Problem 47 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, \(R=300 \Omega, X_{C}=300 \Omega\) and \(X_{L}=500 \Omega .\) The average power consumed in the resistor is 60.0 \(\mathrm{W}\) . (a) What is the power factor of the circuit? (b) What is the rms voltage of the source?

Step-by-Step Solution

Answer

(a) Power factor is approximately 0.832. (b) RMS voltage of the source is about 161.13 V.

Step 1: Calculate the Impedance of the Circuit

The impedance \( Z \) of the series \( L-R-C \) circuit is given by the formula: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the given values: \( Z = \sqrt{(300 \ \Omega)^2 + (500 \ \Omega - 300 \ \Omega)^2} \). Simplifying, \( Z = \sqrt{90000 + 40000} = \sqrt{130000} \). So, \( Z = 360.55 \ \Omega \).

Step 2: Determine the Power Factor

The power factor \( \text{PF} \) is defined as \( \text{PF} = \frac{R}{Z} \). Using the resistor value \( R = 300 \ \Omega \) and the calculated impedance \( Z = 360.55 \ \Omega \), we have \( \text{PF} = \frac{300}{360.55} \approx 0.832 \).

Step 3: Calculate the RMS Current

The average power \( P \) consumed in the resistor is given by \( P = I_{rms}^2 \cdot R \). Rearranging the formula for the rms current: \( I_{rms} = \sqrt{\frac{P}{R}} \). Substituting \( P = 60.0 \ W \) and \( R = 300 \ \Omega \), we have \( I_{rms} = \sqrt{\frac{60}{300}} = \sqrt{0.2} = 0.447 \ A \).

Step 4: Find the RMS Voltage of the Source

The rms voltage \( V_{rms} \) across the entire circuit can be calculated using the relationship \( V_{rms} = I_{rms} \cdot Z \). Substituting \( I_{rms} = 0.447 \ A \) and \( Z = 360.55 \ \Omega \), we get \( V_{rms} = 0.447 \ A \times 360.55 \ \Omega \approx 161.13 \ V \).

Key Concepts

Impedance Calculation

In an L-R-C series circuit, the impedance is a crucial factor that determines the opposition to the flow of alternating current. Unlike simple resistance, impedance takes into account the combined effects of resistance (R), inductive reactance (X_L), and capacitive reactance (X_C). This is because inductors and capacitors introduce additional phase shifts between voltage and current. The formula for calculating impedance (Z) is \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
Substituting the given values, we have \( Z = \sqrt{(300 \ \Omega)^2 + (500 \ \Omega - 300 \ \Omega)^2} = \sqrt{90000 + 40000} = \sqrt{130000} \), which simplifies to \( Z = 360.55 \ \Omega \).

• Resistance (R): Opposition due to resistors alone.
• Inductive Reactance (X_L): Causes current to lag voltage.
• Capacitive Reactance (X_C): Causes current to lead voltage.

Understanding these components and their roles will help you grasp why calculating impedance is essential in AC circuits.

Power Factor

The power factor is a measure that describes how effectively electrical power is being converted into useful work energy. In AC circuits, not all the power is used to perform work due to phase differences caused by reactance. With the power factor (PF), you can understand the efficiency of power usage. It is calculated using the equation: \( \text{PF} = \frac{R}{Z} \), where R is the resistance and Z is the impedance of the circuit.
For our example, using \( R = 300 \ \Omega \) and \( Z = 360.55 \ \Omega \), we find \( \text{PF} = \frac{300}{360.55} \approx 0.832 \).

• Efficiency: Higher PF means more of the supplied power is being effectively used.
• Practical Implication: A low PF can lead to higher losses in the system, requiring more current for the same amount of work done.

Grasping power factor helps in optimizing energy usage and minimizing electrical costs.

RMS Voltage

RMS (Root Mean Square) voltage is a concept used to express the effective value of an AC voltage, comparable to a DC voltage that performs the same amount of work. In an L-R-C circuit, the RMS voltage (V_{rms}) across the circuit is determined by multiplying the RMS current and the impedance.
The formula used in our example is \( V_{rms} = I_{rms} \cdot Z \).
Given \( I_{rms} = 0.447 \ \text{A} \) and \( Z = 360.55 \ \Omega \), we calculate \( V_{rms} = 0.447 \ \text{A} \times 360.55 \ \Omega \approx 161.13 \ \text{V} \).

• Equivalent Value: RMS voltage is the effective value that equates AC power to DC power.
• Importance: It's significant in ensuring the devices connected to the circuit can operate safely and efficiently.

Knowing the RMS voltage is essential for the correct sizing of electronic components in circuits.

RMS Current

RMS current is similar to RMS voltage, being the effective value of an AC current that would produce the same power as a DC current. It gives you an idea about how much current actually contributes to the power consumption in the resistor.
In an L-R-C circuit, you determine the RMS current (I_{rms}) by rearranging the power equation \( P = I_{rms}^2 \cdot R \) to \( I_{rms} = \sqrt{\frac{P}{R}} \).
For our case, with an average power \( P = 60.0 \text{W} \) and resistance \( R = 300 \ \Omega \), we find \( I_{rms} = \sqrt{\frac{60.0}{300}} = \sqrt{0.2} \approx 0.447 \ \text{A} \).

• Measurement Significance: It helps in predicting the heat loss and efficiency in circuits.
• Real-world Applications: Essential for correct component ratings in power systems to avoid overheating.

Understanding RMS current is crucial in designing and analyzing AC circuits for safe and efficient operation.