Problem 21 - Physics

Question

A circuit consists of an \(85-\Omega\) resistor in series with a \(4.0-\mu F\) capacitor, the two being connected between the terminals of an ac generator. The voltage of the generator is fixed. At what frequency is the current in the circuit one-half the value that exists when the frequency is very large?

Step-by-Step Solution

Answer

The frequency is approximately 548 Hz.

Step 1: Understand the Concept

The problem involves an R-C circuit connected to an AC generator. At high frequencies, the reactance of the capacitor becomes negligible, implying the current is maximal and determined by the resistor alone. We aim to find the frequency where the current is half this maximum value.

Step 2: Recall Capacitive Reactance Formula

The capacitive reactance, \( X_c \), depends on the frequency \( f \) and is given by the formula \( X_c = \frac{1}{2\pi f C} \), where \( C \) is the capacitance and \( f \) is the frequency.

Step 3: Express Impedance in Circuit

The total impedance \( Z \) of an L-C circuit is \( Z = \sqrt{R^2 + X_c^2} \) because the resistor and capacitor are in series. For this R-C circuit, \( Z = \sqrt{R^2 + \left(\frac{1}{2\pi f C} \right)^2} \).

Step 4: Determine Initial High-Frequency Current

At very high frequencies, \( X_c \approx 0 \), so \( Z \approx R = 85 \, \Omega \). The maximum current \( I_{max} \) is \( \frac{V}{R} \), where \( V \) is the voltage from the generator.

Step 5: Set Current to Half of Maximum

We want the current to be half of \( I_{max} \), meaning the impedance \( Z \) must double since \( I = \frac{V}{Z} \). So, set \( \sqrt{R^2 + \left(\frac{1}{2\pi f C}\right)^2} = 2R \).

Step 6: Simplify the Equation

Squaring both sides gives \( R^2 + \left(\frac{1}{2\pi f C}\right)^2 = 4R^2 \). Simplify to find \( \left(\frac{1}{2\pi f C}\right)^2 = 3R^2 \).

Step 7: Solve for Frequency

Isolate \( f \) using \( \frac{1}{2\pi f C} = \sqrt{3}R \). Rearrange to obtain \( f = \frac{1}{2\pi \sqrt{3} R C}\). After substituting \( R = 85 \, \Omega \) and \( C = 4.0 \, \mu F = 4.0 \times 10^{-6} \, F \), calculate \( f \approx \frac{1}{2\pi \times 85 \times \sqrt{3} \times 4.0 \times 10^{-6}} \).

Step 8: Calculate the Frequency

Perform the calculation: \( f \approx 548 \, \text{Hz} \).

Key Concepts

RC Circuit

An RC circuit is a simple combination of a resistor (R) and a capacitor (C) in series or parallel. In our case, they are in series and connected to an alternating current (AC) generator. This setup is quite common in electronic devices where you need to filter or smooth out signals.
The resistor limits the flow of current, providing a fixed opposition called resistance, measured in ohms (\(\Omega\)). Capacitors, on the other hand, store and release energy, affecting how current flows over time. Together, they create a dynamic system that depends on the frequency of the input signal.

• Resistor: Provides constant resistance irrespective of frequency.
• Capacitor: Reactance changes with frequency, influencing the total impedance.

Understanding the basic behavior of an RC circuit is essential, especially how it changes with frequency, since this directly impacts electrical signal processing.

Impedance

In AC circuits, impedance is crucial because it represents the total opposition a circuit presents to the flow of alternating current. Impedance, denoted as \(Z\), is a combination of resistance from resistors and reactance from capacitors and inductors.
In an RC circuit:

• Resistance (R) is easy to compute as it's a fixed value.
• Reactance (X) varies with frequency, especially because of capacitive elements.

The total impedance \(Z\) in an RC circuit is calculated as \(Z = \sqrt{R^2 + X_c^2}\), where \(X_c\) is the capacitive reactance. Impedance not only determines how much current flows through the circuit but also introduces phase shifts between voltage and current, which can be critical in AC applications.

Capacitive Reactance

Capacitive reactance is a measure of how much a capacitor resists the change of voltage, often denoted by \(X_c\). It's closely tied to frequency (\(f\)) and capacitance (\(C\)), calculated as \(X_c = \frac{1}{2\pi f C}\).
This formula tells us that:

• At high frequencies, \(X_c\) is low, making the capacitor pass more AC current.
• At low frequencies, \(X_c\) is high, restricting AC current flow.

This behavior is crucial for frequency selectivity in circuits, allowing capacitors to act as filters, blocking certain frequencies while allowing others. Understanding \(X_c\) helps engineers design circuits that can handle the desired range of frequencies efficiently.

Frequency Analysis

Analyzing how an RC circuit behaves over different frequencies is known as frequency analysis. It involves understanding how the voltage and current waveforms change as frequency varies.
In the exercise, we calculated the frequency at which the current is half its maximum value, illustrating a key concept: the relationship between frequency and impedance. As frequency increases:

• Capacitive reactance decreases, allowing more current to flow.
• Overall impedance may decrease, impacting current magnitude.

At the calculated frequency of approximately 548 Hz, the capacitor's reactance increases total impedance, reducing current flow to half of its maximum. This kind of analysis aids in designing circuits for applications like audio filtering, signal processing, and more.