# Resonant Series RLC Circuit

Resonant RLC series circuits and formulas of the resonant frequency , the cutoff frequencies are developed, the bandwidth and the quality factor are defined and all are used in examples with detailed solutions.    
In what follows, the upper case letter $$I$$ is the complex (polar) form of the real current $$i$$ and the upper case letter $$V_i$$ is the complex (polar) form of the real voltage $$v_i$$.
A resonant series RLC circuit calculator may be used to verify the calculations of the examples below and also for more practice and investigations of these circuits.

## A - Resonant series RLC circuit

Consider the series RLC circuit shown below.

For a circuit supplied by a voltage source of frequency $$f$$, the total impedance $$Z$$ of the series RLC circuitis given by:
$Z = R + j \left(\omega L - \dfrac{1}{\omega C} \right)$
The relationship between the current $$I$$ and the voltage $$V_i$$ is given by
$I = \dfrac{V_i}{Z}$
where $$V_i$$ and $$I$$ are the complex form of the voltage $$v_i$$ and the current $$i$$ respectively.
Using the definition of the magnitude of a complex number, the magnitude $$|Z|$$ is given by
$$|Z| = \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2}$$
If $$V_0$$ is the peak value of the voltage source $$v_i = V_0 \cos (\omega t)$$, then the peak value $$I_0$$ of $$I$$ is given by
$$I_0 = \dfrac{V_0}{ |Z| } = \dfrac{V_0}{ \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} }$$
The frequency of resonance is defined as the frequency for which $$I_0$$ is maximum or when the magnitude of $$Z$$ is minimum.
Since the resistance $$R$$ is independent of the frequency, the minimum value of $$|Z|$$ occurs at $$\omega = \omega_r$$ such that
$$\left(\omega_r L - \dfrac{1}{\omega_r C} \right) = 0$$
Solve the above for $$\omega_r$$ to obtain the resonant frequency
$\omega_r = \dfrac{1}{\sqrt {L C}} \quad \quad (I)$
At the resonant frequency $$\omega = \omega_r$$, we have:
1) $$Z = R$$
For $$V_0$$, the peak value of the voltage source $$v_i$$, the peak value $$I_0$$ of $$I$$ is given by
2) $$I_0 = \dfrac{V_0}{R}$$
Let $$X_L = \omega L$$ and $$X_C = \dfrac{1}{\omega C}$$
3) $$X_L = X_C$$

Example 1
Let $$R=300 \; \Omega$$, $$L = 100 \; mH$$ and $$C = 100 \mu F$$ in the series RLC circuit above.
a) Find the resonant frequency $$\omega_r$$
b) Graph $$|Z|$$, $$X_L = \omega L$$, $$X_C = \dfrac{1}{\omega C}$$ and $$I_0$$ as a function of the frequency $$\omega$$ and discuss the graphs obtained.

Solution to Example 1
a)
The resonant frequency $$\omega_r$$ is given by
$$\omega_r = \dfrac{1}{\sqrt {L C}} = \dfrac{1}{\sqrt{100\times10^{-3} \times 100 \times 10^{-6}}} \approx 316.23$$
b)
Below are shown the graphs of $$|Z|$$, $$X_L$$ and $$X_C$$.
From the graphs, $$|Z|$$ has a minimum value equal to $$R = 300 \; \Omega$$ (point A)
The graphs of $$X_L$$ and $$X_C$$ intersect (point B) and are therefore $$X_L = X_C$$ or $$\left(\omega_r L - \dfrac{1}{\omega_r C} \right) = 0$$ .

In the graph below is shown the current $$I_0$$ and is maximum at the resonant frequency $$\omega_r \approx 316.23$$ (after rounding to 2 decimal places)

## B - Average Power in a Resonant Circuit

The average power $$P_a$$ delivered to the series RLC circuit is given by: $\displaystyle \quad \quad P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta \quad \quad (II)$ where $$\theta$$ is the argument of the impedance $$Z = R + j \left(\omega L - \dfrac{1}{\omega C} \right)$$ and is given by
$$\theta = \arctan \left( \dfrac{ \omega L - \dfrac{1}{\omega C} }{R} \right)$$
Using inverse trigonometric functions properties, we have
$$\tan \theta = \left( \dfrac{ \omega L - \dfrac{1}{\omega C} }{R} \right)$$
$$\theta$$ may be assumed to be an acute angle of a right triangle as shown below. (Use the definition of the tangent of an angle in a right triangle and see that you can get $$\tan \theta$$ as defined above.

We now use the same triangle and calculate the power factor $$\cos \theta$$
The hypotenuse of the triangle is calculated as follows
$$AC = \sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}$$
$$\cos \theta = \dfrac{AB}{AC} = \dfrac{R}{\sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}}$$
Substitute $$\cos \theta$$ and $$|Z|$$ in the formula (II) given above and express the power $$P_a$$ as
$$\displaystyle \quad \quad P_a = \dfrac{V_0^2}{2 \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} }\dfrac{R}{\sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}}$$
Simplify
$\displaystyle \quad \quad P_a = \dfrac{V_0^2 R}{2 \left({R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} \right) } \quad \quad (III)$
At the resonant frequency $$\omega_r = \dfrac{1}{\sqrt {LC}}$$, we have $$\left(\omega_r L - \dfrac{1}{\omega_r C} \right) = 0$$ and therefore the power is maximum and equal to $P_{a max} = \dfrac{V_0^2}{2 \; R} \quad \quad (IV)$

## C - Cutoff Frequencies of a Resonant Circuit and The Quality Factor

We now define the cutoff frequencies as the frequencies $$\omega_c$$ at which the power $$P_a(\omega)$$ in (III) is half the maximum power $$P_{a max}$$ in (IV).
Hence we need to solve the equation
$$P_a (\omega_c ) = \dfrac{1}{2} \left(\dfrac{V_0^2}{2 \; R} \right)$$

$$\dfrac{V_0^2 R}{2 \left({R^2 + \left(\omega_c L - \dfrac{1}{\omega_c C} \right)^2} \right) } = \dfrac{1}{2} \dfrac{V_0^2}{2 \; R}$$
Simplify to
$$\dfrac{ R}{2 \left({R^2 + \left(\omega_c L - \dfrac{1}{\omega_c C} \right)^2} \right) } = \dfrac{1}{4 R}$$
Cross multiply, simply and rewrite the above equation as
$$(\omega_c L - \dfrac {1}{\omega_c C } ) = R^2$$
Solve by extracting the square root to obtain two equations
$$\omega_c L - \dfrac {1}{\omega_c C} = \pm R$$
Multiply all terms by $$\omega_c C$$ and simplify
$$\omega_c^2 L C - 1 = \pm \omega_c R C$$
Rewrite as quadratic equations in standard forms
$$\omega_c^2 L C \pm \omega_c R C - 1 = 0$$
Solve the first quadratic equation $$\quad \omega_c^2 L C + \omega_c R C - 1 = 0$$
to obtain two solutions
$$\omega_{c1} = \dfrac {- R C \pm \sqrt{ (R C)^2 + 4 L C }}{ 2 L C }$$

Solve the second quadratic equation $$\quad \omega_c^2 L C - \omega_c R C - 1=0$$
to obtain two solutions
$$\omega_{c2} = \dfrac {R C \pm \sqrt{ (R C)^2 + 4 L C}}{ 2 L C }$$
We have a total of 4 solutions. Note that the quantity $$\sqrt{ (R C)^2 + 4 L C }$$ is larger that $$RC$$ and therefore only two solutions are valid since the cutoff frequency is a positive quantity.
The cutoff frequencies $$\omega_{c1}$$ and $$\omega_{c2}$$ are the two solutions given
$$\omega_{c1} = \dfrac {- R C + \sqrt{ (R C)^2 + 4 L C }}{ 2 L C }$$

$$\omega_{c2} = \dfrac {R C + \sqrt{ (R C)^2 + 4 L C}}{ 2 L C }$$
We already found the the resonant frequency $$\omega_r = \dfrac{1}{\sqrt{LC}}$$
Use simple algebra to rewrite $$\omega_{c1}$$ and $$\omega_{c1}$$ in terms of $$\omega_r$$
$\omega_{c1} = - \dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \quad \quad (V)$
$\omega_{c2} = \dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \quad \quad (VI)$

Note that
$\omega_{c1} \times \omega_{c2} = \omega_r^2 \quad \quad (VII)$

The bandwidth of the resonant circuit is defined by: $$\Delta \omega = \omega_{c2} - \omega_{c1}$$

The quality factor $$Q$$ is defined by
$$Q = \dfrac{\omega_r}{\Delta \omega}$$
Substitute
$$Q = \dfrac {\omega_r} { \left(\dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} - \left(-\dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \right) \right)}$$
Simplify
$Q = \omega_r \dfrac{L}{R} \quad \quad (VIII)$

Example 2
a) Find the resonant frequency, the cutoff frequencies and the quality factor $$Q$$ for a series RLC circuit with $$R=300 \; \Omega$$, $$L = 100 \; mH$$ and $$C = 100 \; \mu F$$.
b) Graph the average power $$P_a$$ against the common logarithm of the angular frequency $$\omega$$ and check graphically the resonant and cutoff frequencies found analytically in part a) above.

Solution to Example 2
b)
Given
$$L = 100 \; mH = 100 \times 10^{-3} \; H = 0.1 \; H$$
$$C = 100 \; \mu F = 100 \times 10^{-6} \; F = 0.0001 \; F$$
Resonant frequency: $$\omega_r = \dfrac{1}{\sqrt{L C}} = \dfrac{1}{\sqrt{0.00001}} = 316.22776 \approx 316.23$$
$$\omega_r^2 = \dfrac{1}{L C} = \dfrac{1}{0.00001} = 100000$$
$$\dfrac{R}{L} = \dfrac{300}{0.1} = 3000$$
Using the formulas (V) and (VI) above, we obtain the two cutoff frequencies and the quality factor as follows:

$$\omega_{c1} = - 1500 + \sqrt{ 1500^2 + 100000} \approx 32.97$$ rad/s
$$\omega_{c2} = 1500 + \sqrt{ 1500^2 + 100000} \approx 3032.97$$ rad/s
The quality factor $$Q$$ is given by
$$Q = \dfrac{316.23}{3032.97 - 32.97} \approx 0.1054$$
b)
Below is shown the graph of $$P_a$$ against $$Log_{10} (\omega)$$ so that the plot shows some useful symmetries.
From the graph, the maximum power is equal to $$0.001666$$ Watts and occurs at $$Log_{}(\omega_r) = 2.5$$.
Hence $$\omega_r = 10^{2.5} \approx 316.22776$$ rad/s
The half maximum power line (in red) is given by $$y = \dfrac{1}{2}$$ of the maximum of $$P_a = \dfrac{}{} = 0.00083$$ and intersect the graph at the cutoff frequencies such that
$$Log_{10}(\omega_{c_1}) = 1.51481$$ which gives $$\omega_{c_1} = 10^{1.51481} = 32.71975$$ rad/s
$$Log_{10}(\omega_{c_2}) = 3.48356$$ which gives $$\omega_{c_2} = 10^{3.48356} = 3044.80861$$ rad/s
Hence the graph gives values of the resonant and cutoff frequencies that are close to those found analytically in part a).

## D - More Examples with Detailed Solutions

Example 3
A series RLC resonant circuit is to designed so that its has the frequencies $$f_{c_1} = 650$$ Hertz and $$f_{c_2} = 950$$ Hertz as lower and upper cutoff frequencies.
a) Calculate the capacitance of the capacitor $$C$$ and the inductance of the inductor $$L$$ if the resistance of the resistor $$R$$ is equal to $$30 \Omega$$.
b) What is the quality factor of the circuit?

Solution to Example 3
a)
Calculate the angular frequencies.
$$\omega_{c_1} = 2 \pi f_{c_1} = 1300 \pi$$ rad/s
$$\omega_{c_2} = 2 \pi f_{c_1} = 1900 \pi$$ rad/s
Use the formula (VII) developed above
$$\omega_{c_1} \times \omega_{c_2} = \omega_{r}^2$$ to calculate he resonant frequency $$\omega_{r}$$ of the circuit.
$$\omega_{r} = \sqrt {\omega_{c_1} \times \omega_{c_2}} = \sqrt {1300 \pi \times 1900 \pi } = 100 \sqrt{247} \pi = 4937.400$$ rad/s
$$\omega_{c_2} - \omega_{c_1} = \dfrac{R}{L}$$
Hence
$$L = \dfrac{R } {\omega_{c_2} - \omega_{c_1}} = \dfrac{30} {1900 \pi - 1300 \pi } = 0.01591$$ H
$$\omega_{r} = \dfrac{1}{\sqrt {L C} }$$
Hence
$$C = \dfrac{1}{\omega_{r}^2 L} = \dfrac{1}{(100 \sqrt{247} \pi)^2 \times 0.01591} = 2.5783 \times 10^{-6}$$ F

b)
The quality factor is given by
$$Q = \dfrac{\omega_{r}}{\omega_{c_2} - \omega_{c_1}} = \dfrac{100 \sqrt{247} \pi}{1900 \pi - 1300 \pi } = 2.62$$