# single-phase AC voltage controller w/ RL load

Discussion in 'Homework Help' started by notoriusjt2, Oct 19, 2010.

Feb 4, 2010
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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Clearly you must find a β value greater than α at which the function goes to zero. Otherwise if β<α the exponential term is increasing rather than decreasing. Since there are many values that satisfy the condition the trick is to use the first value of β greater than α at which the function goes to zero.

I have β=3.681 rad FYI.

Last edited: Oct 21, 2010
3. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
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so now that i have the equation set up. for wt... what value am i supposed to use

wt=(377)(1/60)=6.28?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I think the rms load current would be given by

$I_{rms}=\sqrt{\frac{1}{\pi}\int^{\beta}_{\alpha}{I_o^2d(\omega t)}}$

So you have to square the current function before doing the integration, etc.

Remember the definition of RMS - "root of the mean of the function squared".

That's a painstaking task to say the least for the given current function. Are you sure you had to do this? I would also be very surprised if this was part of an exam question. There might be some value for the student doing this once in their academic life, if only for the discipline required in keeping track of all the terms during integration and so on - otherwise I would look for alternative computational methods in the routine of working life.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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$I_o=\frac{V_m}{Z}$sin(\omega t-\phi)-sin(\alpha-\phi)e^{\frac{(\alpha-\omega t)}{\omega \tau}}$$

${I_o}^2=\frac{V_m^2}{Z^2}$sin^2(\omega t-\phi)+sin^2(\alpha-\phi)e^{\frac{2(\alpha-\omega t)}{\omega \tau}}-2sin(\omega t-\phi)sin(\alpha-\phi)e^{\frac{(\alpha-\omega t)}{\omega \tau}}$$

As I said, the challenge is to then correctly integrate the latter function over the stated limits.

If you aren't confident with integrating wrt 'ωt' you can do it for 't' with the limits being α/ω to β/ω (rather than α to β)

6. ### t_n_k AAC Fanatic!

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FYI - the answer looks to be closest to (b) 3.0 A. I checked this by simulation rather than by doing the aforementioned mental gymnastics.

7. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
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realistically, is there any real-life application to this? is this worth it for me to sit down and plow through this?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Depends whether your professor / teacher wants a worked answer from you.

With respect to the integration you could at least obtain the general result (without human error) using an online integrator such as that found at Wolfram Mathematica.

http://integrals.wolfram.com/index.jsp

I believe the aforementioned will readily handle your particular equation - you wouldn't need to do the squaring operation on the function for Io - just enclose the Io function and place it within parenthesis with a squared index.

To complete the overall task you then only have to plug in the limits and do the ongoing calculations. This could also be made somewhat easier using (say) Excel where you enter the equation and let the spreadsheet do the calculations at the limit values. And so on ...

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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As I said - should you have even a slight ongoing interest in the matter - it's likely to be a challenge to do the integration by hand without making errors. I've attached a pdf result of the Wolfram site online integration. I tip my proverbial hat to the application programmer who produced the software for this. Quite impressive.

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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Even had trouble just entering the equation correctly ...!

a, b & c are the equation constants (α and so on) and ωt is substituted with the integration variable x.

File size:
123 KB
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