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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

In an \(L-R-C\) series ac circuit, the source has a voltage amplitude of 240 V, \(R\) = 90.0 \(\Omega\), and the reactance of the inductor is 320 \(\Omega\). The voltage amplitude across the resistor is 135 V. (a) What is the current amplitude in the circuit? (b) What is the voltage amplitude across the inductor? (c) What two values can the reactance of the capacitor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

A large electromagnetic coil is connected to a 120-Hz ac source. The coil has resistance 400 \(\Omega\), and at this source frequency the coil has inductive reactance 250 \(\Omega\). (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electrical power of 450 W?

An \(L-R-C\) series circuit has R = 500 \(\Omega\), L = 2.00 H, \(C\) = 0.500 \(\mu\)F, and \(V\) = 100 V. (a) For \(\omega\) = 800 rad/s, calculate \(V_R , V_L, V_C\), and \(\phi\). Using a single set of axes, graph \(v\), \(v_R , v_L\), and \(v_C\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part (a) for \(\omega\) = 1000 rad/s. (c) Repeat part (a) for \(\omega = 1250\space rad/s\).

An \(L-R-C\) series circuit is constructed using a 175-\(\Omega\) resistor, a 12.5-\(\mu\)F capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

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