Problem 46 - University Physics with Modern Physics

Question

An \(L \cdot R-C\) series circuit has \(R=300 \Omega .\) At the frequency of the source, the inductor has reactance \(X_{L}=900 \Omega\) and the capacitor has reactance \(X_{C}=500 \Omega .\) The amplitude of the voltage across the inductor is 450 V. (a) What is the amplitude of the voltage across the resistor? (b) What is the amplitude of the voltage across the capacitor? (c) What is the voltage amplitude of the source? (d) What is the rate at which the source is delivering electrical energy to the circuit?

Step-by-Step Solution

Answer

(a) 150 V, (b) 250 V, (c) 250 V, (d) 75 W.

Step 1: Calculate Current Amplitude

In a series circuit, the same current flows through the resistor, inductor, and capacitor. The voltage across the inductor is given by \( V_L = I \times X_L \). Given that \( X_L = 900 \ \Omega \) and \( V_L = 450 \ \text{V} \), we can calculate the current \( I \) using the formula: \( I = \frac{V_L}{X_L} = \frac{450 \text{ V}}{900 \ \Omega} = 0.5 \ \text{A} \).

Step 2: Find Resistor Voltage Amplitude

The voltage amplitude across the resistor, \( V_R \), can be found using Ohm's Law: \( V_R = I \times R \). With \( R = 300 \ \Omega \) and \( I = 0.5 \ \text{A} \), \( V_R = 0.5 \ \text{A} \times 300 \ \Omega = 150 \ \text{V} \).

Step 3: Find Capacitor Voltage Amplitude

The voltage amplitude across the capacitor, \( V_C \), is calculated using the formula \( V_C = I \times X_C \). With \( X_C = 500 \ \Omega \) and \( I = 0.5 \ \text{A} \), the voltage is \( V_C = 0.5 \ \text{A} \times 500 \ \Omega = 250 \ \text{V} \).

Step 4: Calculate Source Voltage Amplitude

To find the amplitude of the source voltage, we add the phasor voltages across each component. The total impedance is given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the values, \( Z = \sqrt{300^2 + (900 - 500)^2} = 500 \ \Omega \). Hence, the source voltage amplitude is \( V_s = I \times Z = 0.5 \ \text{A} \times 500 \ \Omega = 250 \ \text{V} \).

Step 5: Compute Power Delivered

The rate at which energy is delivered to the circuit by the source, or the average power, is given by \( P = I^2 \times R \). Substituting the values, \( P = (0.5 \ \text{A})^2 \times 300 \ \Omega = 75 \ \text{W} \).

Key Concepts

Inductive Reactance

In an AC circuit, inductive reactance describes how an inductor opposes the change in current. This is due to the inductor's property of storing energy in a magnetic field as current changes. Physically, it's described by the formula \( X_L = 2\pi f L \), where \( X_L \) is inductive reactance, \( f \) is the frequency, and \( L \) is the inductance. Reactance has units of ohms (\( \Omega \)).
For our exercise, the reactance of the inductor is given as \( 900 \ \Omega \).

• When the current flows through the inductor, it lags the voltage by 90 degrees due to this reactance.
• This reactance is crucial for determining the voltage across the inductor when analyzing AC circuits.

In the context of the problem, using the given inductive reactance allows us to find that the current through the circuit is 0.5 A when the voltage across the inductor is 450 V. This mirrors the inductor's natural opposition to current changes.

Capacitive Reactance

Capacitive reactance in an AC circuit refers to how a capacitor resists the change in voltage across it. It happens because the capacitor stores energy in an electric field. The formula for capacitive reactance is \( X_C = \frac{1}{2\pi f C} \), where \( X_C \) is the capacitive reactance, \( f \) is frequency, and \( C \) is capacitance. Like inductive reactance, capacitive reactance is measured in ohms (\( \Omega \)).
In the given exercise, the capacitor has a reactance of \( 500 \ \Omega \).

• This causes the current in a capacitor to lead the voltage by 90 degrees.
• Knowing this reactance helps to determine the voltage across the capacitor, especially in our problems involving series circuits.

For the circuit involved in the problem, with a current of 0.5 A determined from the inductor's reactance, the voltage across the capacitor is calculated to be 250 V. It's important to note the phase relationship between different components to solve such circuits accurately.

Ohm's Law

Ohm's Law is a fundamental principle used to relate voltage, current, and resistance in electrical circuits. It is expressed simply as \( V = I \times R \), where \( V \) represents voltage, \( I \) current, and \( R \) resistance.
In our scenario, Ohm's Law allows us to find the voltage across the resistor. The current, calculated as 0.5 A using the inductor's law, is multiplied by the resistor's value of \( 300 \ \Omega\), yielding a voltage of 150 V.

• This law is essential as it applies to each component in a circuit, not just the whole system.
• It supports our understanding of how individual components affect overall circuit behavior.

By using Ohm's Law, we can also calculate the power dissipated in each resistance as shown in the solution, aiding in a complete analysis of AC circuits.

Phasor Addition

Phasor addition is crucial in analyzing AC circuits because it allows the combination of different voltage or current vectors, considering both magnitude and phase.
In an AC circuit with sinusoidal sources, voltages and currents are often out of phase due to the reactive components - inductors and capacitors.

• Phasors transform time-varying signals into a format that simplifies the mathematics of adding them.
• Instead of directly adding voltages across a capacitor, resistor, and inductor, we consider their phase differences.

The exercise includes calculating the source voltage using phasor addition, which results in a total impedance calculation \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
This allows us to determine the source voltage amplitude, given the current. Understanding phasor addition is vital for combining AC signals correctly and predicting the behavior of AC circuits comprehensively.