Question

(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50-\(\mu\)F capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50-\(\mu\)F capacitor?

Short answer

(a) 169.65 ohms and 1696.5 ohms. (b) 1061.03 ohms and 106.1 ohms. (c) Frequency is 149.91 Hz.

Step-by-step solution

Understand Inductive Reactance

The reactance of an inductor (denoted as \(X_L\)) is calculated using the formula \(X_L = 2\pi fL\), where \(f\) is the frequency and \(L\) is the inductance.

Compute Inductor Reactance for 60 Hz

Substitute \(f = 60 \text{ Hz}\) and \(L = 0.450 \text{ H}\) into the formula: \(X_L = 2\pi \times 60 \times 0.450 = 169.65 \text{ ohms}\).

Compute Inductor Reactance for 600 Hz

Substitute \(f = 600 \text{ Hz}\) and \(L = 0.450 \text{ H}\) into the formula: \(X_L = 2\pi \times 600 \times 0.450 = 1696.5 \text{ ohms}\).

Understand Capacitive Reactance

The reactance of a capacitor (denoted as \(X_C\)) is calculated by the formula \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency and \(C\) is the capacitance.

Compute Capacitor Reactance for 60 Hz

Substitute \(f = 60 \text{ Hz}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the formula: \(X_C = \frac{1}{2\pi \times 60 \times 2.50 \times 10^{-6}} = 1061.03 \text{ ohms}\).

Compute Capacitor Reactance for 600 Hz

Substitute \(f = 600 \text{ Hz}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the formula: \(X_C = \frac{1}{2\pi \times 600 \times 2.50 \times 10^{-6}} = 106.1 \text{ ohms}\).

Equate Reactances for Inductor and Capacitor

Set \(X_L = X_C\) to find the frequency where the reactances are equal. So, \(2\pi fL = \frac{1}{2\pi fC}\). Rearranging gives \(f^2 = \frac{1}{(2\pi)^2 LC}\).

Solve for Frequency

Substitute \(L = 0.450 \text{ H}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the equation: \(f = \frac{1}{2\pi \sqrt{0.450 \times 2.50 \times 10^{-6}}} = 149.91 \text{ Hz}\).

Key concepts

Inductive Reactance

The concept of inductive reactance is crucial in understanding how inductors behave in AC circuits. Inductive reactance, denoted as \(X_L\), represents the opposition that an inductor provides to the change of current. It is calculated using the formula: \[X_L = 2\pi fL\] where \(f\) is the frequency in hertz (Hz) and \(L\) is the inductance in henrys (H).

• The reactance increases with frequency; higher frequencies cause greater opposition.
• This is why inductors are more effective at blocking high-frequency signals than low-frequency signals.

By substituting the given inductance and frequency values into the equation, one can easily calculate the reactance at any desired frequency, as seen in the original problem. This concept is a cornerstone of AC circuit analysis and helps to predict the inductive behavior in systems accurately.

Capacitive Reactance

Capacitive reactance, represented as \(X_C\), describes how a capacitor opposes changes in voltage in AC circuits. The formula for capacitive reactance is: \[X_C = \frac{1}{2\pi fC}\] where \(f\) is the frequency in hertz (Hz) and \(C\) is the capacitance in farads (F).

• Unlike inductors, capacitors decrease their reactance as frequency increases. High frequencies pass through capacitors more easily.
• This characteristic makes capacitors effective as filters in circuits that discriminate against low frequencies, a feature commonly used in electronic components like oscillators and filters.

Understanding capacitive reactance is key to predicting how capacitors will respond at varying frequencies, allowing for better circuit designs.

Frequency Calculation

Calculating the frequency at which inductive and capacitive reactances are equal is an interesting exercise. This frequency is commonly referred to as the resonant frequency in tank circuits. It can be determined by setting \(X_L = X_C\), which results in the formula: \[f^2 = \frac{1}{(2\pi)^2 LC}\] Solving for \(f\) gives: \[f = \frac{1}{2\pi \sqrt{LC}}\] By substituting the values of inductance \(L\) and capacitance \(C\) from the problem, you can find the specific frequency where both reactances match.

• This is a fundamental principle in designing resonant circuits, such as those used in radios and other communication equipment.
• Finding this frequency allows for the optimization of signal strengths in various applications.

Such frequency calculations are integral in ensuring efficient circuit operations and minimizing reactive losses.

Inductor and Capacitor Comparison

Inductors and capacitors are vital components in electronics, each serving distinct yet complementary roles. Understanding their differences aids in constructing effective AC circuits. Here’s a comparison:

• Inductors resist changes in current by storing energy in a magnetic field when current flows through them.
  • They are more effective at high frequencies.
  • Commonly used in tuning circuits, filtering applications, and transformers.
• Capacitors resist changes in voltage by storing energy in an electric field between plates.
  • They are more effective at low frequencies.
  • Typically used in power supply circuits, coupling/decoupling applications, and energy storage.
• Working Together:
  • Both can be combined to form LC circuits that resonate at particular frequencies, making them ideal for filtering signals and tuning applications.

The synergy between inductors and capacitors is key in designing efficient electronic devices, especially in alternating current (AC) applications.