it determines the sharpness of the curve.