Output Voltage Equation for AC Source at 90 Hz?

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Homework Help Overview

The problem involves determining the output voltage equation for an AC source with a given rms voltage and frequency. The context is within the subject area of electrical engineering, specifically focusing on alternating current (AC) circuits.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between rms voltage and peak voltage, with attempts to derive the output voltage equation using sine and cosine functions. There is questioning about the correctness of their approaches and the precision of their calculations.

Discussion Status

Participants have shared their calculations and reasoning, noting minor discrepancies in their results. There is an ongoing exploration of whether the expected form of the equation should involve sine or cosine, and some participants express frustration over perceived strictness in formatting or rounding errors.

Contextual Notes

There is mention of potential issues with how the answer is presented, including concerns about phase angles and the possibility of being penalized for small rounding differences. Participants are also considering reaching out to the instructor for clarification.

Ithryndil
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Homework Statement


The rms output voltage of an AC source is 180 V, and the operating frequency is 90 Hz. Write the equation giving the output voltage as a function of time t. (Use t as necessary.)

Homework Equations


[tex](\Delta v) = (\Delta V_{max})sin\omega t[/tex]
[tex]\Delta V_{max}= \sqrt{2}V_{max}[/tex]
[tex]\omega = 2\pi f[/tex]

The Attempt at a Solution



I thought it was a simple solve for the values of omega, Vmax, and plug those into the first equation, but I am apparently wrong according to the website. I don't think I have messed up. The only other thing I can think of is that they want it in cosine instead of sine. I got the following answer:

Voutput = 255sin(566t)
 
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Vrms = Vp/√2

Vt = Vp*sin(ωt) = √2*Vrms*sin(2*π*f*t) = 254.55*sin(565.48*t)

That's the way I figure it.

It's difficult to imagine that they would be picky about that small a rounding error.
 
I meant that with my one equation...I meant that Vmax = sqrt(2)Vrms. And...I know it must be right, they are just being picky about how it's typed in, yet, I get the same answer you get. It shouldn't be cosine (thought it could be, but would require a phase angle). It also shouldn't be negative as far as I can tell. And, as my answer is within 1% of your answer, it shouldn't be picky about that either. I will just try one other thing and then email my teacher about it.
 

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