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An ac generator produce a peak voltage of 55 V. What is the rms voltage of this generator?

Short Answer

Expert verified
The RMS voltage is approximately 38.89 V.

Step by step solution

01

Understanding Peak Voltage

The peak voltage of an AC generator is the maximum voltage it can produce. It is given as 55 V.
02

Formula for RMS Voltage

RMS voltage (V_{rms}) can be calculated from peak voltage (V_{peak}) using the formula: \[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \] This formula is derived from the fact that RMS voltage is the square root of the mean of the squares of all instantaneous values.
03

Calculate RMS Voltage

Using the formula, substitute the peak voltage value (55 V) to find the RMS voltage: \[ V_{rms} = \frac{55}{\sqrt{2}} \approx 38.89 \text{ V} \].

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Key Concepts

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

AC Generator
An AC generator, also known as an alternating current generator, is a device that converts mechanical energy into electrical energy in the form of alternating current. This transformation is achieved through the process of electromagnetic induction, where a coil of wire rotates in a magnetic field to generate electricity.

Key characteristics of an AC generator:
  • It produces a sinusoidal output voltage, which means the voltage changes in a smooth, continuous wave pattern over time.
  • The frequency of this voltage depends on the speed of the rotor and the number of magnetic poles in the generator.
  • Commonly used in power plants to generate electricity for homes and industries.
An AC generator's voltage continuously fluctuates from zero to its maximum or peak value, both in the positive and negative directions, as it forms a complete cycle. Understanding how it operates helps in comprehending how devices and systems handle alternating current efficiently.
Peak Voltage
Peak voltage is a term used to describe the maximum voltage that an AC generator or any alternating current system can produce. It represents the highest point in the sine wave cycle of AC voltage. In our exercise, the peak voltage is noted as 55 V.

Importance of peak voltage:
  • It indicates the absolute potential difference between the highest point of the waveform and the zero potential line.
  • While peak voltage is important for understanding the extremes of the voltage range, it does not fully represent the effective power of the AC signal.
  • Peak voltage is used in calculating the RMS voltage, which gives a more practical measure of voltage for real-world applications.
Grasping the concept of peak voltage is essential to understand other related metrics like RMS voltage, which smoothens out the peaks and gives a more constant value for analysis.
Voltage Calculation
To calculate various characteristics of voltage generated from an AC generator, different formulas are applied depending on what specific value you want to find. When dealing with RMS (Root Mean Square) voltage, a common formula is used:
r>\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
This formula allows us to convert peak voltage to RMS voltage, providing a practical measure of voltage. In simpler terms, RMS voltage is the equivalent DC voltage that would deliver the same power to a load as the AC voltage does over a complete cycle.
  • This calculation is vital for determining the usable power capacity of the AC system.
  • It is crucial in designing electrical circuits and ensuring the correct functioning of electronic devices.
In our example, substituting 55 V for the peak voltage into the formula gives an RMS voltage of approximately 38.89 V. Understanding this calculation helps in interpreting how much real-world work an AC system can perform.

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Most popular questions from this chapter

\- An rms voltage of \(120 \mathrm{V}\) produces a maximum current of \(2.1 \mathrm{A}\) in a certain resistor. Find the resistance of this resistor.

IP (a) Determine the power factor for an \(R C\) circuit with \(R=4.0 \mathrm{k} \Omega\) and \(C=0.35 \mu \mathrm{F}\) that is connected to an ac generator with an rms voltage of \(24 \mathrm{V}\) and a frequency of \(150 \mathrm{Hz}\) (b) Will the power factor for this circuit increase, decrease, or stay the same if the frequency of the generator is increased? Explain.

An \(R C\) circuit consists of a resistor \(R=32 \Omega\), a capacitor \(C=25 \mu F,\) and an ac generator with an rms voltage of \(120 \mathrm{V}\) (a) At what frequency will the rms current in this circuit be 2.9 A? For this frequency, what are (b) the rms voltage across the resistor, \(V_{\text {rms. } R}\), and (c) the rms voltage across the capacitor, \(V_{\mathrm{rms}} c^{2}\) (d) Show that \(V_{\mathrm{rms}, \mathrm{R}}+V_{\mathrm{rms}, \mathrm{C}}>120 \mathrm{V},\) but that \(\sqrt{V_{\mathrm{rms}, R}^{2}+V_{\mathrm{rms}, \mathcal{C}}^{2}}=120 \mathrm{V}\)

A \(4.40-\mu \mathrm{F}\) and an \(8.80-\mu \mathrm{F}\) capacitor are connected in parallel to a \(60.0-\mathrm{Hz}\) generator operating with an rms voltage of \(115 \mathrm{V}\) What is the rms current supplied by the generator?

\(\mathrm{A}\) "75-watt" lightbulb uses an average power of \(75 \mathrm{W}\) when connected to an rms voltage of \(120 \mathrm{V}\). (a) What is the resistance of the lightbulb? (b) What is the maximum current in the bulb? (c) What is the maximum power used by the bulb at any given instant of time?

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