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

Short Answer

Expert verified
(a) 0.832, (b) 161.1 V.

Step by step solution

01

Calculating the Impedance

To find the impedance, we need to know both the resistance and the reactance. The total reactance \( X \) in the circuit is \( X = X_L - X_C = 500 \Omega - 300 \Omega = 200 \Omega \). The total impedance \( Z \) is given by \( Z = \sqrt{R^2 + X^2} \). So, substitute the known values: \[ Z = \sqrt{300^2 + 200^2} = \sqrt{90000 + 40000} = \sqrt{130000} \approx 360.55 \, \Omega \].
02

Calculating the Power Factor

The power factor (PF) is the ratio of the resistance to the total impedance, given by \( \text{PF} = \frac{R}{Z} \). Plug in the values for \( R \) and \( Z \): \[ \text{PF} = \frac{300}{360.55} \approx 0.832 \].
03

Calculating the RMS Current

The average power \( P \) consumed by the resistor is given by \( P = I^2 R \). Rearrange to find the RMS current \( I \): \[ I = \sqrt{\frac{P}{R}} = \sqrt{\frac{60.0}{300}} = \sqrt{0.2} \approx 0.447 \, \text{A} \].
04

Calculating the RMS Voltage

The RMS voltage \( V \) of the source is given by \( V = I Z \). Use the current \( I \) and impedance \( Z \) we calculated previously: \[ V = 0.447 \times 360.55 \approx 161.1 \, \text{V} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Factor
The power factor in an L-R-C circuit is a measure of how effectively the electrical power is being converted into useful work. It is defined as the ratio of the real power (the power actually used in the circuit) to the apparent power (the total power flowing through the circuit). In simpler terms, the power factor indicates how well the current is aligned with the voltage in the circuit.
For our circuit, the power factor (PF) is calculated using the formula \( \text{PF} = \frac{R}{Z} \), where \( R \) is the resistance and \( Z \) is the total impedance.
  • Resistance \( R = 300 \Omega \)
  • Total impedance \( Z \approx 360.55 \Omega \)
By substituting the values into the formula, we find that the power factor is approximately 0.832, meaning the circuit is fairly efficient but not perfectly so. The closer this value is to 1, the more efficient the circuit is.
Impedance Calculation
Impedance is a crucial concept when analyzing AC circuits. It combines resistance with reactance, which is the opposition to a change in current. Total impedance \( Z \) in an L-R-C circuit can be found with the formula \( Z = \sqrt{R^2 + X^2} \), where \( R \) is resistance and \( X \) is the total reactance derived from inductive and capacitive reactance.
To find reactance:
  • Inductive reactance \( X_L = 500 \Omega \)
  • Capacitive reactance \( X_C = 300 \Omega \)
So, total reactance \( X = X_L - X_C = 200 \Omega \). This results in:
  • \( Z = \sqrt{300^2 + 200^2} = \sqrt{130000} \approx 360.55 \Omega \)
By diligently adding these components, we get the impedance that precisely tells us about the circuit’s opposition to the AC flow.
Average Power Consumption
In an L-R-C series circuit, understanding how much power is being consumed is essential. The average power consumed is the real power used in the circuit, which can be calculated using the formula \( P = I^2 R \). In this formula, \( P \) represents the power, \( I \) is the RMS current, and \( R \) is the resistance.
Given:
  • Average power \( P = 60 \text{ W} \)
  • Resistance \( R = 300 \Omega \)
We can rearrange the formula to find the RMS current \( I \):
  • \( I = \sqrt{\frac{P}{R}} = \sqrt{\frac{60.0}{300}} \approx 0.447 \text{ A} \)
With this information, you can determine how much of the supplied power is doing meaningful work in the circuit.
RMS Voltage
The RMS (Root Mean Square) voltage is vital to understand because it reflects the effective value of the voltage in an AC circuit. It corresponds to the amount of DC voltage that would produce the same power in a resistor as the AC voltage.
In the problem, the RMS voltage \( V \) can be calculated using the impedance \( Z \) and the RMS current \( I \) with the formula \( V = I Z \).
  • RMS Current \( I \approx 0.447 \text{ A} \)
  • Impedance \( Z \approx 360.55 \Omega \)
Therefore, substituting these values gives:
  • \( V = 0.447 \times 360.55 \approx 161.1 \text{ V} \)
This voltage tells us about the potential difference that enables the correct amount of current to flow through the circuit, ensuring the desired power consumption.

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Most popular questions from this chapter

A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm\(^2\), mean radius 9.00 cm, and resistance \(R\) = 2.80 \(\Omega\). Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 Hz?

In an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance X\(_C\) of the capacitor is 200 \(\Omega\) and the voltage amplitude across the capacitor is 600 V. The circuit has \(R\) = 300 \(\Omega\). What is the voltage amplitude of the source?

In an \(L-R-C\) series circuit, the source has a voltage amplitude of 120 V, \(R\) = 80.0 \(\Omega\), and the reactance of the capacitor is 480 \(\Omega\). The voltage amplitude across the capacitor is 360 V. (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

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An \(L-R-C\) series circuit with \(L\) = 0.120 H, \(R\) = 240 \(\Omega\), and \(C\) = 7.30 \(\mu\)F carries an rms current of 0.450 A with a frequency of 400 Hz. (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? (g) In the inductor?

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