Resonant Frequency Formula

Resonant Frequency Formula

Resonant frequency in electricity refers to the specific frequency at which an electrical circuit or system naturally oscillates with maximum amplitude. This phenomenon occurs when the inductive and capacitive reactances within a circuit are equal in magnitude but opposite in phase, effectively canceling each other out. As a result, the impedance of the circuit is minimized, and the circuit can oscillate at its maximum energy transfer rate. Resonant frequency is a crucial concept in various applications, including radio and television broadcasting, where it is used to tune receivers to specific frequencies. Learn more about resonant frequency, definition, formula and examples

What is Resonant Frequency?

Resonant frequency in electricity refers to the specific frequency at which an electrical circuit, typically involving inductive and capacitive components, naturally oscillates with maximum amplitude. This phenomenon occurs when the inductive reactance and capacitive reactance within the circuit are equal in magnitude but opposite in phase, causing them to cancel each other out. As a result, the impedance of the circuit is minimized, and the circuit can oscillate with maximum energy transfer and efficiency.

Resonant Frequency Formula

The formula for resonant frequency is typically applied to LC (inductor-capacitor) circuits, where it determines the frequency at which the circuit naturally oscillates with maximum amplitude. The resonant frequency (\( f_0 \)) is given by:

\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

Where,

\( f_0 \): Resonant frequency (in hertz, Hz)
\( L \): Inductance (in henries, H)
\( C \): Capacitance (in farads, F)

Derivation of the Resonant Frequency Formula

Inductive Reactance (\( X_L \)):
\[ X_L = 2\pi f L \]

Capacitive Reactance (\( X_C \)):
\[ X_C = \frac{1}{2\pi f C} \]

Resonance Condition:
At resonance, the inductive reactance and capacitive reactance are equal:
\[ X_L = X_C \]

Setting Reactances Equal:
\[ 2\pi f_0 L = \frac{1}{2\pi f_0 C} \]

Solving for \( f_0 \):
\[ (2\pi f_0 L)(2\pi f_0 C) = 1 \]
\[ 4\pi^2 f_0^2 LC = 1 \]
\[ f_0^2 = \frac{1}{4\pi^2 LC} \]
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

Applications of Resonant Frequency

The applications of resonant frequency are mentioned below:

• Radio Receivers: Tuned circuits, consisting of inductors and capacitors, are used in radio receivers to select specific frequencies from the broad spectrum of radio waves. By adjusting the circuit to resonate at the desired frequency, the radio can isolate and amplify the signal from a particular station.
• Television Tuners: Similar to radio receivers, TV tuners use resonant circuits to select and demodulate the desired television signals.
• LC Oscillators: Consist of inductors and capacitors and use their resonant frequency to generate periodic oscillating signals. They are fundamental in signal generation for communication systems.
• Antenna Design: Resonance is used to match the impedance of an antenna to the transmission line and the transmitter or receiver. This maximizes power transfer and minimizes signal reflection.
• Transmission Lines: In high-frequency transmission, matching the impedance of different circuit elements is crucial to minimize signal loss and reflection. Resonant circuits can help achieve this matching.

Solved Examples on Resonant Frequency Formula

Example 1: Consider an LC circuit with Inductance \( L = 10 \) mH (0.01 H) and Capacitance \( C = 100 \) μF (0.0001 F)

Solution:

To find the resonant frequency \( f_0 \):

\[ f_0 = \frac{1}{2\pi\sqrt{(0.01)(0.0001)}} \]
\[ f_0 = \frac{1}{2\pi\sqrt{0.000001}} \]
\[ f_0 = \frac{1}{2\pi \times 0.001} \]
\[ f_0 = \frac{1}{0.002\pi} \]
\[ f_0 \approx 159.15 \text{ Hz} \]

Thus, the circuit will resonate at approximately 159.15 Hz.

Example 2: Calculating Resonant Frequency

Given:
Inductance (\( L \)) = 2 millihenries (mH) = 0.002 H
Capacitance (\( C \)) = 0.5 microfarads (µF) = 0.5 × 10\(^{-6}\) F

Solution:

1. Convert the values into the standard units:
\[ L = 0.002 \, \text{H} \]
\[ C = 0.5 \times 10^{-6} \, \text{F} \]

2. Use the formula for resonant frequency:
\[ f_r = \frac{1}{2 \pi \sqrt{LC}} \]

3. Substitute the values into the formula:
\[ f_r = \frac{1}{2 \pi \sqrt{(0.002) (0.5 \times 10^{-6})}} \]

4. Calculate the value inside the square root:
\[ LC = 0.002 \times 0.5 \times 10^{-6} = 1 \times 10^{-9} \]

5. Take the square root:
\[ \sqrt{1 \times 10^{-9}} = 1 \times 10^{-4.5} = 1 \times 10^{-4.5} = 0.00003162 \]

6. Calculate the final frequency:
\[ f_r = \frac{1}{2 \pi \times 0.00003162} \]
\[ f_r = \frac{1}{0.000198} \]
\[ f_r \approx 5033 \, \text{Hz} \]

So, the resonant frequency is approximately 5033 Hz.

Example 3: Finding Inductance for a Desired Resonant Frequency

Given:
Desired resonant frequency (\( f_r \)) = 1 MHz = 1 × 10\(^{6}\) Hz
Capacitance (\( C \)) = 100 picofarads (pF) = 100 × 10\(^{-12}\) F

Solution:

1. Convert the frequency and capacitance into standard units:
\[ f_r = 1 \times 10^{6} \, \text{Hz} \]
\[ C = 100 \times 10^{-12} \, \text{F} \]

2. Rearrange the resonant frequency formula to solve for \( L \):
\[ L = \frac{1}{(2 \pi f_r)^2 C} \]

3. Substitute the known values into the formula:
\[ L = \frac{1}{(2 \pi \times 1 \times 10^{6})^2 \times 100 \times 10^{-12}} \]

4. Calculate the value inside the parentheses:
\[ 2 \pi \times 1 \times 10^{6} \approx 6.2832 \times 10^{6} \]

5. Square this value:
\[ (6.2832 \times 10^{6})^2 \approx 3.948 \times 10^{13} \]

6. Multiply by \( C \):
\[ 3.948 \times 10^{13} \times 100 \times 10^{-12} = 3.948 \times 10^1 = 39.48 \]

7. Take the reciprocal to find \( L \):
\[ L \approx \frac{1}{39.48} \approx 0.0253 \, \text{H} \]

So, the inductance required for a resonant frequency of 1 MHz with a 100 pF capacitor is approximately 0.0253 H, or 25.3 millihenries.