Discussion in 'Homework Help' started by weihong892, Aug 15, 2012.

1. ### weihong892 Thread Starter New Member

Aug 15, 2012
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0

I'm stuck at question 2b(iv)! i try to integrate to find the current flowing into the scr but still can't get the correct answer. i got 10.9, but the answer is 10.32A. please help me if u are able to do it! my exam is coming in 1 more day!

Work out the RMS value of the current flowing through the SCR, T1

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2. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Please post your work so that we can walk through it and see where and what assumptions are being made. You are clearly in the ballpark, so it could be one of several fairly minor differences between how you did things and how the author did things. How are you accounting for the forward voltage drops of the SCRs when they are conducting?

3. ### weihong892 Thread Starter New Member

Aug 15, 2012
5
0
2bi) the firing delay angle
Idc= Vo(dc) / RL
14.6= Vodc/10
Vo(dc)= 14.6 x 10
= 146v
Vo(dc) = ((2Vmax)/∏) cos α
α= 0.788279825
α= 0.788279825 x (180/∏)
= 45

2bii.) The power dissipated in RL
P dissipated = I sq 2 R
= 14.6 sq 2 x 10
= 2131.6w
2biii) The power factor of the circuit
pf = (Vdc + Idc)/(Vs + Idc)
= (146 + 14.6)/(230 + 14.6)
= 0.635

2biv) The RMS value of the current flowing therough the scr, T1

Tti(dc) = (1/∏) intergrate (idc) dθ
= 10.94...

i dunno whether this formula is correct, cause in my lecture notes, there is no formula finding scr current. please help me, thanks.

4. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Making just a quick glance through you work, I now suspect that it is quite possible that you got close only by coincidence.

Because you choose not to track units, you did things that, had you tracked units, you would have known were absolutely wrong.

For instance:

What is 146V added to 14.6A?

I certainly have no idea! It's like asking what you get when you add 3 meters to 2 kg.

You can only add two quantities if they have identical units.

So go back and redo your work, tracking units throughout each step.

5. ### weihong892 Thread Starter New Member

Aug 15, 2012
5
0

this is the formula given in the lecture note to find power factor...

6. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Don't round any of the trig functions. Keep it all in the calculator and take it piece by piece. If the formula is right, chances are you'll get closer to the correct answer.

7. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Does it make any sense to you what you get when you add a voltage to a current?

Go ask your instructor what it means and see if they can give you a meaningful answer.

It sounds like you are trying to do engineering-by-cookie-cutter, meaning that you simply parrot formulas and plug numbers into a calculator without any effort or desire to understand them, where they come from, what they mean, or what their limitations are. I'm sure I'm doing at least a bit of a disservice to you, so please don't take it too personally; it's meant more as a general observation.

Describe, in your own words, what "power factor" means. What does it tell us?

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8. ### WBahn Moderator

Mar 31, 2012
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What good will it do to reduce the roundoff error when evaluating a formula that has a 0% chance of being right?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The most obvious explanation lies in an erroneous transcription from whiteboard to notepaper.

Change the signs from addition to multiplication and the result makes somewhat more sense.

10. ### WBahn Moderator

Mar 31, 2012
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Agreed. Also, it appears that the OP did multiply them (you get 0.657 if you evaluate the expression as given).

I'm concluding that whenver he uses 'dc' he really means 'rms'.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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DC is OK in this instance since the starting premise [from the question statement] is that there is negligible current ripple. In any case I'd be wary of such ad hoc formulae as given by the OP, because they are likely to be applied by students when the underlying assumption / premise isn't true.

12. ### DerStrom8 Well-Known Member

Feb 20, 2011
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That's a general statement. I'm not necessarily referring to this specific problem. In general, when working with trig functions, it's very important not to round the answers until the very end.

13. ### WBahn Moderator

Mar 31, 2012
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Yeah, I got to thinking that that was probably the case a bit later. We have little choice but to interpret a post in the context of the posts before it and it's hard to keep in mind that a given post may be written in response to an earlier post before the subsequent posts have even been seen or read. I've certainly been on both sides of that issue!

I will second your recommendation of keeping results in the calculator as long as possible to minimize roundoff error. If that isn't practicle, then carry enough digits in the intermediate results so that the roundoff can't propogate to the least significant digit that will be retained in the final result -- fairly easy to do with multiplication, division but a lot trickier if addition or subtraction is involved anywhere.

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