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31.30. A series ac circuit contains a \(250-\Omega\) resistor, a \(15-\mathrm{mH}\) inductor, a 3.5\(\mu \mathrm{F}\) capacitor, and an ac power source of voltage amplitude 45 \(\mathrm{V}\) operating at an angular frequency of 360 \(\mathrm{rad} / \mathrm{s}\) , (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

Short Answer

Expert verified
(a) Power factor: 0.3027. (b) Average power to circuit: 0.371 W. (c) Average power: Resistor 0.372 W, Capacitor 0 W, Inductor 0 W.

Step by step solution

01

Calculate Inductive and Capacitive Reactance

First, calculate the inductive reactance \( X_L \) using the formula \( X_L = \omega L \), where \( \omega = 360 \, \text{rad/s} \) and \( L = 15 \, \text{mH} = 0.015 \, \text{H} \). Thus, \( X_L = 360 \times 0.015 = 5.4 \, \Omega \).Next, calculate the capacitive reactance \( X_C \) using the formula \( X_C = \frac{1}{\omega C} \), where \( C = 3.5 \, \mu \text{F} = 3.5 \times 10^{-6} \, \text{F} \). Thus, \( X_C = \frac{1}{360 \times 3.5 \times 10^{-6}} = 795.24 \, \Omega \).
02

Calculate Impedance of the Circuit

The total impedance \( Z \) in the series circuit is given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Here, \( R = 250 \, \Omega \), \( X_L = 5.4 \, \Omega \), and \( X_C = 795.24 \, \Omega \).So, \( Z = \sqrt{250^2 + (5.4 - 795.24)^2} = \sqrt{250^2 + (-789.84)^2} = 826.31 \, \Omega \).
03

Calculate Power Factor

The power factor \( \cos \phi \) is the cosine of the phase angle \( \phi \), which is the ratio of the resistance to the impedance: \( \cos \phi = \frac{R}{Z} \).Using \( R = 250 \, \Omega \) and \( Z = 826.31 \, \Omega \), we find \( \cos \phi = \frac{250}{826.31} = 0.3027 \).
04

Calculate Average Power Delivered to Circuit

The average power \( P \) is given by \( P = V_m \cdot I_m \cdot \cos \phi = \frac{1}{2} V_0 I_0 \cos \phi \), where \( V_0 = 45 \, \text{V} \) is the voltage amplitude. First, find the current amplitude \( I_0 \) using \( I_0 = \frac{V_0}{Z} = \frac{45}{826.31} = 0.0545 \, \text{A} \).Then, compute \( P = \frac{1}{2} \times 45 \times 0.0545 \times 0.3027 = 0.371 \, \text{W} \).
05

Average Power Delivered to Each Component

The average power delivered to the resistor is given by \( P_R = \frac{1}{2} R I_0^2 = \frac{1}{2} \times 250 \times (0.0545)^2 = 0.372 \, \text{W} \).The average power delivered to the capacitor and inductor is zero because in ideal AC circuits, capacitors and inductors only store and return energy, but do not dissipate it. Hence, \( P_C = P_L = 0 \, \text{W} \).

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Key Concepts

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

Power Factor
In alternating current (AC) circuits, the power factor is key to understanding how effectively the circuit uses electricity. It is defined as the cosine of the phase difference \( \phi \) between the voltage and the current. The power factor helps indicate how much of the power is used effectively in doing real work.

When the circuit's current and voltage are perfectly in phase, the power factor is 1. This ideal scenario means all the power is being used efficiently. But, in many AC circuits, especially those containing inductors and capacitors, there can be a phase difference.
  • A good analogy is a cup of coffee where the liquid is the real power (the part that does work) and the foam is the reactive power; the actual usable contents depend on the ratio of coffee to foam.
  • A low power factor, as seen in circuits with high reactance, indicates inefficiency since more current is needed to supply the same amount of resistive power.
For our specific circuit, the power factor was calculated as 0.3027, showing that less than a third of the supplied power is being efficiently used.
Inductive Reactance
Inductive reactance \( X_L \) is a measure of the opposition that an inductor presents to alternating current. It depends on the frequency of the AC signal and the value of the inductance.

The formula \( X_L = \omega L \) shows that the reactance increases with frequency \( \omega \) and inductance \( L \).
  • It is measured in ohms (\( \Omega \)), and it reflects how the energy is stored temporarily in the magnetic field around the inductor.
  • In the problem example, with an angular frequency of 360 rad/s and an inductance of 15 mH, the inductive reactance is calculated as 5.4 \( \Omega \).
This value indicates how significantly the inductor resists the AC, influencing the overall behavior of the circuit.
Impedance Calculation
Impedance \( Z \) is the total opposition that a circuit presents to the flow of alternating current. It combines both resistance and reactance (capacitive and inductive) and is a fundamental property in AC circuit analysis.

Impedance can be expressed as a combination of resistance \( R \), inductive reactance \( X_L \), and capacitive reactance \( X_C \): \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
  • Resistance \( R \) is constant regardless of AC frequency, whereas reactance varies with frequency.
  • For our example, the impedance was found to be approximately 826.31 \( \Omega \), considering the given resistance, inductive, and capacitive reactance.
Calculating impedance is crucial for determining current flow and power distribution in the circuit.
Average Power in AC Circuits
The average power of an AC circuit quantifies the amount of actual power consumed over time. It differs from the instantaneous power, reflecting the power use over a complete cycle.

In a circuit with impedance and a non-zero power factor, average power \( P \) is given by: \[ P = \frac{1}{2} V_0 I_0 \cos \phi \] where \( V_0 \) is the voltage amplitude, \( I_0 \) is the current amplitude, and \( \cos \phi \) is the power factor.
  • The average power tells us how much energy is converted to actual work or dissipated as heat, particularly in the resistive elements of the circuit.
  • For circuits with only capacitive or inductive elements, the average power is often zero as they store but do not consume energy.
In the discussed circuit, the average power delivered was calculated as 0.371 W, which aligns with the role of resistance in consuming power.

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Most popular questions from this chapter

\(\operatorname{An} L-R-C\) series circuit consists of a \(2.50-\mu \mathrm{F}\) capacitor, a 5.00-mH inductor, and a \(75.0-\Omega\) resistor connected across an ac source of voltage amplitude 15.0 V having variable frequency. (a) Under what circumstances is the average power delivered to the circuit equal to \(\frac{1}{2} V_{\mathrm{rms}} I_{\mathrm{rma}} ?\) (b) Under the conditions of part (a), what is the average power delivered to each circuit element and what is the maximum current through the capacitor?

31.9. (a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ?

31.23. In an \(L-R-C\) series circuit, the rms voltage across the resistor is 30.0 \(\mathrm{V}\) , across the capacitor it is 90.0 \(\mathrm{V}\) , and across the inductor it is 50.0 \(\mathrm{V}\) . What is the rms voltage of the source?

31.34. In an \(L-R-C\) series circuit, \(R=400 \Omega, L=0.350 \mathrm{H},\) and \(C=0.0120 \mu F .\) (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 550 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitnde can it have if the maximum capacitor voltage is not exceeded?

31.10. A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

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