Problem 70 - Fundamentals of Physics

Question

A \(45.0 \mathrm{mH}\) inductor has a reactance of \(1.30 \mathrm{k} \Omega\). (a) What is its operating frequency? (b) What is the capacitance of a capacitor with the same reactance at that frequency? If the frequency is doubled, what is the new reactance of (c) the inductor and (d) the capacitor?

Step-by-Step Solution

Answer

(a) 4594 Hz, (b) 26.7 nF, (c) 2600 Ω, (d) 650 Ω.

Step 1: Calculate Operating Frequency

The reactance of an inductor can be calculated using the formula \(X_L = 2 \pi f L\), where \(X_L\) is the inductive reactance, \(f\) is the frequency, and \(L\) is the inductance. Rearrange this formula to solve for \(f\): \(f = \frac{X_L}{2 \pi L}\). Substituting the given values \(X_L = 1300 \ \Omega\) and \(L = 45.0 \ \mathrm{mH} = 45.0 \times 10^{-3} \ \, \mathrm{H}\), we have \(f = \frac{1300}{2 \pi \times 45.0 \times 10^{-3}}\). Calculating this gives \( f \approx 4594 \ \mathrm{Hz}\).

Step 2: Calculate Capacitance for Same Reactance

For a capacitor, the reactance is given by the formula \(X_C = \frac{1}{2 \pi f C}\). We know that \(X_C = X_L = 1300 \ \Omega\). Thus, \(C = \frac{1}{2 \pi f X_C}\). Using the frequency calculated in Step 1, \(f = 4594 \ \mathrm{Hz}\), substitute the values: \(C = \frac{1}{2 \pi \times 4594 \times 1300}\). This results in \(C \approx 26.7 \ \mathrm{nF}\).

Step 3: Double the Frequency for New Inductor Reactance

If the frequency is doubled, the new frequency \(f' = 2f = 2 \times 4594 \ \mathrm{Hz} = 9188 \ \mathrm{Hz}\). The new inductive reactance \(X_L' = 2 \pi f' L = 2 \pi \times 9188 \times 45.0 \times 10^{-3}\). Calculating gives \(X_L' \approx 2600 \ \Omega\).

Step 4: Double the Frequency for New Capacitor Reactance

Using the new frequency, the new capacitive reactance \(X_C' = \frac{1}{2 \pi f' C}\). Substituting \(f' = 9188 \ \mathrm{Hz}\) from Step 3 and \(C = 26.7 \ \mathrm{nF}\), we have \(X_C' = \frac{1}{2 \pi \times 9188 \times 26.7 \times 10^{-9}}\). This results in \(X_C' \approx 650 \ \Omega\).

Key Concepts

Inductive Reactance

In AC circuits, inductors exhibit a property known as inductive reactance, which impedes the flow of alternating current. It is measured in ohms and depends on two main factors: the frequency of the AC signal and the inductance of the coil. The relationship is given by the formula:\[ X_L = 2 \pi f L \]Here, \( X_L \) is the inductive reactance, \( f \) is the frequency in hertz, and \( L \) is the inductance in henries. This equation highlights the direct proportionality between the reactance and both frequency and inductance. This means that as the frequency increases, so does the reactance. Conversely, if the frequency decreases, the reactance also decreases.

• High frequencies lead to higher reactance, acting almost like an open circuit.
• At lower frequencies, the reactance lowers, facilitating current flow.

Understanding this concept is crucial, especially when analyzing circuits that operate at varying frequencies.

Capacitive Reactance

Capacitors, on the other hand, exhibit capacitive reactance, which can also limit the flow of alternating currents. However, unlike inductors, capacitive reactance is inversely proportional to both frequency and capacitance. The equation used to quantify it is:\[ X_C = \frac{1}{2 \pi f C} \]Where \( X_C \) is the capacitive reactance, \( f \) is the frequency, and \( C \) is the capacitance. This formula indicates that as the frequency increases, the capacitive reactance decreases. This is contrary to inductive reactance, reflecting the opposing impact capacitors have in AC circuits.

• At high frequencies, capacitive reactance is low, allowing more current through.
• Conversely, at low frequencies, a capacitor presents high reactance, acting as a barrier to current flow.

These inversely proportional characteristics are vital when comparing performance across components in frequency-dependent applications.

Frequency Calculation

Determining the frequency at which a circuit component, like an inductor or capacitor, operates is essential for analyzing its behavior in AC circuits. The frequency is the number of cycles per second of an AC waveform and significantly affects both inductive and capacitive reactance. Calculating frequency involves rearranging formulas relating to reactance:For inductors:\[ f = \frac{X_L}{2 \pi L} \]For capacitors:\[ f = \frac{1}{2 \pi X_C C} \]These formulas reveal how frequency interplay with reactance and component values. When you know any two of these parameters, you can solve for the third. Getting to grips with this concept helps in the design and analysis of circuits, ensuring optimal operation at desired frequencies. By understanding how to manipulate these equations, you can predict how different components will affect performance as the frequency changes.