Problem 90 - Fundamentals of Physics

Question

What capacitance would you connect across a \(1.30 \mathrm{mH}\) inductor to make the resulting oscillator resonate at \(3.50 \mathrm{kHz} ?\)

Step-by-Step Solution

Answer

The capacitance should be approximately 1.591 nF.

Step 1: Write Down the Formula for Resonant Frequency

The resonant frequency \( f \) of an LC oscillator is given by the formula: \[ f = \frac{1}{2\pi\sqrt{LC}} \] where \( L \) is the inductance, and \( C \) is the capacitance.

Step 2: Rearrange the Formula to Solve for Capacitance

Rearrange the formula to solve for \( C \) in terms of \( f \) and \( L \): \[ C = \frac{1}{(2\pi f)^2 L} \]

Step 3: Substitute Known Values into the Formula

Substitute the given values into the formula: \( L = 1.30 \, \text{mH} = 1.30 \times 10^{-3} \, \text{H} \) and \( f = 3.50 \, \text{kHz} = 3.50 \times 10^3 \, \text{Hz} \).

Step 4: Perform the Calculation

Calculate the capacitance \( C \) using the formula: \[ C = \frac{1}{(2\pi \times 3.50 \times 10^3)^2 \times 1.30 \times 10^{-3}} \]First calculate \((2\pi \times 3.50 \times 10^3)^2\): \( 2\pi \approx 6.2832 \), so\[ (6.2832 \times 3.50 \times 10^3)^2 = (21980)^2 = 4.83604 \times 10^8 \]Now calculate the capacitance:\[ C = \frac{1}{4.83604 \times 10^8 \times 1.30 \times 10^{-3}} \]\[ C \approx \frac{1}{6.287052 \times 10^5} \approx 1.591 \times 10^{-9} \, \text{F} \] which is equivalent to \( 1.591 \, \text{nF} \).

Key Concepts

Resonant Frequency

In an LC oscillator, the resonant frequency is a critical parameter. It's the specific frequency at which the system naturally oscillates with the highest amplitude. This phenomenon occurs when the energy exchange between the inductor (L) and the capacitor (C) is perfectly balanced. At this frequency, the energy stored in the magnetic field of the inductor is equal to the energy stored in the electric field of the capacitor.

The resonant frequency (\( f \)) is determined by the values of inductance (\( L \)) and capacitance (\( C \)) according to the formula:

• \[ f = \frac{1}{2\pi\sqrt{LC}} \]

In this formula:

• \( f \): Resonant frequency in hertz (Hz)
• \( L \): Inductance in henrys (H)
• \( C \): Capacitance in farads (F)

Understanding the resonant frequency is essential because any deviation from this frequency can significantly reduce the oscillator's performance. This is why precise calculation and tuning of an LC circuit are crucial to maintain stability and efficiency.

Capacitance Calculation

Calculating capacitance involves determining the amount of electric charge a capacitor can hold at a specific voltage. In an LC oscillator, knowing the resonant frequency and the inductance allows us to compute the necessary capacitance to achieve desired oscillations.

To find the capacitance for resonance at a specific frequency, we rearrange the resonant frequency formula to solve for \( C \):

• \[ C = \frac{1}{(2\pi f)^2 L} \]

Given values for the calculation include:

• \( L = 1.30 \, \text{mH} \) or \( 1.30 \times 10^{-3} \, \text{H} \)
• \( f = 3.50 \, \text{kHz} \) or \( 3.50 \times 10^3 \, \text{Hz} \)

By substituting these values into the equation, you can find the required capacitance. This calculation ensures the LC circuit resonates at the desired frequency, maximizing signal output and stability.

Inductance

Inductance (\( L \)) is a fundamental property of an inductor. It measures the inductor's ability to store energy in its magnetic field when electrical current flows through it. This property is vital in LC oscillators as it participates in setting the resonant frequency.

Inductance is typically expressed in henrys (H), and its value affects how quickly an oscillating circuit can change the direction of current. In our example, the inductance of \( 1.30 \, ext{mH} \) plays a key role in determining the capacitance needed to achieve a resonant frequency of \( 3.50 \, ext{kHz} \).

• Smaller inductance results in a higher resonant frequency.
• Larger inductance results in a lower resonant frequency.

For effective oscillations, the correct inductance must pair with the calculated capacitance. This ensures the system operates at peak efficiency, producing stable and consistent oscillations.