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- Feb 12, 2008
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Must be that admin Bloke. Pfft.
Gonna have to think that over when more sober...Here's a few that I had - I knocked them up quick - while I was having a fag in the workshop/shed. You'll have to imagine the word "the" appears between 'In' and 'Figure'.
Now thats good advice mate.Tough one m8! I presume from your posts this is an on-line exam, was it open book? regs + OSG ??
From my experience, open book exams do tend to have a lot of questions that seem to be worded double negative or obscurely where you have to read & re-read them..
so that even if you know where it is in the regs..
the bod's at C&G want to know that you understand what you have read..
so they try and twist it..
I know on my 17th ed recently I had about 3q's I had put my quick answer down but kept them flagged..
when I re-read them later. I changed my answers because of the wording of the question.
Anyway.. thats all water under the bridge..
write that one off as a bad day!
Tomorrow is another new day!
Remember... as long as you don't give up...
Nobody fails... some just take a little longer than others to reach the completion!
Isn't that a lorry that transports flowers?Sorry Sheepman:I`ll repeat for the hard of hearing amongst the wizards in the welsh variety:
What.........is........a........florrie"?
Serious question. NOT, I repeat NOT a wind-up.
Does anyone know the formula to work out Question 1 chaps?Here's a few that I had - I knocked them up quick - while I was having a fag in the workshop/shed. You'll have to imagine the word "the" appears between 'In' and 'Figure'.
Okay. I`m looking now whilst posting.Does anyone know the formula to work out Question 1 chaps?Regards,
Admin.
[Q1]Does anyone know the formula to work out Question 1 chaps?Regards,
Admin.
Thank you KME,Okay. I`m looking now whilst posting.My initial thought, on looking at the question:
"depends on the ratio of the transformer".
As that isn`t a valid answer, stand by........
OK part1. M3 would read 10A
IF the "supply is 230v; and the load is drawing 10A@200v; then the answer is below 10A; as P=VI.
therefore, although I`ve been unable to verify this by calc. (but Guinness IS good for you!), the only answer I`d be willing to look at is 8A.
Therefore: C
QED KME POP
Thank you too Spec Loc - for all the time and trouble gone into that.[Q1]=====================================
as KME says in this example you can use ye ode P=IV cuz you already have current & power.
However sometimes those cunning C&G bods put some winding ratios instead...
So any question with transformers I always remember;
VS/VP = NS/NP = IP/IS.
or long hand;
[ voltage at secondary divided by voltage at primary ] =
[ number of windings on secondary divided by number of windings on primary ] =
[ current at primary divided by current at secondary ]
{remember current is a r s e-about-face compared to windings & voltages}
they will normally give you two items either Volts, Current, or Windings.
then one other and ask you to find a fourth!
NOTE on this one we have to assume the supply is 230v as it isn't actually given!
and we are not given any winding ratio's..
all we know is voltage & current on the secondary. :|
e.g.
M2 = 200v
load = 2000VA
M3 = 2000/200=10A
VS/VP = IP/IS
200/230 = IP/10
0.86 = IP/10
IP = 0.86 x 10
M1 = 8A (or just above.. was there a decimal point missing of the answer?)
or is it a cheep digital meter with only one decimal place!!! :^O :^O:^O
=========================================
For any others reading
the [Q3]
Power Factor = Real power(watts)/Apparent power(KVa)
so from table:-
pf [a] = 30w/50kVa = 0.6
pf = 1000w/50kVa = 20
pf [c] = 30w/40kVa = 0.75
pf [d] = 30w/30kVa = 1.0
as with many multi choice two answers are throw away even without calculating..
20!!??? is stooopid cuz pf is NO GREATER THAN 1.0
[d] 1.0 unity power factor ahhh the perfect world.. wot no losses???
you just need to calc a & c. to decide which to pick? as said (A) is our answer here!
=================================================
the [Q2] bit
Capacitors are used to compensate losses with inductive loads.
(Voltage leads Current)
Inductors are used to compensate for capacitive loads
(Current leads voltage)
Low power factors are typically caused by the inductive component of many electrical loads, which require capacitors to correct.
i.e. in the ole impedance triangle..
Z is the hypotenuse.
R is the adjacent
and reactance (Xl - Xc) is the opposite
from this you can get your phase angle & power factor
The impedance to alternating currents increases with inductive loads
and decreases with capacitive loads.
1:24!!!!!Thank you too Spec Loc - for all the time and trouble gone into that.Regards,
Admin.
I`m ok then, `cos I`m not exactly young anymore! :_| :_|Staying up late?Did you know your twice as likely to die younger from any number of illnesses..
(If you have less than 6 hours sleep per night.)
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