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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'.

203PartQuestions.jpg


 
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'.

203PartQuestions.jpg
Gonna have to think that over when more sober...

and more awake!!

Bed is calling...

G-Night m8 & Mrs m8!

 
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! :) :) :)
Now thats good advice mate. :)

 
Of the 7 of us in my class taking 203, 3 passed 1st time, 2 out of 4 on the resit (1 of was those a distinction) .

I have to read the questions very slowly, as I dont like how they are worded at all!

Reconised the 1st 2 questions there. Had 2 on star/delta, selv supply for soldering irons. Strangely I can never seem to remember the questions after the test!

Thanks for the welcome on the other thread, and good luck with the resit!

 
ive got my 203 tomorrow not really looking forward to it, out of 27 that took it last night 12 passed that was the mature group, better get revising anyway

sparkss

 
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.
Isn't that a lorry that transports flowers?

; \

 
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'.

203PartQuestions.jpg
Does anyone know the formula to work out Question 1 chaps?

Regards,

Admin.

 
Does anyone know the formula to work out Question 1 chaps?Regards,

Admin.
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

 
Does anyone know the formula to work out Question 1 chaps?Regards,

Admin.
[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 :D (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! :D

=================================================

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.

 
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 KME,

That is exactly what I worked it out as lastnight In Shed) - but had doubts that I was using the wrong formula becasue i got 10A too.

Regards,

Admin.

 
[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 :D (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! :D

=================================================

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.
Thank you too Spec Loc - for all the time and trouble gone into that.

Regards,

Admin.

 
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.)

:eek:

 
--------------------------------------------------------------------------------

I am revising in "Bursts". The mistake that I make, Personnally - Is that I pick up my notes and - Read, read, read - The first lot sinks in, but the rest doesn't.

Even though, I must say that i do stop for a cigarette, now and again. (Most of the time though - I have a ciggy when I am reading .

 
Re-sat it Tonight and got a Distinction. Yeeee Haaaa.

Am I on top of the world?

Oh Yes - Too right I am. A few tough ones in there again. :p Got through them though. :D

:x

 
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