Problem 30 - College Physics

Question

A series \(\mathrm{RC}\) circuit has a resistance of \(200 \Omega\) and a capacitance of \(25 \mu \mathrm{F}\) and is driven by a \(120-\mathrm{V}, 60\) -Hz source. (a) Find the capacitive reactance and impedance of the circuit. (b) How much current is drawn from the source?

Step-by-Step Solution

Answer

Capacitive reactance is about 106.13 Ω, impedance is about 227.44 Ω, and current drawn is approximately 0.527 A.

Step 1: Calculating Capacitive Reactance

The capacitive reactance \( X_c \) is calculated using the formula \( X_c = \frac{1}{2 \pi fC} \), where \( f \) is the frequency and \( C \) is the capacitance. Substituting the given values \( f = 60 \, \mathrm{Hz} \) and \( C = 25\,\mu\, \mathrm{F} = 25 \times 10^{-6} \, \mathrm{F} \), we find: \[ X_c = \frac{1}{2 \pi \times 60 \times 25 \times 10^{-6}} \approx 106.13 \, \Omega. \]

Step 2: Calculating Impedance

The impedance \( Z \) of the circuit is calculated using the formula \( Z = \sqrt{R^2 + X_c^2} \), where \( R \) is the resistance. Using \( R = 200 \, \Omega \) and \( X_c = 106.13 \, \Omega \), we find:\[ Z = \sqrt{200^2 + 106.13^2} \approx 227.44 \, \Omega. \]

Step 3: Calculating Current

To find the current \( I \) drawn from the source, we use Ohm's Law in the context of impedance, \( I = \frac{V}{Z}, \) where \( V = 120 \, \mathrm{V} \) is the voltage of the source. Using the impedance \( Z \) calculated previously, we find:\[ I = \frac{120}{227.44} \approx 0.527 \, \mathrm{A}. \]

Key Concepts

Capacitive Reactance

In an RC circuit, the capacitive reactance, represented as \( X_c \), is a measure of how much the capacitor resists the change in voltage. This opposition happens because the capacitor stores and releases energy, creating a delay between voltage and current. The formula for capacitive reactance is given by:

• \( X_c = \frac{1}{2 \pi f C} \)

where \( f \) is the frequency and \( C \) is the capacitance of the circuit.
Substituting the values provided in the exercise, \( f = 60 \, \text{Hz} \) and \( C = 25 \, \mu F\) (or \( 25 \times 10^{-6} \, F)\), capacitive reactance can be calculated as \( 106.13 \, \Omega \).

This value indicates how much the capacitor is slowing down the current for a given frequency. The lower the frequency, the higher the capacitive reactance, since there's more time per cycle for the capacitor to charge and discharge.

Impedance

Impedance, represented as \( Z \), is the total opposition that a circuit offers to the flow of alternating current (AC). It combines both resistance and reactance, and is crucial for analyzing AC circuits. The formula to calculate impedance in an RC circuit is:

• \( Z = \sqrt{R^2 + X_c^2} \)

Here, \( R \) represents the resistance, and \( X_c \) is the capacitive reactance. For the given values, \( R = 200 \, \Omega \) and \( X_c = 106.13 \, \Omega \), the impedance is calculated to be roughly \( 227.44 \, \Omega \).

This shows how the circuit's resistance and reactance combine to affect the flow of current. Impedance is crucial in designing circuits that work efficiently with AC power, ensuring that we can control and predict how the circuit behaves.

Ohm's Law

Ohm's Law is a foundational principle for understanding electrical circuits. In the context of AC circuits, it relates the voltage \( V \), current \( I \), and impedance \( Z \) as follows:

• \( I = \frac{V}{Z} \)

This formula implies that the current drawn from a source is directly proportional to the voltage and inversely proportional to the impedance. For the circuit in the exercise, with a source voltage \( V = 120 \, V \) and impedance \( Z = 227.44 \, \Omega \), the current is approximately \( 0.527 \, A \).

Understanding this relationship helps in predicting how changing one parameter (like increasing the resistance or capacitance) will affect the other elements in the circuit, such as current. It's a practical application of Ohm's Law, especially when dealing with AC circuits where impedance plays a crucial role.