Problem 6 - College Physics

Question

A 2.20\(\mu \mathrm{F}\) capacitor is connected across an ac source whose voltage amplitude is kept constant at \(60.0 \mathrm{V},\) but whose frequency can be varied. Find the current amplitude when the angular frequency is (a) \(100 \mathrm{rad} / \mathrm{s} ;\) (b) \(1000 \mathrm{rad} / \mathrm{s} ;\) (c) \(10,000 \mathrm{rad} / \mathrm{s}\) .

Step-by-Step Solution

Answer

Current amplitudes are 0.0132 A, 0.132 A, and 1.32 A for frequencies 100, 1000, and 10000 rad/s, respectively.

Step 1: Understand Capacitive Reactance

The capacitive reactance \(X_C\) is given by the formula \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the angular frequency and \(C\) is the capacitance. This formula represents the effective resistance offered by a capacitor to the flow of alternating current.

Step 2: Calculate Capacitive Reactance for Each Frequency

For \(\omega = 100 \text{ rad/s}\):\[ X_C = \frac{1}{100 \times 2.20 \times 10^{-6}} \approx 4545.45 \, \Omega \]For \(\omega = 1000 \text{ rad/s}\):\[ X_C = \frac{1}{1000 \times 2.20 \times 10^{-6}} \approx 454.55 \, \Omega \]For \(\omega = 10000 \text{ rad/s}\):\[ X_C = \frac{1}{10000 \times 2.20 \times 10^{-6}} \approx 45.45 \, \Omega \]

Step 3: Calculate Current Amplitude Using Ohm's Law

The current amplitude \(I\) is given by \(I = \frac{V}{X_C}\), where \(V\) is the voltage amplitude. Substitute the respective values of \(X_C\) from Step 2 to find the current for each frequency.For \(\omega = 100 \text{ rad/s}\):\[ I = \frac{60}{4545.45} \approx 0.0132 \, \text{A} \]For \(\omega = 1000 \text{ rad/s}\):\[ I = \frac{60}{454.55} \approx 0.132 \, \text{A} \]For \(\omega = 10000 \text{ rad/s}\):\[ I = \frac{60}{45.45} \approx 1.32 \, \text{A} \]

Key Concepts

Ohm's Law in AC Circuits

In AC circuits, Ohm's Law is expressed slightly differently than in DC circuits. In these high frequency environments, instead of simple resistance, you encounter impedance. Impedance can encompass resistance, inductance, and capacitance, which all react to AC differently. Specifically, when dealing with capacitive circuits, we talk about capacitive reactance (\[X_C\]).
Ohm's Law in an AC circuit is represented by the equation for current amplitude:\[I = \frac{V}{X_C}\], where:

• \(I\) is the current amplitude.
• \(V\) is the voltage amplitude.
• \(X_C\) is the capacitive reactance.

Capacitive reactance describes how much a capacitor resists the flow of AC at a certain frequency. As the frequency increases, the reactance decreases, allowing more current to flow. Therefore, understanding Ohm's Law in AC circuits allows us to predict how current amplitude will change in response to changing frequencies.

Angular Frequency

Angular frequency is a crucial concept in AC circuits and wave-related phenomena. It is a measure of how quickly the AC source alternates and is denoted by the symbol \(\omega\).
Mathematically, angular frequency is given by:\[ \omega = 2\pi f \], where:

• \(\omega\) is the angular frequency in radians per second.
• \(f\) is the frequency in cycles per second or hertz.

Angular frequency helps determine the reactance of capacitors and inductors. Specifically for capacitors, the capacitive reactance is calculated as:\[ X_C = \frac{1}{\omega C} \] Understanding angular frequency is vital because it helps explain how different components in AC circuits respond to changes in frequency, impacting the overall behavior of the circuit.

Current Amplitude Calculation

To find the current amplitude in an AC circuit with a capacitor, you must first calculate the capacitive reactance using the given angular frequency and capacitance. Once you have \(X_C\), you can proceed with the calculation of current using Ohm's Law for AC circuits.
Here's the step-by-step approach:

• Calculate the capacitive reactance: \[ X_C = \frac{1}{\omega C} \]
• Substitute \(X_C\) and voltage amplitude \(V\) into the formula: \[ I = \frac{V}{X_C} \]

By following these steps for different angular frequencies, such as \(100\, \text{rad/s}\), \(1000\, \text{rad/s}\), and \(10000\, \text{rad/s}\), you will notice that as the frequency increases, the current amplitude increases as well.
This results from the decreasing capacitive reactance, making it easier for the current to flow as the angular frequency rises. Understanding current amplitude calculation is essential for designing and analyzing AC circuits effectively.