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Series Resonance Circuit | Series Resonance

Series Resonance Circuit | Series Resonance :- A circuit in which an inductor (L), a capacitor (C), and a resistor (R) are connected in series, and at a specific frequency (the resonant frequency) the impedance of the circuit becomes minimum (Z_min) and the current becomes maximum (I_max), is called a series resonance circuit/series resonant circuit.

Condition For Resonance And Resonant Frequency (ω_r)

(Series Resonance Circuit | Series Resonance)

We know that the impedance (Z) of a series RLC circuit,

$\displaystyle Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}$ …..(1)

At low frequencies, the value of inductive reactance (X_L) is small, but the value of capacitive reactance (X_C) is high. As the frequency increases, the value of X_L increases and the value of X_C decreases, and at a specific frequency ω (= ω_r),

$\displaystyle {{X}_{L}}={{X}_{C}}$

This is the condition for resonance.

$\displaystyle \Rightarrow {{\omega }_{r}}L=\frac{1}{{{\omega }_{r}}C}$

$\displaystyle \therefore {{\omega }_{r}}=\frac{1}{\sqrt{LC}}$ …..(2)

$\displaystyle \because {{\omega }_{r}}=2\pi {{\nu }_{r}}$

$\displaystyle \therefore {{\nu }_{r}}=\frac{1}{2\pi \sqrt{LC}}$ …..(3)

Since, at the frequency given by equations (2) and (3), $\displaystyle {{X}_{L}}={{X}_{C}}$ , therefore

$\displaystyle Z={{Z}_{min}}=\sqrt{{{R}^{2}}+{{\left( 0 \right)}^{2}}}=R$ …..(4)

Thus, at this frequency, the impedance of the series RLC circuit becomes minimum and the electric current becomes maximum.

$\displaystyle {{I}_{0}}={{I}_{max}}=\frac{{{E}_{0}}}{{{Z}_{min}}}=\frac{{{E}_{0}}}{R}$ …..(5)

Similarly,

$\displaystyle {{\left( {{I}_{rms}} \right)}_{_{max}}}=\frac{{{E}_{rms}}}{{{Z}_{min}}}=\frac{{{E}_{rms}}}{R}$ …..(6)

This frequency is called the resonant frequency (Resonance Frequency) (ω_r = 2πν_r).

Characteristics Of A Series Resonant Circuit

(Series Resonance Circuit | Series Resonance)

(1) Total reactance (X)

$\displaystyle X={{X}_{L}}-{{X}_{C}}=0$

That is, the total reactance of the series resonance circuit is zero.

(2) Impedance (Z)

$\displaystyle Z={{Z}_{min}}=\sqrt{{{R}^{2}}+{{\left( 0 \right)}^{2}}}=R$

Thus, in a series resonance circuit, the impedance is minimum and it becomes equal to the resistance.

(3) The value of the current in the circuit (I)

Peak value of current, $\displaystyle {{I}_{0}}={{I}_{max}}=\frac{{{E}_{0}}}{{{Z}_{min}}}=\frac{{{E}_{0}}}{R}$

RMS value of current, $\displaystyle {{\left( {{I}_{rms}} \right)}_{_{max}}}=\frac{{{E}_{rms}}}{{{Z}_{min}}}=\frac{{{E}_{rms}}}{R}$

Since the value of impedance is minimum at the time of resonance, hence the value of current in the circuit is maximum.

(4) We know that in a series RLC circuit,

$\displaystyle tan\phi =\frac{{{X}_{L}}-{{X}_{C}}}{R}$

In the state of resonance, X_L = X_C, therefore,

$\displaystyle tan\phi =0$

$\displaystyle \Rightarrow \phi =0$

Thus, in the state of resonance, the resultant voltage and current are in the same phase.

(5) The potential difference across the inductor (L) and capacitor (C) are equal in magnitude and opposite in phase (180º), i.e.,

$\displaystyle \overrightarrow{{{V}_{0L}}}=-\overrightarrow{{{V}_{0C}}}$

Thus, at any instant, the value of the applied voltage is equal to the potential difference across the resistor (R), denoted by $\displaystyle \overrightarrow{{{V}_{0R}}}$ .

Resonance Curve

(Series Resonance Circuit | Series Resonance)

The graph between the current flowing in the circuit (I₀ or I_rms ) and the frequency of the applied AC source (ω = 2πν) is called the resonance curve. In a series RLC circuit, the current flowing is :

$\displaystyle {{I}_{rms}}=\frac{{{E}_{rms}}}{Z}=\frac{{{E}_{rms}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}$ …..(7)

From the above equation, the following graph (resonance curve) is obtained between current (I_rms) and frequency (ω) :-

It is clear from the above curve that initially, as the frequency increases, the current also increases and reaches a maximum value at a certain frequency. This frequency is called the resonance frequency (or resonating frequency). If the frequency increases beyond this point, the current again starts to decrease. Similarly, the following graph is obtained between impedance (Z) and frequency (ω) :

Half Power Points and Half Power Frequencies

(Series Resonance Circuit | Series Resonance)

In a resonance curve, the current and the electric power (P_max) in the circuit attain their maximum values at the resonant frequency. On either side of this frequency, as the frequency decreases or increases, the current and power decrease. There are two such frequencies (ω₁ and ω₂) on either side of the resonant frequency (ω_r) at which the power in the circuit is half of the maximum power (P_max). These frequencies are called half-power frequencies, and the points P₁ and P₂ on the graph are called half-power points.

Current at half-power points P₁ and P₂ :

$\displaystyle P=\frac{{{P}_{max}}}{2}=\frac{{{\left[ {{\left( {{I}_{rms}} \right)}_{_{max}}} \right]}^{2}}\times R}{2}=\frac{{{\left[ {{\left( {{I}_{rms}} \right)}_{_{max}}} \right]}^{2}}}{2}\times R$

$\displaystyle \Rightarrow P={{\left[ \frac{{{\left( {{I}_{rms}} \right)}_{_{max}}}}{\sqrt{2}} \right]}^{2}}\times R$

Thus, from the above equation, at the half-power points, the current has a value equal to $\displaystyle \frac{1}{\sqrt{2}}$ times its maximum value $\displaystyle {{\left( {{I}_{rms}} \right)}_{_{max}}}$ .

Band width (β)

(Series Resonance Circuit | Series Resonance)

The interval between the half-power frequencies is called the bandwidth. Bandwidth is denoted by β.

$\displaystyle \beta ={{\omega }_{2}}-{{\omega }_{1}}=2\pi \left( {{\nu }_{2}}-{{\nu }_{1}} \right)$

Note :-

In a state of resonance, the impedance of a series resonant circuit is minimum, allowing the current of the resonant frequency to flow more than currents of other frequencies. That is, in this circuit, the alternating current corresponding to the resonant frequency is maximum; therefore, it is called an acceptor circuit. To tune a radio or TV, we adjust the circuit’s frequency to match the frequency of the desired station. In a radio, there is a series L-C circuit connected to the antenna. By rotating the tuner knob, the capacitance of the capacitor in the circuit is varied such that the L-C circuit resonates with the frequency of a particular station. Then, the current corresponding to that frequency reaches a maximum, while the current corresponding to other stations’ frequencies becomes negligible. Thus, we can hear the program of the selected station. The process of selecting a specific frequency in this way is called tuning, and the series L-C circuit used for this purpose is called a tuning circuit.
At resonance, the current does not depend on L and C; rather, it depends only on the resistance (R) of the circuit and the applied voltage (E).
To achieve resonance in a circuit, the following parameters can be varied :-

(i) L (ii) C (iii) Source frequency (ν)

When two series RLC circuits with the same resonant frequency ν_r are connected in series, the resonant frequency of the series combination also remains ν_r.
The unit of $\displaystyle \sqrt{LC}$ is seconds.

Example 1.

At what frequency will the potential difference across the resistance be maximum ?

Solution :

Voltage across the resistor becomes maximum at the resonant frequency (ν_r), therefore

$\displaystyle {{\nu }_{r}}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\frac{1}{\pi }\times {{10}^{-6}}\times \frac{1}{\pi }}}=500\text{ }Hz$

Example 2.

A capacitor, a resistor, and an inductor of 40 mH are connected in series to an alternating source of 60 Hz frequency. Determine the capacitance of the capacitor. Also, if the current and voltage are in phase, find the values of and .

Solution :

Since, according to the question, the current and voltage are in phase, the circuit is in a resonant state. Therefore :-

$\displaystyle {{\nu }_{r}}=\frac{1}{2\pi \sqrt{LC}}$

$\displaystyle \Rightarrow C=\frac{1}{4{{\pi }^{2}}\nu _{r}^{2}L}$

$\displaystyle \therefore C=\frac{1}{4{{\pi }^{2}}\times {{(60)}^{2}}\times 40\times {{10}^{-3}}}=176\mu F$

At resonance, E = V_R; therefore,

$\displaystyle X=110V$

and

$\displaystyle I=\frac{E}{R}=\frac{110}{220}=0.5A$

Example 3.

A coil, a capacitor, and an AC source with an RMS value of 24 V are connected in series. When the frequency of the source is varied, a maximum RMS current of 6 A is observed. If this coil is connected to a 12 V battery with an internal resistance of 4 Ω, determine the current flowing through the coil.

Solution :

We know that at resonance the current is maximum, therefore the resistance of the coil is :

$\displaystyle R=\frac{V}{I}=\frac{24}{6}=4\Omega$