Phase 6 science exam papers

Bill Electrical Aprentice

does any one have science exam papers from 2008 onwards
I have from 2004 to 2008

superg

superg wrote: »

PM me your email address

Hey, any chance you have phase 4 papers? Cheers in advance

Worth a shot, thanks again

Tony Grant

I have some info relating to phase 6 if you text me your email add to 086 2379704 i will forward you info.

redeye101

Tony Grant wrote: »

I have some info relating to phase 6 if you text me your email add to 086 2379704 i will forward you info.

Tony,

Sitting the phase 6 exams next week. Anything wild or wonderful in store?

Thanks

dno electrical

can anybody help me i need electrical science exam papers from 2009 onwards phase 6 and any other help would be much appreciated sitting exam next week

fourpoint0

Lads any body have any past papers for electrical science phase 6. Any info on the next exam .
Thanks in advance:eek:

tazmania

You should get them here.
Mod edit,

Links removed at site owners request

I have them up to 2009 if you want to PM me I will send them on to you

fourpoint0

That's great Thanks for that, really appreciate it.

Mod edit,

Links removed at site owners request

brightspark123

Hey can anyone help me??or would have any information on how to repeat my last electrical science exam in austrailia,,i am currently living in Australia at the moment and just relieved my results i passed all apart from my electrical science and just wondering is there any way possible of repeating it over in Australia i got 61 out of 70 surely there has to be something that can be done!!any info would be greatly appreciated thank you..

Tutor Mike

I suggest that you contact your FAS Training advisor.
Best of luck

cianwalsh89

hey i am starting my phase 6 electrical coarse and i was looking for some phase 6 past exam papers to work on if anyone could get back to me it would be much appreciated

sparcocars

If you want to PM me your email address i'll send you past papers. I have some past papers in PDF format

nitroxjunkie

looking for past exam papers for electrical science phase 6 any help greatly appreciated

nitroxjunkie

has anyone got these exams phase 6 electrical science exam papers

sparky48

Hi.
Would it be possible to get Phase 6 exam papers Science & Craft

Golly81

Does any one have phase 6 electrical science past papers from 2004 to 2013. Thanks

Captain Chaos

Golly81 wrote: »

Does any one have phase 6 electrical science past papers from 2004 to 2013. Thanks

The science syllabus changed in 2009 to a newer version of 2.2 so papers older than 09 are of little use now. All of the electronic section changed and the PLC questions.

Golly81

Captain Chaos wrote: »

The science syllabus changed in 2009 to a newer version of 2.2 so papers older than 09 are of little use now. All of the electronic section changed and the PLC questions.

Ok. Do you no where I could get my hands on any papers from 2009 to 2013

Calgary22

Your instructor.

Captain Chaos

Golly81 wrote: »

Ok. Do you no where I could get my hands on any papers from 2009 to 2013

I need them myself, I am appealing my final exam after 3.5 years out of the system. I was given the new exam on my final attempt and failed it as I didn't know it had changed and was told I would sit the old exam. I didn't even know what to do with the new formulas in the book. I had not a hope of passing it as 5 of the questions needed the new formulas to answer them and 2 circuit diagram I had to identify I had never seen before in my life.

Calgary22

what exam mate ya stuck on?

Captain Chaos

Calgary22 wrote: »

what exam mate ya stuck on?

PH6 science. I placed my appeal last week and I'm waiting to hear back. The guy I rang at FAS said the national appeals board only meet up 3 times a year though.

At this stage I don't care how much it costs for appeals, re sits etc I just want it done and get my ph7 signed off.

Calgary22

Yea lad that's a **** of exam:/.. yea i no how ya feel but be warned when ya finish phase 7 ya might not be qualified it all depends if your time makes up 208 weeks or not. a lot of dudes myself included taught that when we finish phase 7 were done were not sadly. as i am short weeks to make up the full 4 years. and i have to wait months to do a test that will allow me to get my papers and by the sounds of the exam its a **** has a fail rate of 80%/.

Golly81

Captain Chaos wrote: »

I need them myself, I am appealing my final exam after 3.5 years out of the system. I was given the new exam on my final attempt and failed it as I didn't know it had changed and was told I would sit the old exam. I didn't even know what to do with the new formulas in the book. I had not a hope of passing it as 5 of the questions needed the new formulas to answer them and 2 circuit diagram I had to identify I had never seen before in my life.

I'm out of the system 8 years now so it's going to be hard.im looking at my old notes now . So your telling me there no good now?

Calgary22

8 Years:o yea mate there no good the system has changed in 09 you be best to ring a lecture in one of the many its round Ireland that offer phase 6 ...

Golly81

Calgary22 wrote: »

8 Years:o yea mate there no good the system has changed in 09 you be best to ring a lecture in one of the many its round Ireland that offer phase 6 ...

I was talking to a lecture yesterday I'm meeting him next week. It looks like if have a lot of work to do. Is the exams a lot hard now?

Calgary22

yea mate it depends of your grasp for the first exam you took. if you found it hard you might find these a lot harder:(

kingden2006

hi golly81 just wondering if you got any phase 6 electrical notes?
anything at all would help cheers.

kingden2006

hi calgary just wondering what is involved in phase 6 of the electrical science,is it harder than phase 4 is?
and if i was to start studying stuff would should i go over and are there any particular books apart from withfield,that i could get,especially a book that would cover plcs.
cheers calgary.

kingden2006

a single phase fluorescent lighting load takes a supply current of 16 amps from 230volt 50hz supply at a power factor of .85 lagging.
a 60uf capacitor is used to improve the power factor of the circuit,determine the new value of supply current.
can anyone show me how to do this one and write the steps in order to achieve this.
cheers folks.

2011

kingden2006 wrote: »

a single phase fluorescent lighting load takes a supply current of 16 amps from 230volt 50hz supply at a power factor of .85 lagging.

Here are a few clues:

Power = V x I x PF = 230 x 16 x 0.85 = 3128 watts

Power = I^2 x R

R = P / (I^2)
R = 3128 / (16^2) = 12.23 ohms = resistance of fluorescent lighting load t

Cos^-1 (0.85) = 31.79° as the circuit is inductive the current is lagging.

Calculate reactive power.
From this work out XL.

Next calculate Xc

Xc = 1/ωC
    = 1 / (2x pi x fC)


Where:
f = 50 Hz
C = 60 x 10^-6

Subtract Xc from XL.

Use Pythagoras theorem to solve.

Remember that as the capacitor is connected in parallel with the lighting load 230V will be applied across it.

arse..biscuits

2011 wrote: »

Here are a few clues:

Power = V x I x PF = 230 x 16 x 0.85 = 3128 watts

Power = I^2 x R

R = P / (I^2)
R = 3128 / (16^2) = 12.23 ohms = resistance of fluorescent lighting load t

Cos^-1 (0.85) = 31.79° as the circuit is inductive the current is lagging.

Calculate reactive power.
From this work out XL.

Next calculate Xc

Xc = 1/ωC
    = 1 / (2x pi x fC)


Where:
f = 50 Hz
C = 60 x 10^-6

Subtract Xc from XL.

Use Pythagoras theorem to solve.

Remember that as the capacitor is connected in parallel with the lighting load 230V will be applied across it.

It's not that complicated.

arse..biscuits

kingden2006 wrote: »

a single phase fluorescent lighting load takes a supply current of 16 amps from 230volt 50hz supply at a power factor of .85 lagging.
a 60uf capacitor is used to improve the power factor of the circuit,determine the new value of supply current.
can anyone show me how to do this one and write the steps in order to achieve this.
cheers folks.

Find the angle which the current is lagging by.
Calculate Xc
Calculate the current through the capacitor (V/Xc)
Capacitor current leads by 90 degrees
You now have two currents and you have to add them using phasor addition

BrianDug

Ok so what you know is:
Incoming supply ampere is 16A
P.F = 0.85
V = 230

So if we focus on this for a minute we can draw a power triangle soley based on this information.
On the top line of our power triangle:

kW = 230 x 16 x 0.85 = 3128W (3.128 kW)
kVA = 230 x 16 = 3680VA (3.680 kVA)
kVAr = by Pythagoras theorem sqrt(kVA^2 - kW^2)
= sqrt(3680^2 - 3128^2) = 1938.56 VAr (1.938 kVAr)
Power Factor = kW/kVA so 3128/3680 = 0.85 (as expected)

We could also draw the power triangle using just the current so in the kVA position we would have 16A, in the kW position we would have 13.6 A (16A x 0.85), and in the kVAr position we would have sqrt(16^2 - 13.6^2) = 8.4285A. Doing it this way is just another way of looking at the power triangle. Then power is of course V x I & so kW = 230 x 13.6A = 3128W, kVA = 230 x 16A = 3680VA etc etc...

Ok so now for adding the capacitor, we know in simple terms that inductive reactance/ballasts/coils etc will cause the current to lag the voltage and the worse the power factor the higher the "wasted current"/kVAr/power so in order to help reduce our reactive power/kVAr we need to add a capacitor that will "suck up" the bad power and help reduce our power factor and wasted power.

We are told our capacitor is 60uf so what we are essentially looking to do is find how much current this capacitor requires. By ohms law I = V/R.
Xc can be thought of as R in your case so it will be I = V/Xc.

To find XL its 1/2*pi*frequency*C = 1/2*pi*50*60x10^-6 = 53.05ohms
So I = V/Xc = 230/53.05ohms = 4.3347A

So the capacitor will "take" 4.3347A from the supply. A way to think of it is XL & XC are always fighting each other and that generally XL will always be bigger than XC.

Looking back at the current results we now know with the inductor/XL is 8.4285A and XC is 4.3347A. These two values are all we need to solve. To find our new Total Current/kVA the formula is

Z = sqrt(r^2 + (XL^2 - XC^2) which is same as
Z = sqt(Ir^2 + (IXL^2 - IXC^2)
Z = sqrt(13.6A^2 + (8.4285A - 4.3347A)^2 = 14.20A

14.20A is the total current been drawn from the system so 230V x 14.20A = 3266VA

PF = R/Z = 3128W/3266VA =0.95 or 13.6A/14.20A = 0.95 etc...
Our system was drawing 3680VA(or 16A) before we added a capacitor. With a capacitor its 3266VA( or 14.20A) so we are using less power now.

Try visualise whats going on and it all becomes clear. I tried to do out as many steps as possible so you can see whats happening. I hope I havent confused you!

arse..biscuits

Current 1 is 16 A, lagging by an angle of (cos^-1 0.85) 31.8 degrees

Xc=1/(2 * pi * f * C) = 53.05 ohms
Current 2 = V/Xc = 230/53.05=4.34 A, leading at an angle of 90 degrees

I1 = 16A @-31.8 degrees
I2 = 4.34A @90 degrees

phasor addition
x = (I1 * cos thita 1) + (I2 * cos thita 2) = (16 * cos -31.8)+(4.34 * cos 90)=13.6
y = (I1 * sin thita 1) + (I2 * sin thita 2) = (16 * sin -31.8)+(4.34 * sin 90)= -4.09

Now do pyth. theorm with these values (you can drop the minus)

Total current = Square root (13.6^2 + 4.09^2) =14.2 A

BrianDug

Ay but do they show you how to do it this way?

Many ways to do it, best way is a method that you can follow step by step and understand. Not good to be doing calculations if you dont know whats happening.

I'm sure you will have plenty of samples done in your notes.

arse..biscuits

It is difficult to explain these types of questions here, face to face is a lot easier. I have been teaching this type of stuff for years and my students are always happy.

2011

BrianDug wrote: »

Many ways to do it, best way is a method that you can follow step by step and understand.

Exactly.
There are many correct ways of solving this, what appears complicated for one person may appear straight forward to another.

kingden2006

thanks lads for the help really appreciated and understood,im starting phase 6 now in january got all credits in phase 4 so its not really the maths the worries me its more the plcs,and the questions on them,they appear to be difficult along with some electronic questions.
has anyone any idea on those,are they difficult??they look difficult.
and everyone i have been talking to tells me yhat phase 6 is easier than phase 4?i can"t imagine that to be true can anyone shed any light on this?
thanks again lads and happy x-mas to you all

kingden2006

sorry for the spelling mistakes lads was in a rush.

johnnyfruitcake

If you got credits in phase 4 you'll breeze phase 6, plc questions are mostly about drawing out ladder diagrams, not very hard at all.

sparcocars

Its not that its easier than phase 4 but its not as big of a jump as it is in phase 4 from phase 2. You still need to put the work in but if you did well in phase 4 then you should be okay. A lot of the stuff looks hard when you look at it now but when its explained to you in class it will begin to make sense. When i started phase 6 i thought i was never going to get the plc's but they actually turn out to be relatively straight forward after a week or two. Its nearly lime learning a new language. I got all credits in phase 4 and then got all credits in phase 6. Just keep the head down and you should be fine.

kingden2006

cheers lads.can anyone answer this question and show me which way its done from start to finish cheers lads.
A 450 dc shunt motor has an armature resistance of 0.185ohms.if the field current is 20 amps and the load current is 160 amps,calculate the efficiency of the motor?(neglect iron and mechanical losses)

and here"s another quick two while im here.
A three phase motor is connected in delta and takes 90 amps from the supply.calculate the current in each winding?

the copper losses of a transformer are 90kw on half load.calculate the copper losses at three quarter full load

arse..biscuits

Armature current =160-20=140A

Power losses
(Field) P=V.I=450x20=9000W
(armature) P=I^2.R=140x140x0.185=3626W
Total = 12,626W

Power input = V.I = 450x160=72000W

Power output =input - losses = 72000 - 12625=59374W

Eff = out/in = 59374/72000=0.8246

82.46%

arse..biscuits

A three phase motor is connected in delta and takes 90 amps from the supply.calculate the current in each winding?


I(phase)=90/root 3=51.96A

arse..biscuits

90kw on half load.calculate the copper losses at three quarter full load

Full load copper losses = 90/0.5^2=360kW

Copper losses at 3/4 of full load = 360x0.75^2=202.5kW

kingden2006

cheers arse..biscuits..gentleman

kingden2006

hers another one for you lads if anybody can answer it.
three single phase loads are connected to a 400volts 50hz supply.calculate the magnitude of current flowing in the neutral.
L1 IS 10 AMPS AT UNITY POWER FACTOR.
L2 IS 14 AMPS AT .92 LEADING
L3 IS 16 AMPS AT .97 LAGGING.?
COULD YOU PLEASE SHOW ME THE STEPS ALSO.
THANKS LADS AND HAPPY NEW YEAR TO YOU ALL

kingden2006

A SINGLE PHASE FLUORESCENT LIGHTING LOAD TAKES A SUPPLY CURRENT OF 24 AMPS FROM A 230 VOLT 50HZ SUPPLY AT A POWER FACTOR OF 0.85 LAGGING.
A POWER FACTOR CORRECTION CAPACITOR TAKING 8 AMPS IS USED TO IMPROVE THE POWER FACTOR OF THE CIRCUIT,DETERMINE THE NEW VALUE OF THE SUPPLY CURRENT AFTER POWER FACTOR CORRECTION?

2011

kingden2006 wrote: »

hers another one for you lads if anybody can answer it.
three single phase loads are connected to a 400volts 50hz supply.calculate the magnitude of current flowing in the neutral.
L1 IS 10 AMPS AT UNITY POWER FACTOR.
L2 IS 14 AMPS AT .92 LEADING
L3 IS 16 AMPS AT .97 LAGGING.?
COULD YOU PLEASE SHOW ME THE STEPS ALSO.
THANKS LADS AND HAPPY NEW YEAR TO YOU ALL

The neutral current is equal to the phasor sum of the 3 currents.

When this is drawn out as shown in the diagram below the solution should become clear.

Does this make sense?
Can you take it form here?