Problem 33 - College Physics

Question

In a series \(R-L-C\) circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of \(120 \mathrm{~V}\) and a frequency of \(1.25 \mathrm{kHz}\). Determine (a) the power supplied by the generator; and (b) the power dissipated in the resistor.

Step-by-Step Solution

Answer

The power supplied by the generator and the power dissipated in the resistor are both approximately 7.35 W.

Step 1: Calculate Inductive and Capacitive Reactance

First, determine the inductive reactance \(X_L\) and capacitive reactance \(X_C\) using the formulas:\[X_L = 2 \pi f L \quad \text{and} \quad X_C = \frac{1}{2 \pi f C}\]where \(f = 1.25\, \text{kHz}\), \(L = 20.0\, \text{mH}\), and \(C = 140\, \text{nF}\). 1. Calculate \(X_L\):\[X_L = 2 \pi \cdot 1.25 \times 10^3 \cdot 20.0 \times 10^{-3} = 157\, \Omega\]2. Calculate \(X_C\):\[X_C = \frac{1}{2 \pi \cdot 1.25 \times 10^3 \cdot 140 \times 10^{-9}} = 908\, \Omega\]

Step 2: Calculate the Circuit Impedance

The total impedance \(Z\) of the series \(R-L-C\) circuit is found using:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]Substitute the values:\[Z = \sqrt{350^2 + (157 - 908)^2}\]\[= \sqrt{350^2 + (-751)^2} \approx 827.07\, \Omega\]

Step 3: Determine the Current in the Circuit

The rms current \(I_{rms}\) flowing through the circuit is calculated using Ohm’s law:\[I_{rms} = \frac{V_{rms}}{Z}\]with \(V_{rms} = 120\, \text{V}\):\[I_{rms} = \frac{120}{827.07} \approx 0.145\, \text{A}\]

Step 4: Calculate the Power Supplied by the Generator

The power supplied by the generator is given by:\[P_{gen} = V_{rms} \cdot I_{rms} \cdot \cos \phi\]Where the power factor \(\cos \phi = \frac{R}{Z}\):\[\cos \phi = \frac{350}{827.07} \approx 0.423\]Then the power is:\[P_{gen} = 120 \cdot 0.145 \cdot 0.423 \approx 7.35\, \text{W}\]

Step 5: Calculate the Power Dissipated in the Resistor

The power dissipated in the resistor is given by:\[P_R = I_{rms}^2 \cdot R\]Substituting the known values:\[P_R = (0.145)^2 \cdot 350 \approx 7.35\, \text{W}\]

Step 6: Verify Consistency in Power Calculations

Notice that the power supplied by the generator equals the power dissipated in the resistor, confirming our calculations. This is consistent with energy conservation, as the power in a purely resistive circuit is entirely dissipated as heat in the resistor.

Key Concepts

Inductive Reactance

Inductive reactance is a measure of how much an inductor opposes the change of current in an AC circuit.
Inductors store energy in the form of a magnetic field and release it into the circuit, causing a timing delay or phase shift between voltage and current.
In a circuit with AC supply, the inductive reactance is represented by the symbol \(X_L\) and is calculated using the formula: \[ X_L = 2 \pi f L \]Where:

• \(f\) is the frequency of the AC source.
• \(L\) is the inductance in henries (H).

The higher the frequency or the inductance, the greater the inductive reactance. This translates to a greater opposition to current flow, affecting how AC circuits behave.

Capacitive Reactance

Capacitive reactance, denoted as \(X_C\), is the extent to which a capacitor opposes the change in current in an AC circuit.
Capacitors store energy in an electric field, temporarily holding charge and then releasing it, creating a phase difference where current leads voltage.
To calculate capacitive reactance, use:\[X_C = \frac{1}{2 \pi f C} \]Where:

• \(f\) is the frequency of the alternating current.
• \(C\) is the capacitance measured in farads (F).

With higher frequency or lower capacitance, the reactance is reduced, allowing more current to pass through the circuit. Capacitive reactance is crucial for understanding how capacitors affect circuit performance.

Circuit Impedance

Circuit impedance is the total opposition a circuit presents to the flow of alternating current.
It combines resistive, inductive, and capacitive effects and is represented by the symbol \(Z\).
In an \(RLC\) circuit, impedance is not just resistance, it also comprises reactance, calculated by:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Where:

• \(R\) is the resistance in ohms.
• \(X_L\) is the inductive reactance.
• \(X_C\) is the capacitive reactance.

Impedance is expressed in ohms (\(\Omega\)) and is essential for determining how much current an AC source can drive through a circuit. It is a key factor in analyzing circuit behavior and efficiency.

Ohm’s Law

Ohm’s Law is a fundamental principle when analyzing electrical circuits, stating that the current through a conductor between two points is directly proportional to the voltage across the two points.
The formula is expressed as:\[ I = \frac{V}{Z} \]Where:

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

For an AC circuit, impedance replaces resistance in the formula. Applying Ohm’s Law is crucial for understanding the relationship between voltage, current, and resistance or impedance, enabling accurate calculations and predictions in circuit analysis.

Power Factor

Power factor is a measure of how effectively electrical power is being used in a circuit.
It represents the cosine of the phase difference \(\phi\) between voltage and current in the circuit.
The formula for power factor \(\cos \phi\) is:\[ \cos \phi = \frac{R}{Z} \]Where:

• \(R\) is the resistance in the circuit.
• \(Z\) is the total impedance.

A power factor of 1 means all the power is being effectively converted into work, while lower anything less indicates wasted power.
Typically, a circuit operates efficiently when the power factor is as close to 1 as possible, minimizing losses and improving performance. Understanding this concept is vital for optimizing electrical power systems.