Problem 10 - College Physics

Question

A \(65 \Omega\) resistor, an \(8.0 \mu F\) capacitor, and a \(35 \mathrm{mH}\) inductor are connected in series in an ac circuit. Calculate the impedance for a source frequency of (a) \(300 \mathrm{~Hz}\) and (b) \(30.0 \mathrm{kHz}\)

Step-by-Step Solution

Answer

The impedance is approximately \(65.007 \Omega\) for (a) and \(6598.47 \Omega\) for (b).

Step 1: Calculate capacitive reactance

First, we determine the capacitive reactance \(X_c\) using the formula: \[ X_c = \frac{1}{2\pi f C} \]where \(f\) is the frequency and \(C\) is the capacitance.For (a) \(f = 300\, \text{Hz}\) and \(C = 8.0 \times 10^{-6}\, \text{F}\):\[ X_c = \frac{1}{2\pi \times 300 \times 8.0 \times 10^{-6}} \approx 66.32 \Omega \]For (b) \(f = 30,000\, \text{Hz}\):\[ X_c = \frac{1}{2\pi \times 30,000 \times 8.0 \times 10^{-6}} \approx 0.664 \Omega \]

Step 2: Calculate inductive reactance

Next, calculate the inductive reactance \(X_L\) using the formula:\[ X_L = 2\pi f L \]where \(L\) is the inductance.For (a) \(f = 300\, \text{Hz}\) and \(L = 35 \times 10^{-3}\, \text{H}\):\[ X_L = 2\pi \times 300 \times 35 \times 10^{-3} \approx 65.97 \Omega \]For (b) \(f = 30,000\, \text{Hz}\):\[ X_L = 2\pi \times 30,000 \times 35 \times 10^{-3} \approx 6598.47 \Omega \]

Step 3: Compute total impedance

The total impedance \(Z\) of the series circuit is found using:\[ Z = \sqrt{R^2 + (X_L - X_c)^2} \]Where \(R = 65\, \Omega\).For (a):\[ Z = \sqrt{65^2 + (65.97 - 66.32)^2} \approx 65.007 \Omega \]For (b):\[ Z = \sqrt{65^2 + (6598.47 - 0.664)^2} \approx 6598.47 \Omega \]

Key Concepts

Capacitive Reactance

Capacitive reactance is a measure of a capacitor's opposition to alternating current (AC). It is denoted by \( X_c \) and is dependent on two factors: the frequency of the AC source \( f \), and the capacitance \( C \) of the capacitor. The formula to calculate capacitive reactance is:

• \( X_c = \frac{1}{2\pi f C} \)

A higher frequency or a smaller capacitance leads to lower capacitive reactance, meaning the capacitor opposes the current less effectively. In the given problem, we illustrated this by calculating \( X_c \) for two different frequencies: 300 Hz and 30 kHz. At 300 Hz, the capacitive reactance was found to be approximately 66.32 Ω, indicating a higher opposition at the lower frequency. Conversely, at 30 kHz, \( X_c \) drastically drops to 0.664 Ω, demonstrating minimal opposition at a higher frequency.

Inductive Reactance

Inductive reactance describes how an inductor impedes the flow of AC, represented by \( X_L \). It is influenced by the frequency \( f \) of the AC source and the inductance \( L \) of the inductor. You can calculate it using the formula:

• \( X_L = 2\pi f L \)

As the frequency increases, the inductive reactance also increases, demonstrating a greater opposition to the current. In our solution, we saw this effect with the inductive reactance computed as approximately 65.97 Ω at 300 Hz and significantly higher at 6598.47 Ω at 30 kHz. This profound increase illustrates how inductors become more effective at impeding higher frequency currents.

Series Circuit Analysis

In a series AC circuit, components are connected end-to-end, meaning the current remains the same across each component, but the voltage can differ. Total impedance, \( Z \), or the overall opposition the circuit poses to the AC, combines resistive, capacitive, and inductive reactances. To find \( Z \) in a series circuit, use the expression:

• \( Z = \sqrt{R^2 + (X_L - X_c)^2} \)

The resistance \( R \) is simply an ohmic resistance, while \( X_L \) and \( X_c \) are the inductive and capacitive reactances. The difference \( X_L - X_c \) accounts for the frequency-based characteristics of inductors and capacitors. By analyzing the circuit's AC impedance, we obtained a total impedance of about 65.007 Ω at 300 Hz and a much higher 6598.47 Ω at 30 kHz, reflecting how frequency influences impedance.

Electrical Impedance

Electrical impedance is the effective resistance that an AC circuit presents to the flow of alternating current. It is a combination of resistive and reactive elements, represented by \( Z \). Impedance not only depends on the pure resistance, denoted by \( R \), but also on the effects of capacitive \( X_c \) and inductive \( X_L \) reactances. Calculating impedance in AC circuits is crucial for understanding how different components influence the behavior of the entire circuit.

• Formula: \( Z = \sqrt{R^2 + (X_L - X_c)^2} \)

Impedance is frequency-dependent, and variations in frequency significantly affect \( X_c \) and \( X_L \). As seen in the solution, although resistance remained constant, the substantial frequency change altered the reactances, subsequently affecting the total impedance.

Resonance Frequency

The resonance frequency in an AC circuit is the frequency at which the inductive reactance \( X_L \) exactly cancels out the capacitive reactance \( X_c \), causing them to equal zero. At this point, the circuit exhibits purely resistive impedance as reactive components are neutralized, resulting in maximum current flow. The formula to discover the resonance frequency \( f_0 \) is:

• \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)

At resonance, the circuit is optimally efficient. Even though the given problem dealt with non-resonant frequencies, understanding that resonance eliminates reactive impedances is vital for optimizing circuits in practical applications and settings, like filters and oscillators, where managing impedance is crucial.