Question

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

Short answer

(a) The capacitive reactance is approximately 1738.8 Ω. (b) The voltage across the resistor is v_R(t) = 1.093 sin(120t) V.

Step-by-step solution

Determine the capacitive reactance

The capacitive reactance of the capacitor can be calculated using the formula: \[ X_C = \frac{1}{\omega C} \]where \( \omega \) is the angular frequency (120 rad/s) and \( C \) is the capacitance (4.80 \mu\text{F} = 4.80 \times 10^{-6} \text{F}).Let's calculate:\[ X_C = \frac{1}{120 \times 4.80 \times 10^{-6}} \approx 1738.8 \, \Omega \]

Calculate the total impedance of the circuit

Since the resistor and capacitor are in series, the total impedance \( Z \) is the vector sum of the resistance and capacitive reactance.The impedance is given by:\[ Z = \sqrt{R^2 + X_C^2} \]Substitute the known values:\[ Z = \sqrt{250^2 + 1738.8^2} \approx 1756.7 \, \Omega \]

Find the current through the circuit

The current \( I \) can be found by Ohm's Law for AC circuits:\[ I = \frac{V_C}{X_C} \]where \( V_C = 7.6 \, \text{V} \).Calculate:\[ I = \frac{7.60}{1738.8} \approx 0.00437 \, \text{A} \]

Derive the voltage across the resistor

The voltage across the resistor \( v_R(t) \) can be derived from the expression for current:\[ v_R(t) = I \cdot R \cdot \sin(\omega t) \]Substitute the known values:\[ v_R(t) = 0.00437 \times 250 \times \sin(120t) \]\[ v_R(t) = 1.093 \sin(120t) \]

Key concepts

Capacitive Reactance

Capacitive reactance is a measure of a capacitor's opposition to alternating current. Unlike resistors, capacitors store and release energy by charging and discharging their electric fields. This makes them capable of impeding AC in unique ways, leading to the concept of reactance. The formula to calculate capacitive reactance \( X_C \) is given by:

• \[ X_C = \frac{1}{\omega C} \]

where:

• \( \omega \) is the angular frequency in radians per second (rad/s).
• \( C \) is the capacitance in farads (F).

In the exercise, with parameters \( \omega = 120 \) rad/s and \( C = 4.80 \times 10^{-6} \) F, the capacitive reactance \( X_C \) is calculated as approximately 1738.8 \( \Omega \). This indicates that the current will "lag" the voltage, meaning it slightly "falls behind" in phase. This property is essential in AC circuit analysis, as it dramatically affects the circuit's overall behavior.

Impedance Calculation

Impedance is a comprehensive measure of how much a circuit resists the flow of electric current. Impedance combines both resistance and reactance.

• For resistors, impedance is purely resistive and remains constant regardless of frequency.
• For capacitors, impedance is frequency-dependent and is described by the reactance.

In AC circuits featuring both resistive (R) and reactive (X) components, total impedance \( Z \) can be found using:

• \[ Z = \sqrt{R^2 + X_C^2} \]

In our specific case:

• \( R = 250 \; \Omega \)
• \( X_C = 1738.8 \; \Omega \)

This results in a total impedance \( Z \approx 1756.7 \; \Omega \). This total impedance tells us how much the entire circuit opposes the flow of alternating current and allows for examination of the current phase relation relative to the voltage.

Ohm's Law for AC Circuits

Ohm's Law is a foundational principle in electrical engineering and physics, applied also in AC circuits. It relates the voltage across an element, the current flowing through it, and the impedance it presents. The formula for AC circuits is analogous to that for DC circuits, but it incorporates impedance instead of just resistance:

• \[ I = \frac{V}{Z} \]

where:

• \( I \) is the current in amperes (A),
• \( V \) is the voltage in volts (V),
• \( Z \) is the impedance in ohms (\( \Omega \)).

In the exercise, the current \( I \) flowing through the circuit is calculated using the voltage across the capacitor \( V_C = 7.6 \; V \) and capacitive reactance \( X_C \). This results in \( I \approx 0.00437 \; A \). This formula underlines that higher impedance leads to lower current, providing a key understanding in designing or analyzing AC circuits.

Voltage Calculation across Resistors

Calculating the voltage across resistors in AC circuits often involves finding the current first, as seen in this exercise. The voltage change across a resistor \( v_R(t) \) can be expressed as:

• \[ v_R(t) = I \cdot R \cdot \sin(\omega t) \]

This equation means that the voltage across the resistor depends on:

• The current \( I \) flowing through the circuit,
• The resistance \( R \),
• The sinusoidal variation of the current caused by the AC supply, reflected by \( \sin(\omega t) \).

In our calculation, substituting the known values \( I \approx 0.00437 \; A \), \( R = 250 \; \Omega \), and \( \omega = 120 \; \text{rad/s} \), gives:

• \[ v_R(t) \approx 1.093 \sin(120t) \]

This means the resistor has a sinusoidal voltage with an amplitude of 1.093 V, demonstrating the dynamic interactions between current, impedance, and resistance in such circuits.