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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

The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

In an \(L-R-C\) series circuit, R = 300 \(\Omega\), \(L\) = 0.400 H, and \(C\) = 6.00 \(\times\) 10\(^{-8}\) F. When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 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?

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A resistor with \(R = 300\space \Omega\) and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

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