/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A series \(R-L-C\) circuit is co... [FREE SOLUTION] | 91Ó°ÊÓ

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A series \(R-L-C\) circuit is connected to a \(120 \mathrm{~Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{~V}\). The circuit has a resistance of \(75.0 \Omega\) and an impedance of \(105 \Omega\) at this frequency. What average power is delivered to the circuit by the source?

Short Answer

Expert verified
The average power delivered is approximately 43.52 W.

Step by step solution

01

Identify Given Values

First, collect all the information given in the problem. We have: \( V_{\text{rms}} = 80.0 \, \text{V} \), resistance \( R = 75.0 \, \Omega \), and impedance \( Z = 105 \, \Omega \).
02

Understand the Relationship Between Variables

In an \( R-L-C \) circuit, the average power \( P_{\text{avg}} \) delivered by the source is given by the formula: \( P_{\text{avg}} = V_{\text{rms}}^2 \frac{R}{Z^2} \).
03

Substitute the Known Values into the Formula

Using the formula and the given values, substitute into the equation: \( P_{\text{avg}} = (80.0)^2 \frac{75.0}{105^2} \).
04

Calculate the Average Power

Perform the calculations: - Calculate \( (80.0)^2 = 6400 \).- Calculate \( 105^2 = 11025 \).- Compute the fraction \( \frac{75.0}{11025} \approx 0.0068 \).- Finally, compute \( 6400 \times 0.0068 \approx 43.52 \, \text{W} \).
05

Conclusion

The average power delivered to the \( R-L-C \) circuit by the source is approximately \( 43.52 \, \text{W} \).

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Key Concepts

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

Average Power Calculation in RLC Circuits
The average power calculation in an RLC circuit is an essential concept that helps determine how much power is consumed from the AC source. In such circuits, the average power is calculated using the root mean square (RMS) voltage, resistance, and impedance.
To find the average power \( P_{\text{avg}} \) delivered, we use the formula:\[ P_{\text{avg}} = V_{\text{rms}}^2 \times \frac{R}{Z^2} \]
where:
  • \( V_{\text{rms}} \) is the RMS voltage of the source.
  • \( R \) is the resistance in the circuit.
  • \( Z \) is the total impedance, which accounts for resistance, inductance, and capacitance.
The key to successful average power calculation is correctly substituting these values and conducting the arithmetic operations accurately.
In RLC circuits, actual power used by the circuit is less than what could be computed using only RMS voltage and resistance, due to the presence of reactance components. Therefore, using impedance ensures the calculations account for these components.
This step effectively provides the true power consumed across the resistive elements of the circuit.
Impedance in Circuits
Impedance is a central concept when analyzing AC circuits like the series RLC circuit. It is similar to resistance in DC circuits but also considers the effects of inductance and capacitance.
Impedance, denoted by \( Z \), is a complex quantity and can be expressed using both magnitude and phase angle:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where:
  • \( R \) stands for resistance.
  • \( X_L = 2\pi fL \) is the inductive reactance.
  • \( X_C = \frac{1}{2\pi fC} \) is the capacitive reactance.
Impedance influences the circuit’s current flow from the AC source.
As the frequency of the AC source changes, so do the reactances \( X_L \) and \( X_C \). This affects the total impedance, and thus how much current and power flow through the circuit.
In practical applications, engineers and technicians often have to measure impedance to ensure circuit components function correctly and efficiently.
AC Circuit Analysis
AC circuit analysis is the process of evaluating circuits powered by alternating current (AC) sources. In these circuits, both voltage and current vary sinusoidally over time at a given frequency. Unlike DC circuits, where current flows in one direction, AC circuits have current reversing direction regularly.
The main goals of AC circuit analysis are to determine:
  • The voltages across, and currents through, different components.
  • How impedance affects circuit behavior.
  • Power distribution within the circuit.
To carry out AC circuit analysis, important principles such as Ohm’s law for AC \(( V = IZ )\) and the concepts of phase angle and power factor become crucial.
AC circuit analysis also depends on understanding how circuit components—resistors, capacitors, and inductors—respond to AC signals.
These responses are dictated by the inductive and capacitive reactances as well as the resistive components, which altogether form impedance. Thus, analyzing how these components interact with the AC source allows for comprehensive circuit behavior prediction.
Mastering AC circuit analysis enables effective design and troubleshooting across a range of electrical systems, from household appliances to complex industrial machinery.

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Most popular questions from this chapter

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?

A \(1500 \Omega\) resistor is connected in series with a \(350 \mathrm{mH}\) inductor and an ac power supply. At what frequency will this combination have twice the impedance that it has at \(120 \mathrm{~Hz}\) ?

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

A \(200 \Omega\) resistor is in series with a \(0.100 \mathrm{H}\) inductor and a \(0.500 \mu \mathrm{F}\) capacitor. Compute the impedance of the circuit and draw the phasor diagram (a) at a frequency of \(500 \mathrm{~Hz}\), (b) at a frequency of \(1000 \mathrm{~Hz}\). In each case, compute the phase angle of the source voltage with respect to the current and state whether the source voltage lags or leads the current.

An electric motor is being powered with a voltage amplitude of \(310 \mathrm{~V}\) at \(60.0 \mathrm{~Hz}\). The motor draws a current amplitude of \(10.0 \mathrm{~A}\). Find (a) the root-mean-square voltage, (b) the root-mean- square current, and (c) the average power consumed by the motor.

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