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\- The current and voltage outputs of an operating ac generator have peak values of \(2.5 \mathrm{~A}\) and \(16 \mathrm{~V}\), respectively. (a) What is the average power output of the generator? (b) What is the effective resistance of the circuit it is in?

Short Answer

Expert verified
(a) Average power output is approximately 20.02 W. (b) Effective resistance is approximately 6.39 Ω.

Step by step solution

01

Understand the Problem

We are tasked with finding the average power output of an AC generator with given peak current and voltage, and then determining the effective resistance of the circuit. We need to use the formulas for RMS (root mean square) values and the relationships involving power, current, voltage, and resistance.
02

Calculate the RMS Current and Voltage

The RMS (effective) value of current is given by the peak value divided by \(\sqrt{2}\).\[ I_{rms} = \frac{I_{peak}}{\sqrt{2}} = \frac{2.5}{\sqrt{2}} \approx 1.77 \text{ A} \] The RMS value of voltage is also obtained similarly.\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{16}{\sqrt{2}} \approx 11.31 \text{ V} \]
03

Calculate the Average Power Output

Using the formula for average power output in an AC circuit, \( P_{avg} = I_{rms} \times V_{rms} \). Substitute the RMS values calculated.\[ P_{avg} = 1.77 \text{ A} \times 11.31 \text{ V} \approx 20.02 \text{ W} \]
04

Determine the Effective Resistance

Using Ohm's law, \( R = \frac{V_{rms}}{I_{rms}} \), substitute the known RMS values to find resistance.\[ R = \frac{11.31 \text{ V}}{1.77 \text{ A}} \approx 6.39 \Omega \]

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Key Concepts

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

RMS Current
When dealing with alternating current (AC), it's important to consider the Root Mean Square (RMS) value rather than the peak value. This is because RMS values provide a more useful representation of the current's ability to do work, much like the way a DC equivalent would. The RMS current is calculated using the formula:\[I_{rms} = \frac{I_{peak}}{\sqrt{2}}\]This formula results from the statistical means by which AC current varies sinusoidally through time. Because it averages out the varying values over each cycle, it tells us the effective current that truly represents the power-delivering capability of an AC circuit. By using the peak current value of 2.5 A, the calculation yields an RMS current of approximately 1.77 A.
RMS Voltage
Similar to the RMS current, the Root Mean Square (RMS) voltage signifies the effective voltage in an AC circuit. RMS voltage gives a realistic measure of the AC voltage by equating its power-providing capacity to an equivalent DC voltage. The RMS voltage is calculated with:\[V_{rms} = \frac{V_{peak}}{\sqrt{2}}\]Here, the peak voltage value of 16 V is divided by \(\sqrt{2}\), providing an RMS voltage value of approximately 11.31 V. This value is practical for understanding and comparing the power dynamics in AC circuits, ensuring we account for how voltage swings periodically.
Average Power Output
The average power output in an AC circuit is vital for determining how much power the circuit can deliver. The formula for average power output is:\[P_{avg} = I_{rms} \times V_{rms}\]Inserting the previously calculated RMS current and voltage values gives us:\[P_{avg} = 1.77 \, \text{A} \times 11.31 \, \text{V} \approx 20.02 \, \text{W}\]This calculation allows us to find the real power dissipated by the circuit. This power, measured in watts, is what's truly harnessed and used, lending insightful understanding into its efficiency and performance.
Effective Resistance
The concept of effective resistance is derived from Ohm's Law, where resistance is the ratio of voltage to current. In AC circuits, we use RMS values to calculate effective resistance, ensuring that the measurements are realistic and consistent with real-world conditions. The formula applied is:\[R = \frac{V_{rms}}{I_{rms}}\]By using our RMS voltage of 11.31 V and RMS current of 1.77 A, we find the effective resistance to be approximately 6.39 Ω. Understanding effective resistance provides insight into the opposition a circuit offers to the flow of current, which is crucial for designing efficient circuits and ensuring safe operation.

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Most popular questions from this chapter

The current in a \(60-\Omega\) resistor is given by \(I=(2.0 \mathrm{~A}) \sin (380 t) .\) (a) What is the frequency of the current? (b) What is the rms current? (c) How much average power is delivered to the resistor? (d) Write an equation for the voltage across the resistor as a function of time. (e) Write an equation for the power delivered to the resistor as a function of time. (f) Show that the rms power obtained in part (e) is the same as your answer to part (c).

A single 50 -mH inductor forms a complete circuit when connected to an ac voltage source at \(120 \mathrm{~V}\) and \(60 \mathrm{~Hz}\) (a) What is the inductive reactance of the circuit? (b) How much current is in the circuit? (c) What is the phase angle between the current and the applied voltage? (Assume negligible resistance.)

A tuning circuit in a radio receiver has a fixed inductance of \(0.50 \mathrm{mH}\) and a variable capacitor. (a) If the circuit is tuned to a radio station broadcasting at \(980 \mathrm{kHz}\) on the AM dial, what is the capacitance of the capacitor? (b) What value of capacitance is required to tune into a station broadcasting at \(1280 \mathrm{kHz}\) ?

A series RLC radio receiver circuit with an inductance of \(1.50 \mu \mathrm{H}\) is tuned to an FM station at \(98.9 \mathrm{MHz}\) by adjusting a variable capacitor. When the circuit is tuned to this station, (a) what is its inductive reactance? (b) What is its capacitive reactance? (c) What is its capacitance?

A series RL circuit has a resistance of \(100 \Omega\) and an inductance of \(100 \mathrm{mH}\) and is driven by a \(120-\mathrm{V}, 60-\mathrm{Hz}\) source. (a) Find the inductive reactance and the impedance of the circuit. (b) How much current is drawn from the source?

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