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Chapter 31: Problem 72

An \(L-R-C\) series circuit draws 220 \(\mathrm{W}\) from a \(120-\mathrm{V}\) (rms), 50.0 -Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacistor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

Short Answer

Expert verified
(a) R 鈮� 20.5 惟, (b) C 鈮� 106 碌F, (c) P 鈮� 392.4 W

Step by step solution

01

Identify Given Values and Relevant Equations

We know the real power \( P = 220 \, \mathrm{W} \), the rms voltage \( V_{rms} = 120 \, \mathrm{V} \), frequency \( f = 50.0 \, \mathrm{Hz} \), and power factor \( \text{pf} = 0.560 \). Use the formula for power in an AC circuit: \( P = V_{rms} \cdot I_{rms} \cdot \cos(\theta) \) where \( \cos(\theta) \) is the power factor and \( I_{rms} \) is the rms current.
02

Calculate RMS Current

Rearrange the power formula to find \( I_{rms} \): \[ I_{rms} = \frac{P}{V_{rms} \cdot \cos(\theta)} = \frac{220}{120 \times 0.560} \approx 3.27 \, \mathrm{A} \]
03

Determine Net Resistance R

Use Ohm's law for AC circuits \( V_{rms} = I_{rms} \cdot Z \), and solve it for impedance \( Z \): \( Z = \frac{V_{rms}}{I_{rms}} \). Then, the resistance \( R \) can be found using \( R = Z \cdot \cos(\theta) \). Calculate, \[ Z = \frac{120}{3.27} \approx 36.7 \, \Omega \] and \[ R = 36.7 \times 0.560 \approx 20.5 \, \Omega \]
04

Calculate Required Capacitance for Unity Power Factor

To achieve a unity power factor, the impedance must be purely resistive, meaning the inductive reactance \( X_L \) and capacitive reactance \( X_C \) must cancel out. Calculate total reactance \( X = \sqrt{Z^2 - R^2} \), then use \( X_C = \frac{1}{2 \pi f C} = X \).\[ X = \sqrt{36.7^2 - 20.5^2} \approx 30.1 \]\[ \frac{1}{2 \pi \times 50 \times C} = 30.1 \Rightarrow C \approx 106 \times 10^{-6} \mathrm{F} \]
05

Calculate New Power with Unity Power Factor

The new power is calculated using the same formula as before but with a power factor of unity: \( P = V_{rms} \cdot I_{rms} \cdot 1 \). Hence the new power is \[ P = 120 \times 3.27 \approx 392.4 \, \mathrm{W} \]

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Key Concepts

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

Power Factor Correction
Power factor correction is a crucial aspect in enhancing the efficiency of AC circuits. It involves adjusting the power factor of the electrical system to make it as close to 1 as possible. A power factor of 1, or unity, means that the circuit is entirely resistive, and all the power supplied by the source is effectively used for work.
In the case of an L-R-C circuit, the power factor is influenced by the presence of reactive components like inductors and capacitors. These elements do not consume power but create a phase difference between the voltage and current. The power factor correction aims to eliminate this phase difference, often by adding capacitors or inductors to balance out the effect of the circuit's reactance.
  • With a low power factor, more current is required to deliver the same amount of real power as compared to a circuit with a higher power factor.
  • This can lead to higher losses in the electrical system and increased load on generators and transformers.
  • Correcting the power factor can reduce energy costs and increase the electrical system's overall stability and efficiency.
Impedance in AC Circuits
Impedance is a term used in AC circuits to describe the total opposition a circuit presents to the flow of alternating current. It is a complex quantity, involving both resistance, which dissipates real power, and reactance, which stores and releases energy.
In our L-R-C circuit, impedance is crucial because it combines all elements of resistance and reactance to define the circuit's behavior. While resistance (R) is associated with energy dissipation, inductive reactance (\(X_L = 2\pi fL \)) relates to energy storage in a magnetic field, and capacitive reactance (\(X_C = \frac{1}{2\pi fC} \)) is tied to energy storage in an electric field.
  • Impedance in an AC circuit can vary with frequency, meaning as the frequency changes, the circuit's response to AC signals can alter significantly.
  • This property makes devices like filters and tuners possible.
  • Along with phase angle (\(\theta \)), impedance determines how much current flows through the circuit for a given voltage.
Calculation of Reactance
Reactance is the measure of opposition that inductors and capacitors present to the change in current in AC circuits. Like resistance, reactance affects how currents behave in the circuit, but unlike resistance, it does not dissipate energy.
The total reactance in our circuit can be calculated using the difference between inductive reactance (\(X_L = 2\pi fL \)) and capacitive reactance (\(X_C = \frac{1}{2\pi fC} \)). These values affect the total impedance (\(Z\)) of the circuit, where \[X = X_L - X_C\] represents the net reactive factor.
  • The goal is to manage these two kinds of reactance such that they cancel each other out, leading to a purely resistive impedance.
  • This cancellation is crucial for achieving a power factor of \(1\), optimizing efficiency.
  • The calculated values for reactance are key to designing circuits that respond predictably to changes in frequency.

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Most popular questions from this chapter

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

off to Europe! You plan to take your hair dryer to Europe, where the electrical outlets put out 240 \(\mathrm{V}\) instead of the 120 \(\mathrm{V}\) seen in the United States. The dryer puts out 1600 \(\mathrm{W}\) at 120 \(\mathrm{V}\) . (a) What could you do to operate your dryer via the \(240 \mathrm{V}\) line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at 240 \(\mathrm{V} ?\)

In an \(L-R-C\) series 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.

A coil has a resistance of 48.0\(\Omega .\) At a frequency of 80.0 \(\mathrm{Hz}\) the voltage across the coil leads the current in it by \(52.3^{\circ} .\) Determine the inductance of the coil.

The \(L \cdot R-C\) Parallel Circuit. A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega\) . Let the source voltage be given by \(v=V \cos \omega t .\) (a) Show that the instantaneous voltages \(v_{R}, v_{L},\) and \(v_{C}\) at any instant are each equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C},\) where \(i\) is the current through the source and \(i_{R},\) \(i_{L},\) and \(i_{C}\) are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L},\) and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L},\) and \(i_{C}\) In a phasor diagram, show the phases of these four currents with respect to \(v\) . (c) Use the phasor diagram of part (b) to show that the current amplitude \(I\) for the current \(i\) through the source is given by \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\) (d) Show that the result of part (c) can be written as \(I=V / Z,\) with \(1 / Z=\sqrt{1 / R^{2}+(\omega C-1 / \omega L)^{2}}\)
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