Currect flowing in a circuit & Heater resistance questions

[a1]

PS - Good Luck in your 203 exam Tomorrow Searley.

 

[Andy™]

So Much to do and so little time to do it in - I am Nervous. :_| The college are changing the 2330 from a two year course to a three year course form September.

maybe get Admin2 to ban you from the forum so there are no distractions until its done

 

[a1]

I have hardly been on here Andy Mate. :_|

 

[Andy™]

I have hardly been on here Andy Mate. :_|

it will all be worth it once you finish though. sometimes gotta make sacrifices to get it done

 

[mikey-c]

Good luck with the exam Admin/WW. I'm sure you'ld do well

Started the assignment write up last week, but exams arent til june.

As for the questions, isnt it I = P / (V x pf) so 11.4A

 

[Andy™]

Good luck with the exam Admin/WW. I'm sure you'ld do well Started the assignment write up last week, but exams arent til june.

As for the questions, isnt it I = P / (V x pf) so 11.4A

well, thats bit of an easier way than my method!

 

[SPECIAL LOCATION]

Q2 Need to work out the current firstAMPERES = POWER / VOLTS

3KW(3000) / 230V = 13A

Then for the resistance

RES = VOLTS / AMPERES

230V / 13A = 17.6ohm

There is no need to work out the current... ?:|

why not just go straight to resistance? :|

230v X 230v/3000w = Resistance.

52900/3000 = 17.6ohm

If you can remember how to transpose the main formula it can save time in exams.

i.e. because we know

P=VxI

and I=V/R

therefore P=(VxV)/R

so R=(VxV)/P

HTHGuiness Drink

 

[Apache]

There is no need to work out the current... ?:| why not just go straight to resistance? :|

230v X 230v/3000w = Resistance.

52900/3000 = 17.6ohm

If you can remember how to transpose the main formula it can save time in exams.

i.e. because we know

P=VxI

and I=V/R

therefore P=(VxV)/R

so R=(VxV)/P

HTHGuiness Drink

or even

R = V

 

[Jono Pashley]

Sorry folks but this is defiantly level 2 stuff, some elements will transfer to level 3 but these are level 2 questions all the same.

As for exams i have...

ECS Health & Safety This Wednesday

203 the following Wednesday

206 on Saturday 23rd

205 on June 17th

Gonna be a hectic few months i think.

 

[SPECIAL LOCATION]

Sorry folks but this is defiantly level 2 stuff, some elements will transfer to level 3 but these are level 2 questions all the same.As for exams i have...

ECS Health & Safety This Wednesday

203 the following Wednesday

206 on Saturday 23rd

205 on June 17th

Gonna be a hectic few months i think.

Concur with point 1.

and good luck with points 2, 3, 4 & 5!

Guiness DrinkGuiness DrinkGuiness DrinkGuiness Drink

 

[golfpaul]

so much easier once your qualified....then you just have to go through yearly assessments if you want to be part of a scam

:^O

 

[SPECIAL LOCATION]

or evenR = V

 

[Ed work]

Thank you

 

[a1pacino]

Hi folks please correct me if this is wrong but could the formula for question 1 be

I = P/V x PF

I = 2500/230 x .95

I = 10.32 Amps

is there a difference in formulas when we have "the power dissapated" or when "power is consumed"

Pat

my calc gave me this:

2500/(230 x 0.95)=11.44a

 

[Apache]

my calc gave me this:2500/(230 x 0.95)=11.44a

you are using the brackets wrong. The calculator gives priority to the brackets.

Type in to your calculator

(2500/230)X0.95 and see what you get

you are dividing 2500 by the the product of 230 X 0.95

do you see?

 

[a1pacino]

when i do it this way:

2500/230=10.86

10.86 x 0.95=10.32a

when i do it the way we have been taught in college using fraction button (casio fx-83es) it comes out:

230x0.95=218.5

2500/218.5=11.44a

Also same question is q.14 on paper 2 on admins 301 questions and the answer sheet says it is 11.44a

Confused now as to what is correct!

can anyone clear this one up? ?:|

 

[Apache]

when i do it this way 2500/230=10.86 then 10.86 x 0.95 = 10.32awhen i do it the way we have been taught in college using fraction button (casio fx-83es) it comes out:

230x0.95-218.5 then 2500/218.5=11.44a

Also same question is q.14 on paper 2 on admins 301 questions and the answer sheet says it is 11.44a

Confused now as to what is correct! ?:|

Just think slowly thorough what we are trying to achive

I = P/V x PF

The current is the watts/volts (in effect a ratio) multiplied by the power factor ie 95% of the ratio (the product of P/V)

You HAVE to calculate the P/V first!

Another way of explaining it:

you have 95% (0.95) of the watts and 95% of the volts. If you multiply P by 0.95 then multiply V by 0.95 you get

2375/218.5 = 10.87a

Am I making a word of sense?

 

[a1pacino]

yes you are Apache it is much appreciated as well,

there is a mistake on the 301 answer sheet they must have done it the same way I did!

College has taught me to do it that way its lucky as i would have done it that way in the exam and lost a mark as both of those answers are there!

 

[a1pacino]

i found this example in scaddan book:

if a motor has power rating of 1kw at 230v and the motor windings cause PF of 0.6 then as

PF=kW/kva

kVA=kW/PF

so 1/0.6=1.667kVA or 1667VA

and since

I=VA/Volts

I=1667/230

=7.25a

so using above method for original question:

A single-phase circuit dissipates 2.5kw of power and operates at a power factor of 0.95. The current flowing in the circuit will be:

kVA=kW/PF

2500/0.95=2631VA

and since

I=VA/Volts

I=2631/230=11.4a

Now i'm back to being confused again! ?:|

Apache I am in no way doubting your god given talents and your method does seem to make sense to me as well but are you 100% sure your way is correct?

 

[Apache]

Apache I am in no way doubting your god given talents and your method does seem to make sense to me as well but are you 100% sure your way is correct?

No!

but if the formula given above is correct

I = P/V x PF

then I am!