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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
The average power delivered is approximately 60.95 W.

Step by step solution

01

Identify Known Values

First, list the known values given in the problem:- The root mean square voltage, \(V_{\mathrm{rms}} = 80.0 \text{ V}\).- The resistance, \(R = 75.0 \Omega\).- The impedance, \(Z = 105 \Omega\).
02

Use Power Formula

To find the average power delivered to the circuit, we use the formula:\[P = \frac{V_{\mathrm{rms}}^2}{Z}\]where \(P\) is the average power, \(V_{\mathrm{rms}}\) is the RMS voltage, and \(Z\) is the impedance.
03

Substitute Values Into the Formula

Substitute the known values into the power formula:\[P = \frac{(80.0 \text{ V})^2}{105 \Omega}\]Calculate the square of the voltage, \((80.0 \text{ V})^2 = 6400 \text{ V}^2\).
04

Calculate Average Power

Now, divide the squared voltage by the impedance:\[P = \frac{6400 \text{ V}^2}{105 \Omega} \approx 60.95 \text{ W}\]Therefore, the average power delivered to the circuit is approximately \(60.95 \text{ W}\).

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Key Concepts

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

Average Power Calculation
Understanding how to calculate average power in an L-R-C circuit can help you understand energy consumption by the circuit. The average power, also known as true power, represents the actual electrical power used by the circuit.
In an AC (Alternating Current) circuit, power continuously fluctuates, unlike in a DC (Direct Current) circuit where power remains constant. To find the average, or real power consumed, we use the formula:
  • \(P = \frac{V_{\text{rms}}^2}{Z}\), where \(P\) is the power in watts, \(V_{\text{rms}}\) is the root mean square voltage, and \(Z\) is the total impedance of the circuit.
This formula calculates power using the effective voltage across the circuit and the total opposition (impedance) to current flow. It tells us how much power is actually "used" or "consumed" over time.
From the given example that used \(V_{\text{rms}}=80.0 \, \text{V}\) and \(Z = 105 \, \Omega\), you can see the average power computed as \(60.95 \, \text{W}\). Be mindful of the impedance value, as it affects the power delivered by the AC source significantly.
Impedance in Circuits
Impedance is a crucial concept in AC circuits as it combines both resistance and reactance (inductive and capacitive). It represents the total opposition that a circuit presents to the flow of alternating current. Unlike resistance, which only applies to direct current, impedance applies to alternating current and can have both magnitude and phase.In simple terms, impedance tells us how much the circuit resists the flow of AC at any frequency. It is measured in ohms (\(\Omega\)) and can be calculated using the formula:
  • \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)
where:
  • \(R\) is the resistance
  • \(X_L\) is the inductive reactance
  • \(X_C\) is the capacitive reactance
In an L-R-C circuit, this total impedance is essential for determining how much power the circuit uses and its overall performance.
The given impedance value in the example was \(105 \, \Omega\), which marked the combined opposition to current flow due to resistance and reactance in the circuit. Because impedance provides not only magnitude but also phase information, it plays a vital role in power calculations and system behavior analysis.
RMS Voltage in AC Circuits
Root mean square (RMS) voltage is a way to express the effective voltage of an alternating current (AC) source. In AC circuits, the voltage continuously changes with time, making it necessary to have a value that represents its effective amplitude.The RMS voltage is particularly useful because it allows you to compare AC to a direct current (DC) voltage. It essentially represents the equivalent DC voltage that would produce the same power in a resistive load. The RMS value is derived from the waveform and is calculated as:
  • \(V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}\) for a sinusoidal waveform.
This calculation gives us a steady voltage value that makes it easier to compute average power, as seen in the exercise example with \(V_{\text{rms}} = 80.0 \, \text{V}\).
In essence, RMS voltage accounts for the fact that the voltage varies over time, and provides a single value that can be used for straightforward calculations in power and energy equations.

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Most popular questions from this chapter

You want to double the resonance angular frequency of an \(L-R-C\) series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

A series \(\mathrm{RC}\) circuit with \(R=10 \Omega\) has an impendance with phase angle of \(45^{\circ}\) at frequency \(f=500 \mathrm{~Hz}\). Find the frequency for which the magnitude of the impedance is (a) twice that off, (b) one-half that at \(f\).

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?

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{~V}, R=80.0 \Omega\), and the reactance of the capacitor is \(480 \Omega .\) The voltage amplitude across the capacitor is \(360 \mathrm{~V}\). (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(12.0 \mathrm{~V}\) (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?

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