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\(\operatorname{An} L-R-C\) series 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 at this frequency of \(105 \Omega\). What average power is delivered to the circuit by the source?

Short Answer

Expert verified
Perform the above calculations and the value obtained in Step 2 would be the average power delivered to the circuit by the source.

Step by step solution

01

Calculation of RMS current

Using Ohm’s law, the root mean square (rms) current (\(I_{rms}\)) can be determined using the formula \(I_{rms}=V_{rms}/Z\). Given, \(V_{rms}=80.0 \mathrm{~V}\) and \(Z=105 \Omega\), thus \(I_{rms}=80.0 \mathrm{~V}/105 \Omega\).
02

Computing the Average Power

The average power (\(P_{avg}\)) can be calculated using the formula \(P_{avg}=I_{rms}^2 \times R\). Given \(R=75.0 \Omega\), thus \(P_{avg}=I_{rms}^2 \times 75.0 \Omega\). Here, \(I_{rms}^2\) would be the square of the value obtained in Step 1.

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Key Concepts

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

Impedance in an L-R-C Circuit
In an L-R-C circuit, the impedance is a crucial concept. It denotes the total opposition a circuit offers to the flow of alternating current (AC). Impedance, denoted by \( Z \), is a combination of resistance \( R \), inductive reactance \( X_L \), and capacitive reactance \( X_C \). These reactances add a layer of complexity as they depend on the frequency of the AC source.
Remember that in an L-R-C circuit:
  • Resistance \( R \): Represents opposition to current due to resistors.
  • Inductive Reactance \( X_L \): Opposition due to inductors, given by \( X_L = 2\pi f L \).
  • Capacitive Reactance \( X_C \): Opposition due to capacitors, with \( X_C = 1/(2\pi f C) \).
The impedance is not just a simple sum; it’s a vector sum because reactances can cancel each other out in part. Therefore, the formula for impedance in an L-R-C circuit is:\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]This combination affects not only the amplitude of the current but also its phase relative to the voltage.
Understanding Ohm's Law in AC Circuits
Ohm’s law is pivotal in electrical engineering and understanding basic electrical circuits. It relates voltage (\( V \)), current (\( I \)), and resistance (\( R \)) in simple circuits using the formula \( V = IR \).
In AC circuits, especially those involving L-R-C combinations, we replace resistance with impedance. Hence, the formula modifies to \( V = IZ \), where \( Z \) is the impedance.
  • This means the current is determined not just by the voltage but by the total impedance.
  • In practical terms, a higher impedance at a given voltage results in a lower current.
  • The root mean square (RMS) values are used because they give a representation of the effect of AC as DC would do.
For our exercise, we have the RMS voltage \( V_{rms} \) and impedance \( Z \). Thus, by rearranging Ohm’s law for AC, we calculate the RMS current as: \[I_{rms} = \frac{V_{rms}}{Z}\] This fundamental concept helps in understanding the behavior and analysis of complex AC circuits.
Average Power Calculation in AC Circuits
Calculating the average power in an AC circuit like an L-R-C circuit involves taking into account the resistance and the RMS current. Power in electrical terms often refers to the rate of energy transfer.
In AC circuits, the average power can be calculated using:\[P_{avg} = I_{rms}^2 \times R\] Here, \( R \) is the resistance in the circuit. It signifies how much of the current actually does work (e.g., lights up bulbs or heats elements).
  • Only the resistive part of the impedance contributes to the power used.
  • Any reactive part like inductive or capacitive reactance will not dissipate power, instead they store and return it.
  • This calculation illustrates how effective power needs knowledge of not just current but the resistive opposition in the circuit.
By squaring the RMS current and multiplying by the resistance, you find the average power, emphasizing practical use of power rather than just theoretical calculations.

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Most popular questions from this chapter

A series circuit has an impedance of \(60.0 \Omega\) and a power factor of 0.720 at \(50.0 \mathrm{~Hz}\). The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

A capacitance \(C\) and an inductance \(L\) are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If \(L=5.00 \mathrm{mH}\) and \(C=3.50 \mu \mathrm{F}\), what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

An \(L-R-C\) series circuit has \(R=400 \Omega, L=0.600 \mathrm{H},\) and \(C=5.00 \times 10^{-8} \mathrm{~F}\). The voltage amplitude of the source is \(80.0 \mathrm{~V}\). When the ac source operates at the resonance frequency of the circuit, what is the average power delivered by the source?

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

Analyzing an \(L-R-C\) Circuit. You have a \(200 \Omega\) resistor, a \(0.400 \mathrm{H}\) inductor, a \(5.00 \mu \mathrm{F}\) capacitor, and a variable- frequency ac source with an amplitude of \(3.00 \mathrm{~V}\). You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of \(400 \mathrm{rad} / \mathrm{s} ?\) At this frequency, will the source voltage lead or lag the current?

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