Problem 60 - Fundamentals of Physics

Question

In a series oscillating \(R L C\) circuit, \(R=16.0 \Omega, C=\) \(31.2 \mu \mathrm{F}, L=9.20 \mathrm{mH},\) and \(\mathscr{C}_{m}=\mathscr{E}_{m} \sin \omega_{d} t\) with \(\mathscr{E}_{m}=45.0 \mathrm{~V}\) and \(\omega_{d}=3000 \mathrm{rad} / \mathrm{s} .\) For time \(t=0.442 \mathrm{~ms}\) find \((\mathrm{a})\) the rate \(P_{g}\) at which energy is being supplied by the generator, (b) the rate \(P_{C}\) at which the energy in the capacitor is changing, (c) the rate \(P_{L}\) at which the energy in the inductor is changing, and (d) the rate \(P_{R}\) at which energy is being dissipated in the resistor. (e) Is the sum of \(P_{C}, P_{L}\) and \(P_{R}\) greater than, less than, or equal to \(P_{g} ?\)

Step-by-Step Solution

Answer

The power supplied by the generator (Pg) is greater than the sum of powers in the capacitor, inductor, and resistor (Pc + Pl + Pr).

Step 1: Calculate Voltage and Current

First, identify the voltage across the generator using \( \mathscr{C}_{m} = \mathscr{E}_{m} \sin(\omega_d t) \). For \( t = 0.442 \text{ ms} \), we have:\[ \mathscr{C}_{m} = 45.0 \sin(3000 \cdot 0.442 \times 10^{-3}) \]This calculates to \( \mathscr{C}_{m} \approx 45.0 \sin(1.326) \approx 44.9 \text{ V} \). Next, find the current \( i(t) \) using the impedance of the circuit \( Z = \sqrt{R^2 + (\omega_d L - \frac{1}{\omega_d C})^2} \). Compute the respective values for \( L, C, \omega_d \).

Step 2: Calculate Impedance (Z)

Calculate the impedance of the circuit, which incorporates resistance (R), inductance (L), and capacitance (C).\[ Z = \sqrt{R^2 + \left( \omega_d L - \frac{1}{\omega_d C} \right)^2 } \]Plug in the given values:\[ Z = \sqrt{16.0^2 + (3000 \times 0.0092 - \frac{1}{3000 \times 31.2 \times 10^{-6}})^2 } \]After evaluation, \( Z \approx 30.1 \Omega \).

Step 3: Calculate the Current (i)

Using the peak voltage and the impedance, calculate the peak current in the circuit:\[ I_m = \frac{\mathscr{E}_m}{Z} = \frac{45.0}{30.1} \approx 1.495 \text{ A} \]Then find the instantaneous current at \( t = 0.442 \text{ ms} \) using:\[ i(t) = I_m \sin(\omega_d t - \delta) \]where \( \delta = \tan^{-1}\left(\frac{\omega_d L - \frac{1}{\omega_d C}}{R} \right) \). Substitute to find \( i(0.442 \text{ ms}) \approx 1.49 \text{ A} \).

Step 4: Calculate Power Supplied by Generator (Pg)

The power supplied by the generator at time \( t \) is given by:\[ P_g = \mathscr{C}_{m} i(t) \cos(\omega_d t) \]Substituting values:\[ P_g = 44.9 \times 1.49 \times \cos(1.326) \approx 67.1 \text{ W} \]

Step 5: Calculate Power in Capacitor (PC)

The rate of energy change in the capacitor is:\[ P_C = \frac{1}{2} C \mathscr{C}_m^2 \omega_d \sin(2 \omega_d t) \]Calculating at \( t \):\[ P_C = \frac{1}{2} \times 31.2 \times 10^{-6} \times 44.9^2 \times 3000 \times \sin(2 \times 1.326) \approx -2.84 \text{ W} \]

Step 6: Calculate Power in Inductor (PL)

Use the formula for power in the inductor:\[ P_L = - \frac{1}{2} L i^2 \omega_d^2 \cos(2 \omega_d t) \]Substitute the values at \( t \):\[ P_L = - \frac{1}{2} \times 9.20 \times 10^{-3} \times (1.49)^2 \times 3000^2 \times \cos(2 \times 1.326) \approx 2.82 \text{ W} \]

Step 7: Calculate Power in Resistor (PR)

The power dissipated in the resistor is simply the instantaneous power:\[ P_R = i(t)^2 R \]\[ P_R = (1.49)^2 \times 16.0 \approx 35.6 \text{ W} \]

Step 8: Compare Sum of Powers

Now sum up the powers \( P_C, P_L, P_R \) and compare with \( P_g \):\[ P_T = P_C + P_L + P_R = -2.84 + 2.82 + 35.6 \approx 35.6 \text{ W} \]Since \( P_g = 67.1 \text{ W} \), \( P_T < P_g \).

Key Concepts

Oscillating Circuit

An oscillating circuit, often referred to as an RLC circuit, is a fundamental element in electrical engineering. It consists of a resistor (R), inductor (L), and capacitor (C) connected in series or parallel. These circuits are essential for understanding AC electricity.

An RLC circuit can exhibit a range of behaviors depending on its configuration and the driving frequency. When the circuit is disturbed from its equilibrium position and released, it will oscillate. This oscillation is caused by the energy being alternately stored in the inductor and the capacitor.

• In an inductive coil, energy is stored as a magnetic field.
• In a capacitor, energy is stored as an electric field.

The natural oscillation frequency, \(\omega_n\), is influenced by both the inductance and capacitance, given by: \[ \omega_n = \frac{1}{\sqrt{LC}} \] Understanding this concept is key as it forms the basis for filtering specific frequencies, an important application in radio communications.

Energy Transfer

In an RLC circuit, energy continuously transfers between the inductor and capacitor. This dynamic transfer is a defining characteristic of oscillating circuits. At any given moment, a portion of the energy is stored in either the electric field of a capacitor or the magnetic field of an inductor.

The rate at which this energy transfer occurs is influenced by the frequency of the oscillation and the quality of the circuit. When energy is flowing from one component to another, it experiences minimal losses if the circuit is ideal.

• The energy stored in the capacitor is computed using: \( E = \frac{1}{2} C V^2 \)
• The energy in the inductor is: \( E = \frac{1}{2} L i^2 \)

This energy shifting is responsible for the sinusoidal response observed in oscillators and is pivotal in designing circuits that need to maintain a high level of energy conservation.

Electrical Impedance

Electrical impedance, denoted as Z, is the measure of opposition that a circuit presents to a current when a voltage is applied. In an RLC circuit, the impedance is not just resistance; it also includes reactance from both the inductor and capacitor.

Impedance in RLC circuits can be calculated using the formula: \[ Z = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} \] Here, \( \omega = 2\pi f \) is the angular frequency, where \( f \) is the frequency of the applied voltage.

The phase difference between voltage and current arises because of this impedance, making AC analysis considerably complex when compared to DC. Understanding impedance is crucial for:

• Designing circuits for specific frequency responses
• Matching impedances in different components to maximize power transfer

It effectively controls how an AC signal is processed through a circuit, influencing both the magnitude and the phase of currents and voltages.

Power Dissipation

Power dissipation in an RLC circuit typically refers to the energy lost through the resistor. This loss manifests as heat and represents the portion of power that does not return to the circuit.

In a purely resistive part, the power dissipation can be calculated using: \( P = i^2 R \) or alternatively, \( P = \frac{V^2}{R} \), depending on which quantities are known.

In practical terms, power dissipation is undesirable as it suggests inefficiency. However, managing power loss is essential for the safe and effective design of electronic systems. Key considerations include:

• Ensuring components can handle the dissipation levels to prevent overheating
• Designing heat sinks or employing ventilation when necessary

In many designs, especially in high-frequency applications like radio transmitters, minimizing power dissipation ensures more of the input power is retained for transmission or further processing. Efficient management of power dissipation can dramatically improve the performance and reliability of electronic circuits.