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Chapter 31: Problem 18

A resistor with \(R=300 \Omega\) and an inductor are connected in series across an ac source that has voltage amplitude 500 \(\mathrm{V}\) . The rate at which electrical energy is dissipated in the resistor is 216 \(\mathrm{W}\) (a) What is the impedance \(Z\) of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

Short Answer

Expert verified
(a) Impedance \( Z = 589.32 \, \Omega \) (b) Voltage across the inductor \( V_L = 430.51 \, \mathrm{V} \) (c) Power factor \( \cos(\phi) = 0.509 \)

Step by step solution

01

Calculate the current in the circuit

The power dissipated in the resistor can be calculated using the formula for power in resistive circuits: \[ P = I^2 R \]Given that \(P = 216 \, \mathrm{W}\) and \(R = 300 \, \Omega\), we can solve for the current \(I\):\[ I^2 = \frac{P}{R} = \frac{216}{300} = 0.72 \]\[ I = \sqrt{0.72} = 0.8485 \, \mathrm{A} \]
02

Calculate the impedance of the circuit

The total impedance \(Z\) in a circuit with voltage amplitude \(V_m\) and current amplitude \(I\) is given by:\[ Z = \frac{V_m}{I} \]Given \(V_m = 500 \, \mathrm{V}\) and \(I = 0.8485 \, \mathrm{A}\):\[ Z = \frac{500}{0.8485} = 589.32 \, \Omega \]
03

Calculate the amplitude of the voltage across the inductor

For a series R-L circuit, \(V_m^2 = V_R^2 + V_L^2\), where \(V_R = IR\) and \(V_L\) is the amplitude of the voltage across the inductor. Solve for \(V_L\):\[ V_R = I \cdot R = 0.8485 \, \mathrm{A} \times 300 \, \Omega = 254.55 \, \mathrm{V} \]\[ V_m^2 = V_R^2 + V_L^2 \500^2 = 254.55^2 + V_L^2 \250000 = 64722.7 + V_L^2 \V_L^2 = 250000 - 64722.7 = 185277.3 \V_L = \sqrt{185277.3} = 430.51 \, \mathrm{V} \]
04

Calculate the power factor

The power factor (PF) is the cosine of the phase angle \(\phi\), given by:\[ \cos(\phi) = \frac{R}{Z} \]Substituting the values for \(R\) and \(Z\):\[ \cos(\phi) = \frac{300}{589.32} = 0.509 \]

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Key Concepts

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

Resistor-Inductor Series Circuit
In AC circuits, understanding the behavior of components such as resistors and inductors is crucial. When a resistor and an inductor are connected in series, they form what is commonly referred to as an RL circuit. This type of circuit exhibits distinct characteristics because it combines both resistive and inductive properties. The resistor limits current flow, producing heat as electrical energy is dissipated. On the other hand, the inductor resists changes in current flow due to its property of inductance, which stores energy in a magnetic field.
  • Resistor (R): Converts electrical energy into heat.
  • Inductor (L): Temporarily stores energy as a magnetic field.
These components, when combined, affect the total impedance of the circuit and influence current flow, making the analysis of AC circuits slightly complex compared to DC circuits.
Electrical Power Dissipation
In electrical circuits, power dissipation refers to the process by which electrical energy is converted to heat. In resistors, this phenomenon is common and can be calculated using the formula: \[ P = I^2 R \]Here, \(P\) is the power dissipation, \(I\) is the current, and \(R\) is the resistance. This formula shows that power dissipation is directly proportional to the square of the current and the resistance.
  • Increased Current: Results in higher power dissipation.
  • Resistance: The higher the resistance, the more power dissipated.
Understanding power dissipation is vital to prevent overheating in electrical components, ensuring they operate within safe limits.
Circuit Impedance
Circuit impedance is a fundamental concept in AC circuit analysis. Impedance encompasses both resistance and reactance in a circuit, representing the total opposition that the circuit presents to the flow of alternating current. It is measured in Ohms (\(\Omega\)) and denoted by the symbol \(Z\). The impedance in a series resistor-inductor circuit is calculated using:\[ Z = \frac{V_m}{I} \]Where \(V_m\) is the voltage amplitude and \(I\) is the current amplitude. Impedance dictates how much of the input voltage affects the circuit, hence playing a critical role in circuit design and analysis.
Power Factor Calculation
The power factor is a measure of how effectively electrical power is being used in a circuit. It is expressed as the cosine of the phase angle \(\phi\) between the voltage and current:\[ \text{Power Factor (PF)} = \cos(\phi) = \frac{R}{Z} \]A power factor of 1 indicates that all the power is being effectively converted into work, with no reactive losses. In contrast, a lower power factor signifies inefficiency where more power is being wasted as heat or stored and released by inductive or capacitive components, leading to reactive power.
  • Unity Power Factor: Ideal, indicating minimum losses.
  • Lagging Power Factor: Suggests an inductive load with poor efficiency.
Optimizing power factor is crucial for improving efficiency in electrical systems.

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Most popular questions from this chapter

\(\mathrm{A} 200-\Omega\) resistor, a \(0.900-\mathrm{H}\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor are connected in series across a voltage source that has voltage amplitude 30.0 \(\mathrm{V}\) and an angular frequency of 250 \(\mathrm{rad} / \mathrm{s}\) . (a) What are \(v, v_{R}, v_{L},\) and \(v_{C}\) at \(t=20.0 \mathrm{ms}\) ? Compare \(v_{R}+\) \(v_{L}+v_{C}\) to \(v\) at this instant. (b) What are \(V_{R}, V_{L},\) and \(V_{C} ?\) Compare \(V\) to \(V_{R}+V_{L}+V_{C} .\) Explain why these two quantities are not equal.

In an \(L-R-C\) series circuit the magnitude of the phase angle is \(54.0^{\circ},\) with the source voltage lagging the current. The reactance of the capacitor is \(350 \Omega,\) and the resistor resistance is 180\(\Omega .\) The average power delivered by the source is 140 \(\mathrm{W} .\) Find (a) the reactance of the inductor; (b) the rms current; (c) the rms voltage of the source.

An \(L-R-C\) series circuit consists of a \(50.0-\Omega\) resistor, a \(10.0-\mu \mathrm{F}\) capacitor, a \(3.50-\mathrm{mH}\) inductor, and an ac voltage source of voltage amplitude 60.0 \(\mathrm{V}\) operating at 1250 \(\mathrm{Hz}\) . (a) Find the current amplitude and the voltage amplitudes across the inductor, the resistor, and the capacitor. Why can the voltage amplitudes add up to more than 60.0 \(\mathrm{V}\) (b) If the frequency is now doubled, but nothing else is changed, which of the quantities in part (a) will change? Find the new values for those that do change.

An \(L-R-C\) series circuit has \(C=4.80 \mu \mathrm{F}, L=0.520 \mathrm{H}\) and source voltage amplitude \(V=56.0 \mathrm{V}\) . The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{V},\) what is the value of \(R\) for the resistor in the circuit?

In an \(L-R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=\) 0.0360\(\mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?
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