Problem 31 - Fundamentals of Physics

Question

(a) At what frequency would a \(6.0 \mathrm{mH}\) inductor and a \(10 \mu \mathrm{F}\) capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same \(L\) and \(C\).

Step-by-Step Solution

Answer

Frequency is approximately 650 Hz and both reactances are about 24 ohms.

Step 1: Identify Reactance Formulas

The reactance of an inductor (\(X_L\)) is given by \(X_L = \omega L\), and the reactance of a capacitor (\(X_C\)) is given by \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency.

Step 2: Set Reactances Equal

To find the frequency where their reactances are equal, set \(\omega L = \frac{1}{\omega C}\). This equation will help find the angular frequency where the reactances are equal.

Step 3: Solve for Angular Frequency \(\omega\)

Rewriting the equation \(\omega L = \frac{1}{\omega C}\) gives \(\omega^2 = \frac{1}{LC}\). Solving for \(\omega\) gives \(\omega = \frac{1}{\sqrt{LC}}\).

Step 4: Convert Angular Frequency to Frequency

The frequency \(f\) is related to angular frequency by \(f = \frac{\omega}{2\pi}\). Substituting \(\omega = \frac{1}{\sqrt{LC}}\) gives \(f = \frac{1}{2\pi \sqrt{LC}}\).

Step 5: Calculate the Frequency

Substitute \(L = 6.0 \times 10^{-3} \mathrm{H}\) and \(C = 10 \times 10^{-6} \mathrm{F}\) into the equation: \(f = \frac{1}{2\pi \sqrt{6.0\times 10^{-3} \cdot 10\times 10^{-6}}}\), to calculate the frequency.

Step 6: Calculate Reactance

Using \(\omega = \frac{1}{\sqrt{LC}}\), substitute back into either reactance formula \(X_L = \omega L\) or \(X_C = \frac{1}{\omega C}\) to find the reactance value.

Step 7: Verify Natural Frequency

The calculated frequency from Step 5 is the resonant or natural frequency for the circuit, as it satisfies the equation for the angular frequency \(\omega = \frac{1}{\sqrt{LC}}\) used for resonant frequency in LC circuits.

Key Concepts

Reactance

In an LC circuit, reactance is a measure of how much the inductor and capacitor oppose the flow of alternating current, due to the elements' capacitance and inductance, respectively. Understanding reactance is vital when analyzing how these components behave in AC circuits.

For an inductor, reactance is defined by the formula:

• \( X_L = \omega L \)

where \( \omega \) is the angular frequency and \( L \) is the inductance. This indicates that the reactance increases with higher frequency, which makes it harder for the alternating current to pass.

For a capacitor, the formula is:

• \( X_C = \frac{1}{\omega C} \)

Here, \( C \) represents capacitance. The reactance decreases as frequency goes up, allowing for more current flow in the circuit. At a certain frequency, commonly known as the resonant frequency, both the inductor and capacitor will have exactly the same reactance, perfectly canceling their effects on the circuit. This concept is integral to understanding LC circuits and their applications.

Natural Frequency

Natural frequency is a fundamental concept in understanding how LC circuits oscillate. LC circuits naturally oscillate at a frequency determined by their inductance \( L \) and capacitance \( C \). This is the frequency at which the circuit naturally likes to "swing" without any external influence.

The formula for calculating the natural frequency \( f_0 \) is:

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

This is derived from the relationship between the angular frequency \( \omega \) and the natural frequency, where \( \omega = \frac{1}{\sqrt{LC}} \).

When the LC circuit is at its resonant or natural frequency, energy flows back and forth between the inductor and the capacitor. The result is minimal energy loss, making it ideal for many applications such as radio transmitters and filters where precision frequency is essential.

Angular Frequency

Angular frequency \( \omega \), measured in radians per second, is closely connected to the frequency of oscillation in LC circuits. It is a measure of how many "radians" of a wavefront pass a given point per second.

The relationship between angular frequency \( \omega \) and regular frequency \( f \) (which you might encounter more often) is expressed as:

• \( \omega = 2\pi f \)

Angular frequency is used to easily relate to the physical properties of the circuit, such as in the reactance formulas for inductors and capacitors. By knowing \( \omega \), you get insight into the circuit's behavior at different frequencies.

In the context of this particular exercise, the angular frequency that results in equal reactance of the inductor and capacitor is \( \omega = \frac{1}{\sqrt{LC}} \). This means at this particular \( \omega \), or equivalently at the resonant frequency, both reactances will balance out, allowing the circuit to oscillate freely and efficiently.