Problem 13 - Physics

Question

A 30.0 -mH inductor has a reactance of \(2.10 k \Omega\). (a) What is the frequency of the ac current that passes through the inductor? (b) What is the capacitance of a capacitor that has the same reactance at this frequency? The frequency is tripled, so that the reactances of the inductor and capacitor are no longer equal. What are the new reactances of (c) the inductor and (d) the capacitor?

Step-by-Step Solution

Answer

(a) 11.17 kHz; (b) 6.75 pF; (c) 6.30 kΩ; (d) 700 Ω.

Step 1: Use Reactance Formula for Inductor

The reactance for an inductor is given by the formula \( X_L = 2 \pi f L \), where \( X_L \) is the inductive reactance, \( f \) is the frequency, and \( L \) is the inductance. We need to find the frequency \( f \) when \( X_L = 2.10 \times 10^3 \Omega \) and \( L = 30.0 \times 10^{-3} \) H.

Step 2: Solve for Frequency

Re-arranging the formula \( X_L = 2 \pi f L \) to solve for \( f \), we get \( f = \frac{X_L}{2 \pi L} \). Substituting the known values, \( f = \frac{2.10 \times 10^3}{2 \pi \times 30.0 \times 10^{-3}} \). Calculate this to find \( f \).

Step 3: Calculate Frequency

Calculating \( f = \frac{2.10 \times 10^3}{2 \pi \times 30.0 \times 10^{-3}} \), we find that the frequency \( f \approx 11.17 \) kHz.

Step 4: Determine Capacitance for Same Reactance

The reactance of a capacitor is given by \( X_C = \frac{1}{2 \pi f C} \). Setting this equal to the inductive reactance \( X_L = 2.10 \) kΩ, we have \( 2.10 \times 10^3 = \frac{1}{2 \pi \times 11.17 \times 10^3 \times C} \). Solve this equation for \( C \).

Step 5: Solve for Capacitance

Re-arranging gives \( C = \frac{1}{2 \pi \times 11.17 \times 10^3 \times 2.10 \times 10^3} \). Calculate \( C \).

Step 6: Calculate Capacitance

Calculating \( C \approx \frac{1}{148,234,926} = 6.75 \times 10^{-12} \) F, or 6.75 pF.

Step 7: Determine New Reactance of Inductor with Tripled Frequency

If the frequency is tripled, then \( f = 3 \times 11.17 \times 10^3 \) Hz. Use \( X_L = 2 \pi f L \) to find the new reactance \( X_L \).

Step 8: Calculate New Reactance of Inductor

With the new frequency, \( f = 33.51 \times 10^3 \) Hz. Substituting into \( X_L = 2 \pi f L \), we have \( X_L = 2 \pi \times 33.51 \times 10^3 \times 30.0 \times 10^{-3} \). Calculate this value.

Step 9: Determine New Reactance of Capacitor with Tripled Frequency

At the new frequency, use \( X_C = \frac{1}{2 \pi f C} \). Substituting in \( f = 33.51 \times 10^3 \) and the calculated capacitance \( C = 6.75 \times 10^{-12} \) F, calculate the new reactance of the capacitor.

Step 10: Calculate New Reactance of Capacitor

With \( f = 33.51 \times 10^3 \) Hz and \( C = 6.75 \times 10^{-12} \) F, \( X_C = \frac{1}{2 \pi \times 33.51 \times 10^3 \times 6.75 \times 10^{-12}} \). Calculate to find the new reactance \( X_C \).

Key Concepts

Inductance

Inductance in an electrical circuit is the characteristic of a conductor by which a change in current flowing through it induces an electromotive force (emf) in both the conductor itself and neighboring conductors. Think of it as the circuit's "reluctance" to change the current flowing through it. This is particularly important in AC circuits, where current constantly changes direction. The inductor, often a coil of wire, responds by resisting these changes through inducing its own voltage.
The term "inductive reactance" refers to how much an inductor opposes the current flow at given AC frequencies. It's represented by a simple formula:

• \( X_L = 2 \pi f L \)

where,

• \( X_L \) is the inductive reactance, measured in ohms (Ω)
• \( f \) is the frequency of the alternating current in hertz (Hz)
• \( L \) is the inductance in henrys (H)

The higher the frequency, the greater the inductive reactance, which means the circuit resists more against the AC current. This is why inductors are used in circuits to filter high-frequency signals or block them altogether.

Capacitance

Capacitance is the ability of a component or circuit to collect and store energy in the form of an electrical charge. A capacitor typically consists of two conductive plates separated by an insulating material called the dielectric. In AC circuits, capacitors play a vital role due to their ability to store and release energy.
The opposition that a capacitor presents to the flow of AC current is known as capacitive reactance. This is inversely proportional to both frequency and capacitance, given by the formula:

• \( X_C = \frac{1}{2 \pi f C} \)

where,

• \( X_C \) represents capacitive reactance, in ohms (Ω)
• \( f \) is the frequency, the rate at which the AC signal alternates, measured in Hz
• \( C \) is the capacitance, measured in farads (F)

Higher frequencies result in lower capacitive reactance, which allows AC signals to pass more easily. Conversely, for low frequencies, the reactance increases, resisting the flow of current more. Capacitors are essential for filtering specific frequencies and are widely used in tuning applications, such as in radios.

AC Circuit Analysis

Analyzing AC circuits involves understanding how alternating current behaves differently compared to direct current. AC circuits often include components like resistors, inductors, and capacitors, each affecting the current and voltage in unique ways.
In AC circuit analysis, one crucial aspect is understanding the reactance and impedance of the circuit. Here's a simple explanation:

• **Reactance** encompasses two types: inductive (from inductors) and capacitive (from capacitors). It reflects the opposition these components provide to alternating currents due to their energy storage and release properties.
• **Impedance** is the overall opposition faced by AC in a circuit and is a combination of resistance and reactance. It's represented by a complex number, factoring both magnitude and phase difference between current and voltage.

A key tool for understanding these elements is the phasor diagram, which visually represents the relationships between different voltages and currents in an AC circuit. By converting time-varying signals into rotating vectors (phasors), engineers can simplify calculations and understand the phase relationships between components quickly.
For anyone studying AC circuits, mastering these concepts is critical for designing circuits that can efficiently handle the varying conditions of AC signals.