Resonant Series RLC Circuit

Table of Contents

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
\[ Q = \omega_r \dfrac{L}{R} \quad \quad (VIII) \]

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
\( Q = \dfrac{\omega_{r}}{\omega_{c_2} - \omega_{c_1}} = \dfrac{100 \sqrt{247} \pi}{1900 \pi - 1300 \pi } = 2.62\)

More References and Links

• Resonant Series RLC Circuit Calculator
• RLC series circuits
• Series RLC Circuit Impedance Calculator
• Parallel RLC Circuit Impedance Calculator
• RLC Series Current Graphing Calculator