Problem 3 - Physics

Question

A capacitor is connected across the terminals of an ac generator that has a frequency of \(440 \mathrm{~Hz}\) and supplies a voltage of \(24 \mathrm{~V}\). When a second capacitor is connected in parallel with the first one, the current from the generator increases by 0.18 A. Find the capacitance of the second capacitor.

Step-by-Step Solution

Answer

The capacitance of the second capacitor is approximately \(2.72 \mu F\).

Step 1: Understand the Problem

We are given a system with a capacitor connected to an AC generator. When a second capacitor is added in parallel, the current increases by 0.18 A. We need to find the capacitance of this second capacitor.

Step 2: Use the Formula for Capacitive Reactance

For a capacitor in an AC circuit, the capacitive reactance is given by the formula \( X_C = \frac{1}{2\pi f C} \), where \( f \) is the frequency and \( C \) is the capacitance. This formula will allow us to find the reactance contributed by each capacitor.

Step 3: Relate Current to Capacitive Reactance

The AC current through a capacitor is related to the voltage and reactance by the formula \( I = \frac{V}{X_C} \). This implies that the change in current when a second capacitor is added is due to the change in total reactance.

Step 4: Calculate Change in Reactance

The change in current \( \Delta I = 0.18 \mathrm{~A} \) is due to the second capacitor. The change in reactance can be calculated using \( \Delta I = \frac{V}{X_{C1}} - \frac{V}{X_{C2}} \), where \( X_{C1} \) and \( X_{C2} \) are the reactances before and after the second capacitor is added.

Step 5: Simplify for the Second Capacitance

Since the capacitors are in parallel, \( X_{C2} = X_C \text{ where } \frac{1}{X_C} = \frac{1}{X_{C1}} + \frac{1}{X_{C2new}} \). Since \( \Delta I = \frac{V}{X_{C2new}} \), we can solve \( X_{C2new} = \frac{V}{\Delta I} \).

Step 6: Compute the Second Capacitor's Capacitance

Using the frequency \( f = 440 \mathrm{~Hz} \) and voltage \( V = 24 \mathrm{~V} \), we first find \( X_{C2new} \) as \( X_{C2new} = \frac{24}{0.18} = 133.33 \Omega \). Then, using \( X_{C2new} = \frac{1}{2\pi \cdot 440 \cdot C_{new}} \), solve for \( C_{new} \): \( C_{new} = \frac{1}{2\pi \cdot 440 \cdot 133.33} \approx 2.72 \times 10^{-6} \mathrm{~F} \).

Key Concepts

Capacitive Reactance

Capacitive reactance is a crucial concept when dealing with capacitors in AC circuits. It essentially describes how a capacitor opposes the flow of alternating current (AC) at a certain frequency. The formula for capacitive reactance is given by:

• \( X_C = \frac{1}{2\pi f C} \)

Here, \( X_C \) is the capacitive reactance, \( f \) is the frequency of the AC supply, and \( C \) is the capacitance of the capacitor. Reactance is measured in ohms, just like resistance.
Reactance decreases with increase in capacitance or frequency. This means higher capacitance or frequency results in lower opposition to current flow, hence more current passes through the circuit.
Understanding capacitive reactance helps us analyze how a capacitor behaves in an AC circuit, particularly how it impacts current and voltage relationships.

AC Current

Alternating current (AC) is a type of electrical current where the flow of electric charge periodically reverses direction. In most homes and industries, electricity is supplied in the form of AC due to its efficient transmission over long distances.
In an AC circuit involving a capacitor, the current depends on the voltage and the reactance of the capacitor. This is described by the relationship:

• \( I = \frac{V}{X_C} \)

where \( I \) is the current, \( V \) is the voltage, and \( X_C \) is the capacitive reactance.
As the reactance of a capacitor decreases (due to increased capacitance or frequency), the current increases for the same applied voltage. This behavior is pivotal when components are arranged in circuits, especially for controlling current flow through adjusting capacitance.

Parallel Capacitance

When capacitors are connected in parallel, the total or equivalent capacitance increases. This is because the overall surface area available for storing charge increases.When we add capacitors in parallel, their combined capacitance is calculated by simple addition:

• \( C_{total} = C_1 + C_2 + ... + C_n \)

This means that parallel capacitors collectively hold more charge, leading to a lower overall capacitive reactance. As a result, the circuit can conduct more AC current. Adding a capacitor in parallel, as discussed in the exercise, increases the total current drawn from the source due to this lower reactance.
In practical applications, parallel capacitors can help achieve desired current or voltage levels attainable with standard capacitor values.

Electrical Frequency

Frequency in an electrical circuit defines how often the current changes direction per second and is measured in Hertz (Hz). It plays a fundamental role in determining the behavior of capacitive reactance in AC circuits.
The capacitive reactance is inversely proportional to the frequency. Hence, as frequency increases, the capacitive reactance decreases, which leads to an increase in current flowing through the capacitor. This property is crucial in applications like tuning radios or filtering signals, where different frequencies need to be handled differently by the circuit components. For example, a high capacitive reactance might be undesirable in high-frequency applications, such as in RF circuits, because it would impede the current too much compared to low-frequency circuits.