Root Mean Square Voltage Multiplied by \sqrt{}2: Explained

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Homework Help Overview

The discussion revolves around the relationship between peak voltage and root mean square (r.m.s.) voltage in the context of an AC generator, particularly when the frequency of rotation is altered.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore how doubling the frequency affects the peak voltage and its relationship to the r.m.s. value. Questions arise regarding the transition from division to multiplication in the context of voltage calculations.

Discussion Status

Some participants are actively questioning the implications of frequency changes on peak voltage and r.m.s. values, while others are attempting to clarify their understanding of the relationships involved. There is an indication of productive engagement, but no explicit consensus has been reached.

Contextual Notes

Participants are considering the definitions and relationships between peak voltage and r.m.s. voltage, as well as the effects of frequency changes, which may not be fully established in the original problem statement.

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Homework Statement



An AC generator produces a voltage of peak value V. The frequency of rotation of the coil of the generator is doubled. The r.m.s. value of the voltage produced is

Homework Equations



The answer is V multiplied by [tex]\sqrt{}2[/tex] but I don't understand why.

The Attempt at a Solution



I know that Vrms = Vo/[tex]\sqrt{}2[/tex]so how does a division become a multiplication?
 
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Still looking at this. Does doubling the frequency mean V peak becomes 2 V peak?

In which case 2 V / [tex]\sqrt{}2[/tex] = V[tex]\sqrt{}2[/tex]

Is that right?
 
Initially the peak voltage is V. What does doubling the frequency of rotation do to the voltage produced? So what is the new peak voltage in terms of V? Then work out the rms value for this new peak voltage.

edit: I was too late. Yes, I think that is right : )
 
Thank you! :smile:
 

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