Problem 27 - College Physics

Question

A circuit with a single capacitor is connected to a \(120-\) \(\mathrm{V}, 60\) -Hz source. (a) What is its capacitance if there is a current of \(0.20 \mathrm{~A}\) in the circuit? (b) What would be the current if the source frequency were halved?

Step-by-Step Solution

Answer

(a) Capacitance is approximately \(4.42 \, \mu\mathrm{F}\). (b) New current is \(0.10 \, \mathrm{A}\) when frequency is halved.

Step 1: Understand the problem

We have a capacitor connected to a source, and we need to find its capacitance given a specific current and voltage. Then, we need to determine the current when the frequency of the source is halved.

Step 2: Identify the formula for capacitive reactance

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. We know the voltage \( V \), the current \( I \), and the frequency \( f \), and we need to find \( C \).

Step 3: Relationship between current and voltage

In an AC circuit, the relationship between the current \( I \), voltage \( V \), and capacitive reactance \( X_C \) is \( I = \frac{V}{X_C} \). We can solve for \( X_C = \frac{V}{I} \).

Step 4: Calculate capacitive reactance

Given \( V = 120 \, \mathrm{V} \) and \( I = 0.20 \, \mathrm{A} \), calculate \( X_C = \frac{V}{I} = \frac{120}{0.20} = 600 \, \Omega \).

Step 5: Determine capacitance from capacitive reactance

Using the formula for capacitive reactance, \( X_C = \frac{1}{2\pi f C} \), solve for \( C \): \( C = \frac{1}{2\pi f X_C} \). Substitute \( f = 60 \, \mathrm{Hz} \) and \( X_C = 600 \, \Omega \) to find \( C \).

Step 6: Calculate capacitance

Substitute the values into the formula: \( C = \frac{1}{2 \pi \times 60 \times 600} \approx 4.42 \, \mu\mathrm{F} \).

Step 7: Effect of halving the frequency

When frequency is halved, \( f = 30 \, \mathrm{Hz} \). The new capacitive reactance \( X'_C = \frac{1}{2\pi f C} \). Calculate \( X'_C \) and then find the new current \( I' = \frac{V}{X'_C} \).

Step 8: Calculate new capacitive reactance and current

Calculate \( X'_C = \frac{1}{2 \pi \times 30 \times 4.42 \times 10^{-6}} \approx 1200 \, \Omega \). Then calculate \( I' = \frac{120}{1200} = 0.10 \, \mathrm{A} \).

Key Concepts

Capacitance Calculation

In electrical engineering, calculating the capacitance of a capacitor in a circuit is essential to understand the behavior of AC circuits. A capacitor's capacitance is a measure of its ability to store electrical charge. The relationship between capacitance, frequency, and capacitive reactance is given by the formula \( X_C = \frac{1}{2\pi f C} \), where \( X_C \) is the capacitive reactance, \( f \) is the frequency of the AC source, and \( C \) is the capacitance. This equation is paramount since it establishes how the reactance changes with frequency as well as capacitance.

To find the capacitance if a 120-V AC source with a frequency of 60 Hz produces a current of 0.20 A, first determine the capacitive reactance using \( X_C = \frac{V}{I} = \frac{120}{0.20} = 600 \Omega \). Then solve for capacitance \( C \) using \( C = \frac{1}{2\pi f X_C} \). Substituting the known values, \( C = \frac{1}{2\pi \times 60 \times 600} \approx 4.42 \, \mu\mathrm{F} \).

Understanding this process allows you to effectively determine the capacitance in various AC circuits and analyze how it influences the behavior of electrical systems.

AC Circuits

Alternating Current (AC) circuits operate by continuously changing the direction of the current flow, typically in a sinusoidal manner. This characteristic significantly affects components within the circuit, such as capacitors and inductors. The frequency of the AC source, often measured in Hertz (Hz), plays a crucial role in defining the circuit's overall behavior.

In AC circuits, capacitors are key elements due to their ability to influence the phase relationship between voltage and current. Capacitors introduce reactance into the circuit, known as capacitive reactance \( X_C \), which affects how the circuit responds to different frequencies. This reactance decreases with increasing frequency, which is why capacitors are often used in filter circuits.

The AC circuit in this exercise demonstrates how halving the frequency of the source impacts the capacitive reactance, and consequently, the current. When the frequency was reduced from 60 Hz to 30 Hz, the current also decreased. This occurs because the capacitive reactance \( X_C = \frac{1}{2\pi f C} \) increases as frequency decreases, thereby reducing the current \( I = \frac{V}{X_C} \) through the circuit.

Voltage and Current Relationship

In AC circuits, the relationship between voltage and current is interlinked through the concept of reactance. Reactance, similar to resistance in DC circuits, is the opposition a circuit presents to the current, but it's specifically due to capacitive or inductive elements rather than resistive ones. For capacitive circuits, this relationship is defined by \( I = \frac{V}{X_C} \), where \( I \) is the current, \( V \) is the voltage, and \( X_C \) is the capacitive reactance.

This equation indicates that for a fixed voltage, the current is inversely proportional to the capacitive reactance. As shown in the exercise, when the frequency is halved, increasing the reactance, the current decreases. So, at 60 Hz, with a reactance of 600 \( \Omega \), the current was 0.20 A. When the frequency was halved to 30 Hz, the capacitive reactance doubled to 1200 \( \Omega \), reducing the current to 0.10 A.

This clear relationship between voltage, current, and reactance is fundamental to predicting how an AC circuit will behave under different frequencies and is crucial for designing and analyzing circuits efficiently.