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Discuss OHMS law where do these formulas come from in the Australia area at ElectriciansForums.net

swc

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Hi
I currently studying 2330 Lv3 to become an electrician. I've never understood where these formulas come from. I've always managed to get by just taking them for granted. Never really did use them at Lv2. With a 50% failure rate nationwide it's crucial for me to be able to remember these formulas.

Code:
R x I (squared)
V(squared) / R
square root(P x R)
p/ I (squared)
V(squared)/p


I understand what's squared and square root means, I don't understand how the formula becomes to be V(squared) /p for example
I'm assuming they're transposed some how from a formula. I would appreciated it if you could shed some light on this. Driving me nuts.
 
I know all about writing for long exams (4nrs was the longest, god knows how the examiner marked it as my hand writing is bad to start with then degrades).

All examinations are dumbed down and to a certain extent the teaching reflects that. You have lecturers (mostly) who are either there as they can't get work as a spark (for whatever reason) and are bitter and twisted or on easy street as its their last gig.

From what I've read getting someone in the middle who isn't worn down by the sheer monotony of teaching the same thing to a mostly disinterested bunch of people is becoming less and less so it becomes a vicious circle of falling pass rates and easier exams to show "progress"
 
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scan0001.jpg

look at that
 
Scotsparky, that's laminated and stuck to the inside lid of my top box, ive tried posting it many times but it was a doc.x or something which wasn't recognised, ive got 3 phase calcs along the bottom also.
 
R x I (squared)
V(squared) / R
square root(P x R)
p/ I (squared)
V(squared)/p

Archy Styrigg is spot on with his answer in post #4 apart from "3rd" where he's mixed up √P x R with √P ÷ R.

Thought I'd just give a bit of an explanation about what he's said.

1. R x I² = P -- This is just (R x I) x I

R x I gives you voltage (V). You then multiply this by current (I) to give you power (P) i.e P = V x I


2. V² ÷ R = P -- What's happening here is the same as (V ÷ R) x V

V ÷ R gives you current (I). You then multiply this by voltage (V) to give you power (P) i.e P = I x V


3. √P x R = V -- This is just working backwards from V² ÷ R

V² ÷ R gives you power (P) so P x R will give you V² and √V² gives you voltage (V)


4. P ÷ I² = R -- This is the same as (P ÷ I) ÷ I

P ÷ I gives you voltage (V). You then divide by current (I) to give you resistance (R) i.e R = V ÷ I


5. V² ÷ P = R -- This is just the opposite of V² ÷ R which gives you power (P). V² ÷ P gives you resistance (R). Think of it like 10 ÷ 2 = 5 and 10 ÷ 5 = 2
 
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Current (single phase): I = P / V[SUB]p[/SUB]×cos φ Current (3 phases): I = P / √3 V[SUB]l[/SUB]×cos φ or I = P / 3 V[SUB]p[/SUB]×cos φ
Power (single phase): P = V[SUB]p[/SUB]×I[SUB]p[/SUB]×cos φ Power (3 phases): P = √3 V[SUB]l[/SUB]×I[SUB]l[/SUB]×cos φ or P = √3 V[SUB]p[/SUB]×I[SUB]p[/SUB]×cos φ


Power factor PF = cos φ = R/(R2 + X2)[SUP]1/2[/SUP], φ = power factor angle. For the purely resistive circuit, PF = 1 (perfect).
The apparent power S is calculated according to Pythagoras, the active power P and reactive power Q. S = √(P[SUP]2[/SUP] + Q[SUP]2[/SUP])

thats the bit to paste under Scotsparky's pire wheel and you're ready for print.:28:
 
Current (single phase): I = P / V[SUB]p[/SUB]×cos φ Current (3 phases): I = P / √3 V[SUB]l[/SUB]×cos φ or I = P / 3 V[SUB]p[/SUB]×cos φ
Power (single phase): P = V[SUB]p[/SUB]×I[SUB]p[/SUB]×cos φ Power (3 phases): P = √3 V[SUB]l[/SUB]×I[SUB]l[/SUB]×cos φ or P = √3 V[SUB]p[/SUB]×I[SUB]p[/SUB]×cos φ

You're just being flash now! :)
 

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