Problem 12 - University Physics with Modern Physics

Question

\(\mathrm{A} 250-\Omega\) resistor is connected in series with a \(4.80-\mu \mathrm{F}\) capacitor and an ac source. The voltage across the capacitor is \(v_{C}=(7.60 \mathrm{V}) \sin [(120 \mathrm{rad} / \mathrm{s}) t] .\) (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage \(v_{R}\) across the resistor.

Step-by-Step Solution

Answer

(a) 1736.11 Ω

Step 1: Determine the Capacitive Reactance

The formula for the capacitive reactance \( X_C \) is given by \[ X_C = \frac{1}{\omega C} \]where \( \omega = 120 \) rad/s is the angular frequency and \( C = 4.80 \times 10^{-6} \) F is the capacitance.Substituting the given values:\[ X_C = \frac{1}{120 \times 4.80 \times 10^{-6}} \approx \frac{1}{0.000576} \approx 1736.11 \Omega \].

Key Concepts

Capacitive Reactance

Capacitive reactance, denoted as \( X_C \), is an important concept in AC circuit analysis. It is the opposition to the change of voltage across a capacitor. In simpler terms, it tells us how much a capacitor will "resist" the flow of alternating current. Unlike resistance in a pure resistor, a capacitor's resistance isn't constant but depends on the frequency of the applied AC signal. The formula to calculate capacitive reactance is \[ X_C = \frac{1}{\omega C} \] where \( \omega \) is the angular frequency, and \( C \) is the capacitance.

Every capacitor reacts differently depending on the frequency of the signal passing through it. This is why AC circuit analysis often focuses on this aspect for understanding how circuits behave at different frequencies.

• As the frequency increases, the capacitive reactance decreases.
• If the frequency is very low (close to DC), the reactance becomes very high, making capacitors nearly "open” circuits.

Angular Frequency

Angular frequency, represented by \( \omega \), is a measure of how quickly an AC waveform oscillates. In AC circuits, it is often more insightful than linear frequency. Angular frequency relates closely to the experiences of engineers and physicists working with oscillating systems and simple harmonic motions.

In the formula \( \omega = 2 \pi f \), \( f \) is the linear frequency in hertz. However, for circuit equations, this is typically expressed in radians per second. Angular frequency is used in calculations of capacitive reactance \( X_C \) to better capture how rapid oscillation impacts circuit behavior.

• High angular frequency implies rapid oscillation, leading to decreased reactance in capacitors.
• Low angular frequency results in slow oscillation, increasing reactance.

Voltage Across Components

In AC circuits, understanding the voltage across each component is crucial for a complete analysis. In our circuit example, we have a resistor and a capacitor in series with an AC source. The voltage across the capacitor is given by the equation \( v_{C} = (7.60 \, V) \sin [(120 \, \text{rad/s}) t] \). This form indicates a sinusoidal AC voltage driving the component.

To find the voltage across the resistor \( v_R \), we need to acknowledge the series configuration and the behavior of AC circuits. The total voltage in such a circuit is the vector sum of the voltages across all components. As the voltage across the capacitor is known, Kirchhoff's voltage law helps in determining \( v_R \):

• The equation governing the sum of voltages: \( v_{\text{source}} = v_{C} + v_{R} \)
• Use Ohm's law with current \( i = \frac{v_{C}}{X_C} \) to find the voltage across the resistor \( v_{R} = i \times R \).

Analyzing voltage across components allows understanding of how each part contributes and behaves in the complete AC circuit.