Resonance is an important concept in oscillatory motion. The resonant frequency is the characteristic frequency of a body or a system that reaches the maximum degree of oscillation. In an electrical system, the resonant frequency is defined as the frequency at which the transfer function reaches its maximum value. Thus for a given input, the maximum output can be obtained. It has been proved that the resonance is obtained when the capacitive impedance and the inductive impedance values are equal. In this article, we will discuss the resonant frequency formula with examples. Let us begin an interesting topic!

**Resonant Frequency Formula**

**Concept of Resonance Frequency**

The resonant frequency is the frequency of a circuit under resonant. A Resonant circuit is also known as the LC circuit or tank circuit. This circuit contains an inductor and capacitor attached parallel to each other. The resonant circuits are used to create a particular frequency or to select a particular frequency from a complex circuit.

**The Formula for Resonant Frequency:**

So, the resonant frequency formula is:

\(f_{0}=\frac{1}{2\pi \sqrt{LC}} \)

Where \(f_0\) is the the resonant frequency is denoted as, the inductance is L and the capacitance is C

**Derivation:**

Let us consider a series connection of R, L and C. This series connection is excited by an AC source.

Let us first calculate the impedance Z of the circuit.

\(Z = R + j\omega L – \frac{j}{\omega C}\)

\(Z = R + j( \omega L – \frac{1}{\omega C})\)

With the condition of resonance, the circuit is purely resistive. This means the imaginary part of the impedance Z will be zero during resonance condition or at a resonant frequency. You should always keep this in your mind while calculating the resonant frequency for a given circuit.

This means,

\(\omega L – \frac{1}{\omega C} = 0\)

\(\omega L = \frac{1}{\omega C}\)

so,

\(\omega ^2 = \frac {1}{LC}\)

Also

\(\omega = \frac{1}{2\pi f}\)

i.e. f = \(\frac {1}{2\pi \omega} \)

Substituting the values , we get

\(f_0 = \frac {1}{2\pi}\sqrt (LC) \)

Therefore, Resonant Frequency f_0 for the Resonance Circuit, will be

\(f_{0}=\frac{1}{2\pi \sqrt{LC}} \)

## Solved Examples

Q.1: An electrical circuit is given. Determine the the resonant frequency of this circuit. It has inductance of 25 mH , and capacitance as 5Î¼F?

Answer:

The given parameters in the problem are:

\(L = 50 mH = 50\times 10^{-3} H \)

\(C = 5 \mu F = 5 \times 10^{-6} F \)

The formula for resonant frequency is:

\(f_{0}=\frac{1}{2\pi \sqrt{LC}}\)

Substituting the values in the above formula,

\(f_{0}=\frac{1}{2\pi \sqrt{50\times 10^{-3}Â \times 5 \times 10^{-6} }}\)

\(f_{0}=\frac{1}{2\pi \times 5 \times 10^{-4}} \)

\(f_{0}=\frac{1}{2\times 3.14 \times 5 \times 10^{-4}} \)

\(f_0\) = 318.47 Hz

Therefore resonance frequency will be 318.47 Hz

Q.2: Determine the resonant frequency of a circuit whose value of inductance is 40 mH and capacitance is \(8 \mu F\).

Solution:

L = 40 mH

L = \(40 \times 10^{-3} H\)

C = \(8Â Î¼F = 8 \times 10^{-6} F \)

To calculate the resonant frequency we will apply the formula as below:

\(f_{0}=\frac{1}{2\pi \sqrt{LC}}\)

Substituting the values in the above formula,

\(f_{0}=\frac{1}{2\times 3.14 \sqrt{40\times 10^{-3}Â \times 8 \times 10^{-6} }}\)

Thus after solving we get,

\(f_0 = 281.35 Hz\)

Therefore resonance frequency will be 281.35 Hz

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