Electric Resonance

AC Circuits · Maharashtra HSC

Theory

13.8 Electric Resonance

Resonance is observed in systems that have a tendency to oscillate at a particular frequency, called the natural frequency of oscillation of the system. When such a system is driven by an energy source whose frequency is equal to the natural frequency of the system, the amplitude of oscillation becomes large and resonance is said to occur. Tuning a radio to a station, and a glass shattering at a concert, are both examples.

(a) Series Resonance Circuit

A circuit in which inductance $L$ , capacitance $C$ and resistance $R$ are connected in series, and which admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit.

The impedance of an LCR circuit is

$Z = \sqrt{R^2+\left(\omega L - \dfrac{1}{\omega C}\right)^2}$

At very low frequencies the inductive reactance $X_L = \omega L$ is negligible but the capacitive reactance $X_C = \dfrac{1}{\omega C}$ is very high. As the applied frequency is increased, $X_L$ increases and $X_C$ decreases. At some angular frequency $\omega_r$ they become equal:

$X_L = X_C \qquad \Rightarrow \qquad \omega_r L = \dfrac{1}{\omega_r C}$

$\therefore\ (\omega_r)^2 = \dfrac{1}{LC} \qquad \text{or} \qquad (2\pi f_r)^2 = \dfrac{1}{LC}$

$\therefore\ 2\pi f_r = \dfrac{1}{\sqrt{LC}}$

$\therefore\ f_r = \dfrac{1}{2\pi\sqrt{LC}}$

where $f_r$ is called the resonant frequency.

At this particular frequency, since $X_L = X_C$ , we get $Z = \sqrt{R^2+0} = R$ , which is the least value of $Z$ . Thus when the impedance of an LCR circuit is minimum, the circuit is said to be purely resistive, the current and the voltage are in phase, and hence the current

$i_0 = \dfrac{e_0}{Z} = \dfrac{e_0}{R}$

is maximum. This condition of the LCR circuit is called the resonance condition and this frequency is called the series resonant frequency.

At $\omega = \omega_r$ the peak current is maximum, and its maximum value is inversely proportional to $R$ — for lower $R$ , $i_0$ is large, and vice versa.

Series resonance curve: a graph of rms current against the frequency of the AC. It rises to a sharp maximum at $f_r$ and falls off on either side; a smaller $R$ gives a taller and sharper peak. At resonance the rms current becomes maximum. The circuit at the resonant condition is very useful in radio and TV receivers for tuning in the signal from a desired transmitting station or channel.

Characteristics of a series resonance circuit

Resonance occurs when $X_L = X_C$ .
Resonant frequency $f_r = \dfrac{1}{2\pi\sqrt{LC}}$ .
Impedance is minimum and the circuit is purely resistive.
Current has a maximum value.
When a number of frequencies are fed to it, it accepts only the one frequency $f_r$ and rejects the others; the current is maximum for this frequency. Hence it is called an acceptor circuit.

(b) Parallel Resonance Circuit

A parallel resonance circuit consists of a coil of inductance $L$ and a condenser of capacity $C$ joined in parallel to a source of alternating emf,

$e = e_0\sin\omega t$

In the case of an inductor the current lags behind the applied emf by a phase angle of $\pi/2$ , so the instantaneous current through $L$ is

$i_L = \dfrac{e_0}{X_L}\sin\!\left(\omega t - \dfrac{\pi}{2}\right)$

Similarly, in a capacitor the current leads the emf by $\pi/2$ , so

$i_C = \dfrac{e_0}{X_C}\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$

The total current in the circuit at this instant is

$i = i_C + i_L = \dfrac{e_0}{X_L}\left(-\cos\omega t\right) + \dfrac{e_0}{X_C}\cos\omega t = e_0\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)\cos\omega t$

$\therefore\ i = e_0\left(\omega C - \dfrac{1}{\omega L}\right)\cos\omega t$

The current is minimum when

$\omega C - \dfrac{1}{\omega L} = 0 \qquad \Rightarrow \qquad \omega C = \dfrac{1}{\omega L} \qquad \Rightarrow \qquad \omega^2 = \dfrac{1}{LC}$

$\therefore\ \omega = \dfrac{1}{\sqrt{LC}} \qquad \Rightarrow \qquad 2\pi f_r = \dfrac{1}{\sqrt{LC}} \qquad \Rightarrow \qquad f_r = \dfrac{1}{2\pi\sqrt{LC}}$

The condition for parallel resonance is therefore $\omega C = \dfrac{1}{\omega L}$ , i.e. $X_C = X_L$ — the same condition, and the same resonant frequency, as in the series case.

At the parallel resonant frequency $f_r$ the current is minimum: the circuit allows minimum current to flow through it, and the impedance is maximum. A parallel resonant circuit is very useful in wireless transmission or radio communication and in filter circuits.

Parallel resonance curve: a graph of current against frequency showing a sharp minimum (a dip) at $f_r$ — the mirror image of the series curve.

Characteristics of a parallel resonance circuit

Resonance occurs when $X_L = X_C$ .
Resonant frequency $f_r = \dfrac{1}{2\pi\sqrt{LC}}$ .
Impedance is maximum.
Current is minimum.
When alternating currents of different frequencies are sent through it, it offers a very high impedance to the current of the resonant frequency $f_r$ and rejects it, but allows currents of other frequencies to pass. Hence it is called a rejector circuit.

Do You Know — Both L and C Are Essential

Resonance occurs in a series LCR circuit when $X_L = X_C$ , i.e. $\omega = \dfrac{1}{\sqrt{LC}}$ . For resonance to occur, the presence of both $L$ and $C$ in the circuit is essential. Only then will the voltages across $L$ and $C$ , being 180° out of phase, cancel each other and let the current amplitude become $e_0/R$ , so that the total source voltage appears across $R$ . So we cannot have resonance in an LR or a CR circuit.

Standard Results

$X_L = X_C$ ; $\omega_r = \dfrac{1}{\sqrt{LC}}$ ; $f_r = \dfrac{1}{2\pi\sqrt{LC}}$ — the same for series and parallel, and independent of $R$ .
Series: $Z_{min} = R$ , $i_{max} = e_{rms}/R$ , $\cos\phi = 1$ , acceptor circuit.
Parallel: $Z$ maximum, current minimum, rejector circuit.
At series resonance the voltages across $L$ and across $C$ are equal in magnitude and opposite in phase, so $V_L = V_C = i\,X_L$ ; each can be far larger than the source voltage.

Worked Pattern

$L = 1$ mH, $C = 2\ \mu$ F: $f_r = \dfrac{1}{2\times3.142\times\sqrt{10^{-3}\times2\times10^{-6}}} = 3.558\times10^3$ Hz. With $R = 10\ \Omega$ and $e_0 = 10$ mV, at resonance $i_0 = e_0/R = 10^{-3}$ A, and $V_L = i_0(2\pi f_rL) = 22.36\times10^{-3}$ V $= V_C$ .

$L = 1$ H, $C = 36\ \mu$ F, $R = 12\ \Omega$ , 120 V supply: at resonance $Z = R = 12\ \Omega$ , $i_{rms} = 120/12 = 10$ A, and $f_r = \dfrac{1}{2\times3.142\times6\times10^{-3}} = 26.52$ Hz.

$L = 125$ mH in parallel with $C = 50\ \mu$ F: $f_r = \dfrac{1}{2\pi\sqrt{625\times10^{-8}}} = 63.69$ Hz.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
What is series LCR resonant circuit? Obtain the expression for impedance. Hence state the conditions for series resonance and derive the expression for resonant frequency.
Maharashtra HSC board question — asked in July 2018Answer 1 ↓
2
Assuming expression for impedance in a parallel resonant circuit, state the conditions for parallel resonance. Define resonant frequency and obtain an expression for it.
Maharashtra HSC board question — asked in March 2019Answer 2 ↓
3
State the conditions for current and impedance in parallel resonance circuit.
Maharashtra HSC board question — asked in July 2023Answer 3 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

4
An LCR series combination has R = 10 $\Omega$ , L = 1 mH and C = 2 $\mu$ F. Determine: i. The resonant frequency ii. the current in the circuit, and iii. Voltages across L and C, when an alternating voltage of 10 mV, operating at the resonant frequency, is applied to the series combination.
Maharashtra HSC board question — asked in March 2008Answer 4 ↓
5
An a.c circuit consists of inductor of 125 mH connected in parallel with a capacitor of 50 $\mu$ F. Determine resonant frequency.
Maharashtra HSC board question — asked in October 2014Answer 5 ↓
6
A parallel L-C circuit comprises of a 5 H inductor and 5 $\mu$ F capacitor. Calculate the resonant frequency of the circuit.
Maharashtra HSC board question — asked in February 2020Answer 6 ↓
7
A 0.1 H inductor, a $25\times10^{-6}$ F capacitor and a 15 $\Omega$ resistor are connected in series to a 120 V, 50 Hz AC source. Calculate the resonant frequency.
Maharashtra HSC board question — asked in March 2022Answer 7 ↓
8
A 1000 mH inductor, 36 $\mu$ F capacitor and 12 $\Omega$ resistor are connected in series to 120 V, 50 Hz AC source. Calculate: i. impedance of the circuit at resonance ii. current at resonance iii. resonant frequency
Maharashtra HSC board question — asked in July 2022Answer 8 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
i. A circuit in which inductance L, capacitance C and resistance R are connected in series and the circuit admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit. ii. The impedance (Z) of an LCR circuit is given by, $Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$ iii. At very low frequencies, inductive reactance $X_L=\omega L$ is negligible but capacitive reactance $X_C=\dfrac{1}{\omega C}$ is very high. iv. As we increase the applied frequency then $X_L$ increases and $X_C$ decreases. v. At some angular frequency ( $\omega_r$ ), $X_L=X_C$ i.e., $\omega_rL=\dfrac{1}{\omega_rC}$ $\therefore(\omega_r)^2=\dfrac{1}{LC}\ \text{ or }\ (2\pi f_r)^2=\dfrac{1}{LC}$ $\therefore2\pi f_r=\dfrac{1}{\sqrt{LC}}$ $\therefore f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Where $f_r$ is called the resonant frequency. vi. At this particular frequency $f_r$ , since $X_L=X_C$ we get $Z=\sqrt{R^2+0}=R$ . This is the least value of Z. vii. Thus, when the impedance of an LCR circuit is minimum, circuit is said to be purely resistive, current and voltage are in phase and hence the current $i_0=\dfrac{e_0}{Z}=\dfrac{e_0}{R}$ is maximum. This condition of the LCR circuit is called resonance condition and this frequency is called series resonant frequency.
2.↑ back to the question
i. A parallel resonance circuit is an AC circuit in which a parallel combination of an inductor and capacitor is connected to a source of an alternating e.m.f. ii. Let the alternating emf supplied by the source be, $e=e_0\sin\omega t$ iii. In case of an inductor, the current lags behind the applied emf by a phase angle of $\pi/2$ , then the instantaneous current through L is given by $i_L=\dfrac{e_0}{X_L}\sin(\omega t-\pi/2)$ iv. Similarly in a capacitor, as current leads the emf by a phase angle of $\pi/2$ , we can write $i_C=\dfrac{e_0}{X_C}\sin(\omega t+\pi/2)$ v. Therefore, the total current i in the circuit at this instant is, $i=i_C+i_L$ $=\dfrac{e_0}{X_L}\sin(\omega t-\pi/2)+\dfrac{e_0}{X_C}\sin(\omega t+\pi/2)$ $=\dfrac{e_0}{X_L}(-\cos\omega t)+\dfrac{e_0}{X_C}\cos\omega t$ $=e_0\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)\cos\omega t$ $i=e_0\left(\omega C-\dfrac{1}{\omega L}\right)\cos\omega t$ vi. Therefore, when $\left(\omega C-\dfrac{1}{\omega L}\right)=0$ , i = minimum, $\therefore\omega C=\dfrac{1}{\omega L}$ $\therefore\omega^2=\dfrac{1}{LC}$ $\therefore\omega=\dfrac{1}{\sqrt{LC}}$ $\therefore2\pi f=\dfrac{1}{\sqrt{LC}}$ $\therefore f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Where $f_r$ is called the resonant frequency. vii. Conditions for parallel resonance to occur, $\omega C=\dfrac{1}{\omega L}\text{ i.e., }X_C=X_L.$
3.↑ back to the question
In parallel resonant circuit, current is minimum and impedance is maximum.
4.↑ back to the question
Given: $R=10\ \Omega$ , $L=1$ mH $=10^{-3}$ H, $C=2\ \mu$ F $=2\times10^{-6}$ F, $e_0=10$ mV $=10^{-2}$ V To find: i. Resonant frequency ( $f_r$ ) ii. Current ( $i_0$ ) iii. Voltage across L ( $V_L$ ) iv. Voltage across C ( $V_C$ ) Formulae: i. $f_r=\dfrac{1}{2\pi\sqrt{LC}}$ ii. $i_0=\dfrac{e_0}{R}$ iii. $V_L=i_LX_L$ Calculation: From formula (i), $f_r=\dfrac{1}{2\times3.142\times\sqrt{10^{-3}\times2\times10^{-6}}}$ $=\dfrac{1}{2\times3.142\times\sqrt{2\times10^{-9}}}$ $=3.558\times10^3\text{ Hz}$ From formula (ii), $i_0=\dfrac{10^{-2}}{10}=10^{-3}\text{ A}$ From formula (iii), $V_L=I_0(2\pi f_rL)\qquad(\because I_L=I_0\text{ and }X_L=2\pi f_rL)$ $V_L=10^{-3}\times2\times3.142\times3.558\times10^3\times10^{-3}$ $=22.36\times10^{-3}$ V At resonance, $X_L=X_C$ $\therefore V_L=V_C$ $\therefore V_C=22.36\times10^{-3}$ V Ans: i. The resonant frequency is $3.558\times10^3$ Hz ii. Current in the circuit is $10^{-3}$ A iii. Voltage across L is $22.36\times10^{-3}$ V iv. Voltage across C is $22.36\times10^{-3}$ V.
5.↑ back to the question
Given: $L=125$ mH $=125\times10^{-3}$ H, $C=50\ \mu$ F $=50\times10^{-6}$ F To find: Resonant frequency ( $f_r$ ) Formula: $f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Calculation: Using formula, $f_r=\dfrac{1}{2\times3.14\times\sqrt{125\times10^{-3}\times50\times10^{-6}}}$ $=\dfrac{1}{6.28\times\sqrt{625\times10^{-8}}}$ $=\dfrac{1}{6.28\times25\times10^{-4}}$ $=\dfrac{10^4}{6.28\times25}=\dfrac{400}{6.28}$ $\therefore f_r=63.69\text{ Hz}$ Ans: The resonant frequency in the A.C. circuit is 63.69 Hz.
6.↑ back to the question
Given: $L=5$ H, $C=5\ \mu$ F $=5\times10^{-6}$ F To find: Resonant frequency ( $f_r$ ) Formula: $f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Calculation: Using formula, $f_r=\dfrac{1}{2\times3.142\times\sqrt{5\times5\times10^{-6}}}$ $=\dfrac{1}{6.284\times\sqrt{25\times10^{-6}}}$ $=\dfrac{1}{6.284\times5\times10^{-3}}$ $=\dfrac{1000}{6.284\times5}$ $=\text{antilog}\{\log(1000)-[\log(6.284)+\log(5)]\}$ $=\text{antilog}\{3.000-(0.7983+0.6990)\}$ $=\text{antilog}(1.5027)$ $\therefore f_r=31.82\text{ Hz}$ Ans: The resonant frequency in the A.C. circuit is 31.82 Hz.
7.↑ back to the question
Given: $L=0.1$ H, $C=25\times10^{-6}$ F, $R=15\ \Omega$ , $e_{rms}=120$ V To find: Resonant frequency ( $f_r$ ) Formula: $f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Calculation: From formula, $f_r=\dfrac{1}{2\times3.142\times\sqrt{0.1\times25\times10^{-6}}}$ $=\dfrac{1}{9.9356\times10^{-3}}$ $f_r=100.6$ Hz Ans: The value of resonant frequency is 100.6 Hz.
8.↑ back to the question
Given: $L=1000$ mH $=1$ H, $R=12\ \Omega$ , $C=36\ \mu$ F $=36\times10^{-6}$ F, $e_{rms}=120$ V To find: i. Impedance of the circuit (Z) ii. Current at resonance ( $i_{rms}$ ) iii. Resonant frequency ( $f_r$ ) Formulae: i. $i_{rms}=\dfrac{e_{rms}}{Z}$ ii. $f_r=\dfrac{1}{2\pi\sqrt{LC}}$ Calculation: At resonance, $Z=R$ $\therefore Z=12\ \Omega$ From formula (i), $i_{rms}=\dfrac{120}{12}$ $\therefore i_{rms}=10$ A From formula (ii), $f_r=\dfrac{1}{2\times3.142\times\sqrt{1\times36\times10^{-6}}}$ $=\dfrac{1}{2\times3.142\times6\times10^{-3}}$ $=\dfrac{1}{37.704\times10^{-3}}$ $f_r=26.52$ Hz Ans: i. The Impedance of the circuit is 12 $\Omega$ . ii. The current at resonance is 10 A. iii. The value of resonant frequency is 26.52 Hz.