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31\. 46. A circuit consists of a resistor and a capacitor in series with an ac source that supplies an rms voltage of 240 \(\mathrm{V}\) . At the frequency of the source the reactance of the capacitor is 50.0\(\Omega\) . The rms current in the circuit is 3.00 A. What is the average power supplied by the source?

Short Answer

Expert verified
The average power supplied by the source is approximately 562 W.

Step by step solution

01

Calculate Impedance of the Circuit

The total impedance \( Z \) of the circuit composed of resistance (\( R \)) and reactance (\( X_C \)) in series is calculated by the formula: \( Z = \sqrt{R^2 + X_C^2} \). Given: \( X_C = 50.0 \Omega \) and \( I = 3.00 \) A. We need to find \( R \) first from the current \( I \).
02

Find Resistance Using Ohm’s Law

With the given rms current \( I = 3.00 \) A and the rms voltage of 240 V, use Ohm's Law: \( V = I \cdot Z \). Rearranging gives \( Z = \frac{240}{3.00} = 80 \Omega \). Substitute this into the first equation to solve for \( R \): \( 80 = \sqrt{R^2 + 50^2} \). Solve to find \( R = \sqrt{80^2 - 50^2} = \sqrt{6400 - 2500} = \sqrt{3900} \approx 62.45 \Omega \).
03

Calculate Average Power Supplied

The average power \( P \) supplied by the source is given by \( P = I^2 R \). Substituting the known values, \( P = 3.00^2 \times 62.45 = 9 \times 62.45 \approx 562.05 \, \text{W} \).

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Key Concepts

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

Impedance
In AC circuits, impedance is crucial for analyzing the opposition that a circuit offers to the flow of alternating current (AC). It can be thought of as the AC equivalent of resistance. The total impedance \( Z \) in a circuit that includes both resistive (\( R \)) and reactive (\( X \)) components, such as inductors or capacitors, can be calculated using the formula: \[ Z = \sqrt{R^2 + X^2} \]In our exercise, we are dealing with a capacitive reactance, denoted as \( X_C \), which contributes to the total impedance. When a resistor and capacitor are in series, both their effects are combined to determine how the AC current will flow in the circuit.
It’s important to note that impedance is a complex quantity, combining both real (resistive) and imaginary (reactive) parts. In practical terms, impedance is measured in ohms (\(\Omega\)), just like resistance.
Reactance
Reactance, like resistance, is a property that impedes the flow of electric current, but it only occurs in AC circuits. It arises from the presence of capacitors or inductors, which store energy temporarily in the form of an electric or magnetic field, respectively. In an AC circuit with a capacitor, the reactance is termed capacitive reactance, represented by \( X_C \).
The capacitive reactance can significantly impact the behavior of circuits by introducing a phase difference between the voltage and current. Its value is influenced by both the capacitance value and the frequency of the AC source. The formula to find capacitive reactance is: \[ X_C = \frac{1}{2\pi f C} \]where \( f \) is the frequency and \( C \) is the capacitance. In the exercise at hand, \( X_C = 50 \Omega \) reflects how the capacitor reacts at the given frequency, contributing to the total impedance of the circuit. This reactance reduces as frequency increases, meaning capacitors offer less opposition at higher frequencies.
Ohm's Law
Ohm's Law is a fundamental principle used to calculate relationships between voltage, current, and resistance in electrical circuits. It is formulated as: \[ V = I \cdot R \]where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. In AC circuits, this can be extended to include impedance by substituting resistance with impedance, resulting in: \[ V = I \cdot Z \]In our problem, with an RMS voltage of 240 V and current of 3.00 A, we used this principle to find the total impedance: \[ Z = \frac{V}{I} = \frac{240}{3.00} = 80 \Omega \]
This calculated impedance includes both the resistive and reactive effects in the series circuit. Understanding Ohm's Law in both its DC and AC forms is essential for predicting circuit behavior and solving for unknown quantities like resistance in our exercise.
Average Power Calculation
Calculating the average power in AC circuits allows us to determine the power consumed or provided by a circuit over a complete cycle. The average power \( P \) in an AC circuit is calculated using the formula: \[ P = I^2 \cdot R \]Here, \( I \) is the RMS current and \( R \) is the resistance (not the impedance) in the circuit. Remember, while impedance includes reactance, for power calculations, only the resistive part is used because only resistors dissipate real power as heat.
In the given exercise, when the resistance \( R \) was found to be approximately \( 62.45 \Omega \), the average power was computed to be: \[ P = 3.00^2 \times 62.45 = 9 \times 62.45 \approx 562.05 \, \text{W} \]
This illustrates how AC power calculations depend significantly on accurately determining resistive values to find the power actually converted to work or heat in a circuit.

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Most popular questions from this chapter

31.37. A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}\) (mns) 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? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

31.22. An \(L-R-C\) series circuit is constructed using a \(175-\Omega\) resistor, a \(12.5-\mu \mathrm{F}\) capacitor, and an \(8.00-\mathrm{mH}\) inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 \(\mathrm{V}\) . (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part \((\mathrm{c}),\) how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

31.64. The Resonance width. Consider an \(L-R-C\) series circuit with a \(1.80-\mathrm{H}\) inductor, a \(0.900-\mu F\) capacitor, and a \(300-\Omega\) resistor. The source has terminal rms voltage \(V_{\mathrm{rms}}=60.0 \mathrm{V}\) and variable angular frequency \(\omega .\) (a) What is the resonance angular frequency \(\omega_{0}\) of the circuit? (b) What is the rms current through the circuit at resonance, \(I_{\max 0} ?(\mathrm{c})\) For what two values of the angular frequency, \(\omega_{1}\) and \(\omega_{2},\) is the rms current half the resonance value? (d) The quantity \(\left|\omega_{1}-\omega_{2}\right|\) defines the resonance width. Calculate \(I_{\text { rmid }}\) and the resonance width for \(R=300 \Omega, 30.0 \Omega,\) and \(3.00 \Omega .\) Describe how your results compare to the discussion in Section \(31.5 .\)

31.28. 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 nf 105\(\Omega\) . What average power is delivered to the circuit by the source?

31\. A8. At a frequency \(\omega_{1}\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_{2}=2 \omega_{1},\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega_{3}=\omega_{1} / 3,\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance \(R\) to form a series \(L-R-C\) circuit? what will be the resonance angular frequency of the circuit?

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