Problem 29 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, \(R=300 \Omega, L=0.400 \mathrm{H},\) and \(C=6.00 \times 10^{-8} \mathrm{F} .\) When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 \(\mathrm{A}\) . (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Step-by-Step Solution

Answer

(a) 150 V; (b) Resistor: 150 V, Inductor: 1630 V, Capacitor: 1630 V; (c) 75 W.

Step 1: Calculate the Resonance Frequency

The resonance frequency for an L-R-C series circuit is given by: \[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \]Substitute the given values of \(L = 0.400\, \text{H} \) and \(C = 6.00 \times 10^{-8}\, \text{F} \) into the formula:\[ f_0 = \frac{1}{2 \pi \sqrt{0.400 \times 6.00 \times 10^{-8}}} \]After calculation, this gives:\[ f_0 \approx 1.297 \times 10^4 \text{Hz} \]

Step 2: Calculate Voltage Amplitude of the Source

At resonance, the impedance in the circuit is pure resistive and equal to the resistance \(R\). Using Ohm's law:\[ V = I \times R \]where \(I = 0.500\, \text{A} \) is the current amplitude. Hence:\[ V = 0.500 \times 300 = 150 \text{V} \]Thus, the voltage amplitude of the source is \(150 \text{V} \).

Step 3: Calculate Amplitude of Voltage across the Resistor

The voltage across the resistor, \( V_R \), at resonance is given by Ohm's law:\[ V_R = I \times R \]With \(I = 0.500\, \text{A} \) and \(R = 300 \Omega\):\[ V_R = 0.500 \times 300 = 150 \text{V} \]This means \( V_R \) is also \(150 \text{V}\).

Step 4: Calculate Amplitude of Voltage across the Inductor and Capacitor

At resonance, the voltages across the inductor and capacitor are equal in magnitude but opposite in phase.The voltage amplitudes are:\[ V_L = I \times \omega_0 L \]\[ V_C = \frac{I}{\omega_0 C} \]First, calculate angular frequency \(\omega_0 = 2\pi f_0\):\[ \omega_0 = 2\pi \times 1.297 \times 10^4 \approx 8.15 \times 10^4 \text{rad/s} \]Now compute \( V_L \) and \( V_C \):\[ V_L = 0.500 \times 8.15 \times 10^4 \times 0.400 \approx 1630 \text{V} \]\[ V_C = \frac{0.500}{8.15 \times 10^4 \times 6.00 \times 10^{-8}} \approx 1630 \text{V} \]

Step 5: Calculate Average Power Supplied by the Source

The average power \(P\) supplied by the source is given by:\[ P = I^2 \times R \]where \(I = 0.500\, \text{A} \) and \(R = 300 \Omega \):\[ P = (0.500)^2 \times 300 = 75 \text{W} \]Thus, the average power supplied is \(75 \text{W} \).

Key Concepts

L-R-C Series Circuit

An L-R-C series circuit is a type of electrical circuit that includes an inductor (L), a resistor (R), and a capacitor (C) connected in a single path. These three components play distinct roles that are crucial to the functioning of the circuit.

In an L-R-C series circuit, the inductor generates a magnetic field when electric current flows through it. The resistor controls the current flow by providing resistance, which is the opposition to current. The capacitor stores energy in an electric field between its plates and releases it as needed.

A significant aspect of the L-R-C series circuit is resonance. This occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other out, resulting in the total impedance being equal to just the resistance of the circuit. Thus, the circuit acts primarily as a resistive circuit at this frequency. This special point is called the resonant frequency.

Ohm's Law

Ohm's law is a fundamental principle in electrical engineering and physics. It relates the voltage (V), current (I), and resistance (R) in a circuit through the equation: \[ V = I \times R \].

This law indicates that the voltage across a resistor is proportional to the current flowing through it, with resistance being the constant of proportionality. In the context of an L-R-C series circuit, Ohm's law is used to determine various voltages in different parts of the circuit.

• At resonance, the impedance is purely resistive, meaning the overall impedance equals the resistance \(R\).
• The voltage across the resistor is calculated using the current amplitude and resistance: \(V_R = I \times R\).
• Likewise, the source voltage amplitude is derived similarly at resonance, where the contribution of reactance is nullified.

Average Power Calculation

The average power in a resistive circuit is calculated as \[ P = I^2 \times R \]. This formula stems from Ohm's law and the basic definition of power, which is the rate at which electrical energy is consumed or transformed in the circuit.

Within an L-R-C circuit at resonance, the total impedance equals the resistance, so:

• The current is in phase with the voltage, leading to maximum power transfer.
• The average power can be computed using the given current amplitude and resistance.
• This calculation provides the actual useful power utilized by the circuit as opposed to reactive power stored in other components.

These principles help in understanding energy consumption and efficiency in AC circuits.

Impedance at Resonance

Impedance, often denoted as \(Z\), is the complex quantity that combines both resistance and reactance (combination of inductance and capacitance) in AC circuits. In the case of resonance in an L-R-C series circuit, impedance simplifies significantly.

At resonance:

• The inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)) are equal and opposite, so they cancel each other out (\(X_L = X_C\)).
• The impedance \(Z\) thus equals the resistance \(R\) alone, removing any phase difference between voltage and current.
• This results in a condition where the apparent power equals the real power, as the circuit behaves purely resistive with minimum impedance.

This condition is pivotal for maximizing energy transfer and efficiency in applications involving AC circuits, such as radio receivers and various signal processing equipment.