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Resonance in series LCR circuit

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Theory of series resonance in an LCR circuit

In alternating current (a.c.) theory, all voltages, currents and impedances may be treated as complex quantities. In the series LCR circuit, therefore, the a.c. source (input) voltage $\mathbf{v_{\big in}}$ and the a.c current $\mathbf{i}$ are complex and related as follows

\begin{eqnarray*} \mathbf{v_{\big in}} & = & R \; \mathbf{i} + j\omega L \; \mat... ...athbf{i} \; \Big[ R + + j (\omega L - \frac{1}{\omega C}) \Big] \end{eqnarray*}

The corresponding impedance is also complex and is given by
\begin{eqnarray*} \mathbf{Z} & = & \frac{\mathbf{v_{\big in}}} {\mathbf{i}} \\ & = & R + j (\omega L - \frac{1}{\omega C}) \end{eqnarray*}

The magnitude of the impedance is
\begin{eqnarray*} Z & = & \sqrt{ R^{2} + \Big(\omega L - \frac{1}{\omega C} \Big)^{2} } \end{eqnarray*}

We can then express $\mathbf{Z}$ in terms of its magnitude as follows
\begin{eqnarray*} \mathbf{Z} & = & Z\; e^{\;\big{j \phi} \\ & = & Z\; (\cos\phi + j \sin\phi) \end{eqnarray*}

$\big\phi$ being the phase angle. From the above equations, we have,
\begin{eqnarray*} \cos\phi & = & \frac{R}{Z} \\ \sin\phi & = & \frac{ \omega L ... ...\phi & = & \frac{ \omega L - \frac{1}{\big\omega \big{C}} } {R} \end{eqnarray*}

If we express the time-dependent complex source-voltage $\mathbf{v_{\big in}}$ as $V_{0}e^{\;\big{ j \omega t}}$, then
\begin{eqnarray*} \mathbf{i} & = & \frac{\mathbf{v_{\big in}}} {\mathbf{Z}} \\ ... ... t -\phi)}} \\ & = & I_{0} \; e^{\;\big{j ( \omega t -\phi)}} \end{eqnarray*}

$V_{0}$ and $\big{I_{0}}$ being the magnitude of the source-voltage and current respectively.

Thus we see that the current lags the source-voltage (or alternatively the source-voltage leads the current) by an angle $\big\phi$.

Now, if $\omega L > \frac{1}{\big {\omega C}}$ or alternatively $ \omega > \frac{\big{1}}{\sqrt{\big{LC}}} $, then $\big\phi$ would be +ve and the current would indeed lag the source-voltage.

On the other hand, if $\omega L < \frac{1}{\big {\omega C}}$ or alternatively $ \omega < \frac{\big{1}}{\sqrt{\big{LC}}} $, then $\big\phi$ would be -ve and the current would then lead the source-voltage.

Most importantly, if $\omega L = \frac{1}{\big {\omega C}} $ or alternatively $ \omega = \frac{\big{1}}{\sqrt{\big{LC}}} = \omega_{0} \; $, then $\big\phi$ would be zero and the current and the source-voltage would be `in-phase'. Then $Z$ turns out to be equal to $R$ and the circuit effectively becomes resistive. Moreover, this value of $Z$ happens to be the minimum, since $Z$ is expressed as the square-root of the sum of two squares as shown above. The corresponding magnitude of current, $I_{0}$, becomes maximum, equalling $\big\frac{\big{V_{0}}}{\big{R}}$, and we have Resonance'. The corresponding linear frequncy, viz., $f_{0}= \big{\frac{1}{\big{2\pi}}} \omega_{0} = \big{\frac{1}{\big{2\pi}}} \frac{\big{1}}{\sqrt{\big{LC}}} $ is called the resonant-frequency.

In order to graphically follow the time-evolution of the voltages and currents, one needs to plot the waveform profiles. The waveforms are obtained by considering any of the two real parts, viz., the sine or cosine part of the corresponding complex number. For example, since the source-voltage can be written as
\begin{eqnarray*} \mathbf{v_{\big in}} & = & V_{0}e^{\;\big{ j \omega t}} \\ & = & V_{0}(\cos \omega t + j \sin \omega t ), \end{eqnarray*}

the expression for the corresponding waveform may be taken to be $ v_{\big in} (t) = V_{0}\sin \omega t $. Similarly, the current waveform may be expressed as $ i (t) = I_{0} \sin (\omega t-\phi)$.

Let us now examine the voltages across the circuit components, $R$, $L$ and $C$. The voltage across $R$ is given by
\begin{eqnarray*} \mathbf{v_{R}} & = & R\;\mathbf{i} \\ & = & R\;I_{0} \; e^{\;\big{j ( \omega t -\phi)}} \end{eqnarray*}

The voltage across $L$ is given by
\begin{eqnarray*} \mathbf{v_{L}} & = & j \omega L\;\mathbf{i} \\ & = & e^{\fra... ...{0} \; e^{\;\big{j ( \omega t + \frac{\big\pi}{\big{2}} -\phi)}} \end{eqnarray*}

The voltage across $C$ is given by
\begin{eqnarray*} \mathbf{v_{L}} & = & \frac{\big{1}}{\big{ j \omega C}} \;\math... ... \; e^{\;\big{j ( \omega t - \frac{\big{\pi}}{\big{2}} -\phi)}} \end{eqnarray*}

The corresponding real voltages are given by
\begin{eqnarray*} v_{R} (t) & = & R\;I_{0} \; \sin ( \omega t -\phi ) \\ v_{L} ... ... \;I_{0} \; \sin ( \omega t - \frac{\big{\pi}}{\big{2}} -\phi ) \end{eqnarray*}

From the above equations it is clear that the $v_{R}$ is `in-phase' with the current, $v_{L}$ leads the current as also $v_{R}$ by $\frac{\big{\pi}}{\big{2}}$ radians or $90^{\circ}$, and, $v_{C}$ lags the current as also $v_{R}$ by $\frac{\big{\pi}}{\big{2}}$ radians or $90^{\circ}$. $v_{L}$, therefore, leads $v_{C}$ by $\frac{\big{\pi}}{\big{2}} - ( - \frac{\big{\pi}}{\big{2}} ) = \pi $ radians or $180^{\circ}$ or in other words $v_{L}$ and $v_{C}$ are `out-of-phase'.

It is interesting to note the phase-relationships among the different voltages. At the resonant-frequency, the waveform for $v_{R}$ would have the maximum amplitude. This is expected, as at resonance, the current in the circuit, $I_{0}$, is maximum. Interestingly, at this frequency, $v_{L}$ and $v_{C}$ are of the same amplitude, viz., $\sqrt{\frac{\big{L}}{\big{C}}}\;I_{0}$. Since, they are also `out-of-phase', they mutually cancel each other leaving only $v_{R}$ which then equals the source-voltage $v_\big{in}$.

The ratio of the amplitude of the voltage across $L$ or $C$ (which are equal) at resonance to the amplitude of the source-voltage at resonance is called the voltage-magnification $(\gamma)$. The name suggests that the value of the ratio may be much greater than 1 and that is indeed the case.
\begin{eqnarray*} \gamma & = & \frac {\omega_{\big{0}} L I_{\big{0}}} {R I_{\big... ...} L} {R} \\ & = & \sqrt{\frac{\big{L}}{\big{C}}} \frac {1} {R} \end{eqnarray*}

Alternatively
\begin{eqnarray*} \gamma & = & \frac {\frac{\big{1}}{\big{ \omega_{\big{0}} C}} ... ...} C R}} \\ & = & \sqrt{\frac{\big{L}}{\big{C}}} \frac {1} {R} \end{eqnarray*}



One may draw the phase-versus-frequency plots pertaining to the phase angle $\big\phi$, the phase angle for $v_{R}$ (which is $ - \big\phi $), the phase angle for $v_{L}$ (which is $ \frac{\big\pi}{\big{2}} - \big\phi $) and the phase angle for $v_{C}$ (which is $ - \frac{\big\pi}{\big{2}} - \big\phi $).

From the definition of $\big\phi$ above and the phase-angle versus frequency plots, we may observe that when the linear-frequency of the source, $\big f$, is equal to zero, $\big\phi = -90^{\circ}$. When $\big f$ is equal to the resonant-frequency $\big f_{0}$, $\big\phi = 0$. And when $\big f$ is equal to $\infty$, $\;\big\phi = +90^{\circ}$.

We may also observe, that as long as $f < \big {f_{0}}$, or alternatively, $\omega L < \frac{1}{\big\omega \big{C}}$, $\big\phi$ is negative. On the other hand, if $f > \big {f_{0}}$, or alternatively, $\omega L > \frac{1}{\big\omega \big{C}}$, $\big\phi$ is positive.

Besides, at all frequencies, the phase angle for $v_{L}$ is $90^{\circ}$ plus the phase-angle for $v_{R}$ while the phase angle for $v_{C}$ is $90^{\circ}$ less than the phase-angle for $v_{R}$, just as prescribed by theory.

Earlier, we talked of resonance at $f = \big {f_{0}}$, when the amplitude of the current, viz., $I_{0}$, and therefore the amplitude of $v_{R}$, becomes maximum. Strictly speaking, we should refer to this resonance as current-resonance. We also have voltage-resonance at two other frequencies, $\big {f_{0}_{L}}$ and $\big {f_{0}_{C}}$, at which the respective amplitudes of $v_{L}$ and $v_{C}$ become maximum. These frequencies can be found by taking the derivatives of the respective amplitudes with respect to the frequency and equating them to zero. They turn out to be:
\begin{eqnarray*} \big {f_{0}_{L}} & = & \big{\frac{1}{\big{2\pi}}} \frac{\big{1}}{\sqrt{\big{LC}}} \; \sqrt{\frac{2L} {2L-R^{2}C} } \end{eqnarray*}

\begin{eqnarray*} \big {f_{0}_{C}} & = & \big{\frac{1}{\big{2\pi}}} \; \sqrt{\frac{1}{L C} -\frac{R^{2}}{2L^{2}} } \end{eqnarray*}



We may also draw the current resonance curve and voltage resonance curves (for $L$ and $C$). These resonance curves are the response of the respective amplitudes with frequency of the source. It is interesting to observe that $\big {f_{0}_{C}}$ lies to the left of $\big {f_{0}}$ and $\big {f_{0}_{L}}$ lies to the right of $\big {f_{0}}$, in agreement with their formulae. We may also plot the magnitude of impedance, $Z$, as a function of the frequency. As expected, $Z$ is minimum and $I_{0}$ is maximum at the current-resonance frequency, $\big {f_{0}}$.

If we follow the current-resonance curve, there are two frequencies, $\big {f_{1}}$ and $\big {f_{2}}$ ( $\big {f_{1}} < \big {f_{2}}$ ), symmetrically placed on either side of $\big {f_{0}}$, for which the current-amplitude equals $\frac{\big 1} {\big \sqrt{2}}$ times the maximum value of the current-amplitude $I_{0}$. The corresponding power at these two frequencies is therefore half of the power at $\big {f_{0}}$. $\big {f_{1}}$ and $\big {f_{2}}$ are therefore called 'half-power frequencies'. $\big {f_{2}} - \big {f_{1}}$ is called the bandwidth ($\Delta$) of the circuit. An expression for the bandwidth can be obtained as follows:
\begin{eqnarray*} Z \quad at \quad \big {f_{1}}, \big {f_{2}} & = & \frac {\big ... ...\big I_{0} \quad at \quad \big {f_{0}}} } \\ & = & \sqrt {2} R \end{eqnarray*}

We also have,
\begin{eqnarray*} Z \quad at \quad \big {f_{1}}, \big {f_{2}} & = & \sqrt{ R^{2} + \Big(\omega L - \frac{1}{\omega C} \Big)^{2} } \end{eqnarray*}

Equating the last two equations, we have,
\begin{eqnarray*} \sqrt {2} R & = & \sqrt{ R^{2} + \Big(\omega L - \frac{1}{\ome... ...^{2} \\ or R & = & \pm \Big(\omega L - \frac{1}{\omega C} \Big) \end{eqnarray*}

The $+$ sign corresponds to the upper half-power frequency $\big {f_{2}}$ or $\big {\omega_{2}}$, since at this frequency, $\omega L > \frac{1}{\big\omega \big{C}}$. So we may write
\begin{eqnarray*} R & = & + \Big(\omega_{2} L - \frac{1}{\omega_{2} C} \Big) \end{eqnarray*}

The $-$ sign corresponds to the lower half-power frequency $\big {f_{1}}$ or $\big {\omega_{1}}$, since at this frequency, $\omega L < \frac{1}{\big\omega \big{C}}$. So we may write
\begin{eqnarray*} R & = & - \Big(\omega_{1} L - \frac{1}{\omega_{1} C} \Big) \\ R & = & \Big(\frac{1}{\omega_{1} C} - \omega_{1} L \Big) \end{eqnarray*}

We may solve the last two equations for $\big {\omega_{1}}$ and $\big {\omega_{2}}$ and therefrom obtain $\big {f_{1}}$ and $\big {f_{2}}$.
\begin{eqnarray*} \big {f_{1}} & = & \frac{-R + \sqrt{R^{2}+4L/C}} {4\pi L } \\ \big {f_{2}} & = & \frac{+R + \sqrt{R^{2}+4L/C}} {4\pi L } \end{eqnarray*}

The bandwidth ($\Delta$) is therefore given by
\begin{eqnarray*} \big {f_{2}} - \big {f_{1}} & = & \frac{R}{2\pi L} \end{eqnarray*}

.The Quality (Q) factor of the circuit is defined as the ratio of the (current) resonant frequency to the bandwidth.
\begin{eqnarray*} Q & = & \frac{\big {f_{0}}}{\Delta} \end{eqnarray*}

The smaller is the value of $\Delta$, greater is the value of $Q$, and sharper is the (current) resonance. $Q$ is also called the selectivity of the circuit. The higher is the value of $Q$, the easier it is to select or accept the resonant frequency $\big {f_{0}}$ from nearby frequencies. It may be worth mentioning that another name of the series resonant circuit is acceptor circuit.

Phasors
All a.c voltage and currents can be pictorially represented by rotating vectors called phasors. The amplitude of the a.c. voltage or current equals the length of the phasor while its frequency equals the rate of rotation of the phasor in a counter-clockwise sense from a fixed reference direction which is conventionally taken as the x-axis.

How do we represent a phasor mathematically?
The complex notation of an a.c. quantity facilitates expressing a phasor mathematically. To see how, lets take the example of the a.c. voltage across $R$ which in complex form reads $\mathbf{v_{R}} = V_{R_{0}} \; e^{\;\big{j ( \omega t -\phi)}} $. The amplitude $V_{R_{0}}$ and the initial phase $ - \big\phi $ are the only quantities of importance since the angular frequency of the source, $\omega$, always remains constant. Hence, we may separate out $\omega$ as follows:
\begin{eqnarray*} \mathbf{v_{R}} & = & V_{R_{0}} \; e^{\;\big{j ( \omega t -\phi... ...ega t} \\ & = & \mbox{\boldmath$V_{R}$} \; e^{\big{j \omega t} \end{eqnarray*}

$\mbox{\boldmath$V_{R}$} = V_{R_{0}} \; e^{- j \big \phi}$ is the exponential form of the phasor corresponding to the complex time-dependent voltage across $R$. It can also be expressed in the rectangular form, viz., $ V_{R_{0}} \;(\cos\phi - j \sin\phi)$ or in polar form, viz., $V_{R_{0}} \;\angle \big\phi$. Pictorially, the phasor is denoted at time $t=0$ by an arrow of length proportional to $V_{R_{0}}$ making an angle $ - \big\phi $ with the horizontal reference line. As t increases, the phasor rotates in a counter-clockwise sense with the frequency $\omega$. The projection of the arrow onto the vertical `y' axis as it rotates in time gives us the instanteneous sinusoidal variation of the voltage, which turns out to be $ V_{R_{0}}\;\sin \omega t$.

Calculation of power in a series LCR circuit
. The instanteneous power $P\;(t)$ delivered to the impedance $Z$ is given by
\begin{eqnarray*} P\;(t) & = & v_{\big {in}}(t)\; i (t) \\ & = & V_{0} \sin \o... ... - V_{\big {r.m.s.}}I_{\big {r.m.s.}}\cos(2\omega t -\big\phi)) \end{eqnarray*}

$P\;(t)$ comprises of a cosinusoidal component $ - V_{\big {r.m.s.}}I_{\big {r.m.s.}}\cos(2\omega t -\big\phi))$ oscillating symmetrically about a constant value $ V_{\big {r.m.s.}}I_{\big {r.m.s.}}\cos\big\phi $. This constant value is therefore called the
Average power
. Let us denote it by $P_{\big {avg}}$. It may be noted that this value for the average-power could also have been arrived at by integrating $P\;(t)$ over a time-period and then dividing by the value of the time-period.

We may express $\cos\big\phi$ in terms of the resistance,$R$, and the magnitude of the impedance,$Z$, as has been shown before, to be able to write
\begin{eqnarray*} P_{\big {avg}} & = & V_{\big {r.m.s.}}I_{\big {r.m.s.}}\cos\bi... ...xtrm Since} \quad V_{\big {r.m.s.}} = I_{\big {r.m.s.}}\;Z \par \end{eqnarray*}

$P_{\big {avg}}$ is maximum when $\big\phi$ equals $0^{\circ}$. This happens when $Z = R$, the circuit being effectively resistive. On the other hand $P_{\big {avg}}$ acquires the minimum value of zero when $\big\phi$ equals $\pm 90^{\circ}$, the circuit being effectively reactive in this case.

The term, $\cos\big\phi$ which equals $\frac{P_{\big {avg}}} {V_{\big {r.m.s.}}I_{\big {r.m.s.}}}$ is called the power-factor. Since $P_{\big {avg}}$ is always $+ve$ and so are $V_{\big {r.m.s.}}$ and $I_{\big {r.m.s.}}$, the power-factor varies only between $0$ and $1$.

One may draw and compare the waveforms for $P\;(t)$, $P_{\big {avg}}$, $v_{\big{in}(t)}$ and $i(t)$. $P\;(t)$ is observed to vary sinusoidally about $P_{\big {avg}}$ with a frequency, twice that of the frequency of the source, as predicted by theory. We may note that since $P\;(t) = v_{\big {in}}(t)\; i (t)$ and $v_{\big {in}}(t) = 0 $ at $t=0$, so $P\;(t) = 0$ at $t=0$, as can be seen from the plots.

It is of interest to gain insight into the physical interpretation of the waveforms. We must remember that we are concerned with the steady state and not the transient state of the circuit. In this steady state, the inductor and capacitor receive energy from the a.c. voltage-source, store it respectively in their magnetic and electric fields during a portion of each half-cycle of the voltage-source, and return all of this energy to the source during the remaining portion of each half-cycle. On the other hand, the energy received by the resistor from the a.c. voltage-source is never returned. The whole of it is dissipated in the resistor as heat or some other form.

The net flow of energy to a passive circuit as the one considered here, in the steady state and in one cycle, is therefore always positive or zero.

As long as the $P\;(t)$ curve is positive, it implies that energy is flowing from the source to the passive elements $L$, $C$ and $R$. Whereas, when the $P\;(t)$ curve is negative, it implies that energy is being returned to the source from the passive elements $L$ and $C$. A careful look at the waveforms reveal that this occurs during the time the curves for $v_{\big {in}} (t)$ and $i(t)$ have opposite polarities.

At the (current) resonant-frequency, $\big {f_{0}}$, the negative portion of the $P\;(t)$ - curve disappears. This signifies that all the energy is dissipated in the resistor and none of it is returned. This is to be expected, since at this frequency, the circuit is essentially resistive, as discussed before.

On the other hand, it may also be observed that if the frequency of the source is made such that the phase-difference between the source-voltage and current waveforms becomes $90^{\circ}$, then the circuit essentially becomes reactive and then the positive and negative excursions of the $P\;(t)$-curve become equal and $P_{\big {avg}}$ becomes zero.

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