Cant remember studying this but if anyone can help me out that that would be sweet

[lofty84]

Hi im reviseing for my 203 and this question has got me stumped its a bit hard to draw it out exactly as tthe lines do not match up but its basically two lines that form a right angle with an arrow head on the end of them.

Its quite hard trying to remember things as if you failed an exam you just moved on as the resists weremonths away. Haveing got through everything else it would be nice to get this one under my belt so its just the 301, 302, and 303 to do but I dont remember looking at things like this.




_ _ _ _ _ _ _ _ _ _ > I ref
|_|
|
|
|
V

V (this is the letter V)

the possible answers are

a. purely resistive
b.purely capacitive
c.resistor and inductor in series
d.purely inductive

the answer is b (purely capacitive) but why is this is the picture supposed to be showing me something

thankyou

 

[ian.settle1]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Hi im reviseing for my 203 and this question has got me stumped its a bit hard to draw it out exactly as tthe lines do not match up but its basically two lines that form a right angle with an arrow head on the end of them.

Its quite hard trying to remember things as if you failed an exam you just moved on as the resists weremonths away. Haveing got through everything else it would be nice to get this one under my belt so its just the 301, 302, and 303 to do but I dont remember looking at things like this.




_ _ _ _ _ _ _ _ _ _ > I ref
|_|
|
|
|
V

V (this is the letter V)

the possible answers are

a. purely resistive
b.purely capacitive
c.resistor and inductor in series
d.purely inductive

the answer is b (purely capacitive) but why is this is the picture supposed to be showing me something

thankyou

It is showing you that the current leads the voltage by 90 degrees as a pure cpactive circuit

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Heres another that I cant remember howto do

A transformer takes aload of 25 KVAat a power factor of 0.8.calculate the input power in KW

a.31.25 KW
b.200 KW
c.20 KW
d.312.5KW

The answers c. 20 KW

Ive worked out that 0.8 x 25 is 20 but how come you do it this way instead of say doing 25000/0.8

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

I cant beleive this lol ten questions in and heres the third one im stuck with.

if a 8 kw motorwith a power factor of 0.9isconnected across a 400v supply, calculate the current taken by the load ?

a.20 A
b.22.2 A
c.2.88 MA
d.2.22 A

the answers 22.2A and I guessed right but I wouldnt mind knowing how to do the calculation

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

This isnt looking good but at least ive given myself a chance by starting the revision early

How many numbers of turns are needed on a transformer if the primary windings are 1000 and the primary and secondary currents are 5 A and 20 A respectivly.

the options are

a. 200 turns
b.250 turns
c.100 turns
d. 50 turns

the answers b and i did come to that conclusion but was the way i did it right.

I said 1000/5 = 200

and 1000/20= 50

I then addedthem together

 

[ian.settle1]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Heres another that I cant remember howto do

A transformer takes aload of 25 KVAat a power factor of 0.8.calculate the input power in KW

a.31.25 KW
b.200 KW
c.20 KW
d.312.5KW

The answers c. 20 KW

Ive worked out that 0.8 x 25 is 20 but how come you do it this way instead of say doing 25000/0.8

How do you get the input lower than the output think you will find the answer is A

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Im back with another .lets hope it doesnt continue at this rate as I have 140 practice questions.

ok

what is the kva of aload given that the true power is 100kW and the ractive power is 25 kVAr

options are

a. 175 kVA
b. 150 kVA
c. 200 kVA and
d. 125 KVA

the answers d (125 kVA) but surely you dont add the 25 kVAr to the 100 Kw and what with 100 added to 25 being 125 just stick a kVA on the end

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

another

what is the total impedance of a circuit with a 4 (ohms sign) resistor and a 10 mH inductor

no idea on how to work this one out

4 (ohms)
7 (ohms)
5 (ohms)
6 (ohms)


the answers 5 ohms but ive no idea how to work that out

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

How do you get the input lower than the output think you will find the answer is A

It deffinantly said the answer was C mate just went back and double checked. Do you think thats wrong then Ian. Thanksfor haveng a look for me

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

This isnt looking good but at least ive given myself a chance by starting the revision early

How many numbers of turns are needed on a transformer if the primary windings are 1000 and the primary and secondary currents are 5 A and 20 A respectivly.

the options are

a. 200 turns
b.250 turns
c.100 turns
d. 50 turns

the answers b and i did come to that conclusion but was the way i did it right.

I said 1000/5 = 200

and 1000/20= 50

I then addedthem together

I don't know why your way works or even if it does or if it's just a coincidence, but:

Power in = Power out
Therefore the voltage must be inversely proportional to the current, ie the higher the current the lower the voltage to produce the same power, and vice versa.

First find the ratio:
20/5 = 4

Then apply the ratio of power and/or voltage to the number of turns:
1000/4 = 250


I appear to have mislaid the [divided by] button on my keyboard; I'm sure it used to have one.

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

another

what is the total impedance of a circuit with a 4 (ohms sign) resistor and a 10 mH inductor

no idea on how to work this one out

4 (ohms)
7 (ohms)
5 (ohms)
6 (ohms)


the answers 5 ohms but ive no idea how to work that out

With almost all the reactance questions they throw at you you'll have to find XL or XC.
Assuming f (frequency) = 50 HZ

XL = 2[Pi] f L
XL = 2x[Pi]x50x0.01
ZL = 3.14

Now you have the reactance of the coil you can find the total impedance:


Z = [square root] (R[squared] + XL~XC[squared])
Z = [square root] (4[squared] + 3.14[squared])
Z = [square root] 16 + 9.87
Z = 5

Could do with being able to use scientific symbols here...

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

hopeing for a question like that now, nicely explained and very much appreciated

know any of the others

 

[amer]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if i am not mistaken the formula is and the answer is 250 [FONT=&quot]Np Is transpose formula [/FONT]
[FONT=&quot]─ ─ Ns= Np×Ip[/FONT]
[FONT=&quot]Ns Ip ─────[/FONT]
[FONT=&quot] Is[/FONT]

 

[amer]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if i am not mistaken the formula is and the answer is 250 [FONT=&quot]Np Is transpose formula [/FONT]
[FONT=&quot]─ ─ Ns= Np×Ip[/FONT]
[FONT=&quot]Ns Ip ─────[/FONT]
[FONT=&quot] Is[/FONT]

 

[Guest123]

Re: Cant remember studying this but if snyone can help me out that that would be swee

I cant beleive this lol ten questions in and heres the third one im stuck with.

if a 8 kw motorwith a power factor of 0.9isconnected across a 400v supply, calculate the current taken by the load ?

a.20 A
b.22.2 A
c.2.88 MA
d.2.22 A

the answers 22.2A and I guessed right but I wouldnt mind knowing how to do the calculation

I get 12.8A....

 

[Guest123]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Heres another that I cant remember howto do

A transformer takes aload of 25 KVAat a power factor of 0.8.calculate the input power in KW

a.31.25 KW
b.200 KW
c.20 KW
d.312.5KW

The answers c. 20 KW

Ive worked out that 0.8 x 25 is 20 but how come you do it this way instead of say doing 25000/0.8

KW - KVAxPF = 25x0.8 = 20.

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

I cant beleive this lol ten questions in and heres the third one im stuck with.

if a 8 kw motorwith a power factor of 0.9isconnected across a 400v supply, calculate the current taken by the load ?

a.20 A
b.22.2 A
c.2.88 MA
d.2.22 A

the answers 22.2A and I guessed right but I wouldnt mind knowing how to do the calculation

I had to get my notes out for these - reactance I can do, but I struggled with this.

P
I = --------
V x pf



8000
I = -------------
400 x 0.9



I = 22.2A​

 

[Guest123]

Re: Cant remember studying this but if snyone can help me out that that would be swee

What about root 3 (1.732) ???

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

That would give an answer of 38.5A, which wasn't one of the options.
I seem to remember 3 phase supplies have to be treated differently but the question just stated it was a 400v supply with no mention of 3 phase, so I didn't want to over-complicate the issue.

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Thanks to everyone whos tried to help me. Really is much appreciated.

in regards to the sq root of three the 1.732 does that only comeinto play on three thase systems ?

 

[amer]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Thanks to everyone whos tried to help me. Really is much appreciated.

in regards to the sq root of three the 1.732 does that only comeinto play on three thase systems ?

yes you are right only in 3 phase

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

The voltage supplied to the primary winding of a step down transformer is 660v and the output voltage is 230v. If the primary current is 12A the value of the secondary current will be.

Would the answer be 660/230 x 12 = 34.4

 

[amer]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Is= VP*Ip/ Vs 660 * 12 =7920 / 230 =34.4

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

This isnt looking good but at least ive given myself a chance by starting the revision early

How many numbers of turns are needed on a transformer if the primary windings are 1000 and the primary and secondary currents are 5 A and 20 A respectivly.

the options are

a. 200 turns
b.250 turns
c.100 turns
d. 50 turns

the answers b and i did come to that conclusion but was the way i did it right.

I said 1000/5 = 200

and 1000/20= 50

I then addedthem together

The way to do it is with ratios if the number of turns increases 5 times then the voltage increases 5 times. Hower with transformers it's the opposite way around with I and V, if the current increases by , say, 10 times then the voltage decreases by 10 times - and vice versa.

In your example the the current has increased from 5A to 20A (20/5 =4) so it has increased 4 times.
That means the voltage must decrease by 4x . Also that the number of turns has decreased by 4x.
so 1000 turns /4 = your answer (250 turns) hope that helps a bit

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

How do you get the input lower than the output think you will find the answer is A

the input power is in kw and is just the power dissipated in the resistance, whereas the apparent power kVA takes account of the current used to generate magnetic fields as well. (It is the hypotenuse on the triangle so will be bigger than the adjacent)

It is not like saying you are getting out more power out than you are putting in.

The extra current for the apparent power is still there and this is why you have to calculate your cable size on the apparent power not the true power with inductive loads.

This is why there are always so many questions/problems on here when people are working with inductive lighting - they do there calcs with the power (kW) not the apparent power (kVA).

edit: the correct answer is c.

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Thanks to everyone who has got involved and helped me out.

If anyone knows how to tackle some of the harder unanswered ones it would be really helpfull

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Thanks to everyone who has got involved and helped me out.

If anyone knows how to tackle some of the harder unanswered ones it would be really helpfull

couldn't find any that were unanswered...

It is probably worth you remembering this-

" the kVA figure (apparent power) will always be bigger than the kW figure (true power)"

the only exception is if the the power factor was 1 (unity) and then they would be equal.

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Im back with some more which are confuseing me

1. A resistor and n inductor in a series circuit have a resistance of 6 (ohms) and an inductive Reactance of 8 (ohms) respectively. What is the total impedance.

The answer is 10 (ohms) but 8 + 6 is not 10

2. A transformer has a turns ratio of 400:4. If the supply voltage is 400v, what will be the secondary voltage

The answers 4 ,so does this mean that if it said it had a turns ratio of 400:10 and the supply voltage was 400v the secondary voltage would be 10 ??

Thanks

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

heres a weird one

what is the power factor of a circuit with a capacitor connectd in series, given the resistance is 6 ohms and the capacitive reactance is 8 ohms

answers are

0.6 lagging
0.75 leading
0.6 leading
0.75 lagging

completely stumped on this one

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Im back with some more which are confuseing me

1. A resistor and n inductor in a series circuit have a resistance of 6 (ohms) and an inductive Reactance of 8 (ohms) respectively. What is the total impedance.

The answer is 10 (ohms) but 8 + 6 is not 10

2. A transformer has a turns ratio of 400:4. If the supply voltage is 400v, what will be the secondary voltage

The answers 4 ,so does this mean that if it said it had a turns ratio of 400:10 and the supply voltage was 400v the secondary voltage would be 10 ??

Thanks

1.
Z = √(R² + XL²)
Z = √(6² + 8²)
Z = 10

2. The figures given are nice and easy to understand, ie 1 volt per turn, so yes if the turns ration wer 400:10 the secondary voltage would be 10.

 

[Adam W]

Re: Cant remember studying this but if snyone can help me out that that would be swee

heres a weird one

what is the power factor of a circuit with a capacitor connectd in series, given the resistance is 6 ohms and the capacitive reactance is 8 ohms

answers are

0.6 lagging
0.75 leading
0.6 leading
0.75 lagging

completely stumped on this one

pf = R ÷ Z

You have the value for R so you need to find Z first

Z = √(R²+XC²)
Z = √(6²+8²)
Z = 10


pf = R ÷ Z
pf = 6 ÷ 10
pf = 0.6

Remember the word "CIVIL" -
C I V = Capacitive, current (I) leads the voltage (V)
V I L = Inductive (L), current (I) comes after V, therefore lags.

It's a capacitive circuit, so the pf will be 0.6 leading.

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

heres a weird one

what is the power factor of a circuit with a capacitor connectd in series, given the resistance is 6 ohms and the capacitive reactance is 8 ohms

answers are

0.6 lagging
0.75 leading
0.6 leading
0.75 lagging

completely stumped on this one

notice the values 6 ohms and 8 ohms (multiples of 3 and 4) impedance is 10 ohms (multiple of 5)
using pythagorus - if you need it

PF=R/Z
=6/10
= 0.6

and a capacitor means it is leading

edit : apologies novice sparkus didn't see you had answered it on the next page - at least we agree lol

 

[JUD]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Im back with another .lets hope it doesnt continue at this rate as I have 140 practice questions.

ok

what is the kva of aload given that the true power is 100kW and the ractive power is 25 kVAr

options are

a. 175 kVA
b. 150 kVA
c. 200 kVA and
d. 125 KVA

the answers d (125 kVA) but surely you dont add the 25 kVAr to the 100 Kw and what with 100 added to 25 being 125 just stick a kVA on the end

I get 103kVA for this one.

kVA = √(kW² + kVAr²) = √(100² + 25²) = 103kVA

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

I get 103kVA for this one.

kVA = √(kW² + kVAr²) = √(100² + 25²) = 103kVA

Aaah the good old option E) !!

In actual fact it is D)



It is the typical trick typo question - the typist mistook a 7 for a 2! The reactive power is actually 75 kVAr!

I know this because for some reason the examiners have a fascination for 3, 4, 5 triangles - in this case 75, 100, 125 - why i don't know - maybe they don't own calculators.

 

[JUD]

Re: Cant remember studying this but if snyone can help me out that that would be swee

Aaah the good old option E) !!

In actual fact it is D)



It is the typical trick typo question - the typist mistook a 7 for a 2! The reactive power is actually 75 kVAr!

I know this because for some reason the examiners have a fascination for 3, 4, 5 triangles - in this case 75, 100, 125 - why i don't know - maybe they don't own calculators.

Of course. I should have known.

You have to correct the question before you answer it.

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if anyone can help me to answer these id be greatfull as my exams in the morning

the power factor of a resistive and capacative circuit haveing an impedance of 3 ohms and a resistance of 2.5 ohms is

the answers 0.83 leading

 

[JUD]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if anyone can help me to answer these id be greatfull as my exams in the morning

the power factor of a resistive and capacative circuit haveing an impedance of 3 ohms and a resistance of 2.5 ohms is

the answers 0.83 leading

pf = Resistance ÷ Impedance

2.5 ÷ 3 = 0.83 (leading because it's capacitive)

 

[Sintra]

Re: Cant remember studying this but if snyone can help me out that that would be swee

PF=R/Z

PF=2.5/3

PF = 0.83 leading as its capacitive

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

thanks alot guys

 

[lofty84]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if anyone knows why the kva of a load is 125 Kva given that the true power is 100kW and the reactive power is 75Kvar that would be handy.

 

[pushrod]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if anyone knows why the kva of a load is 125 Kva given that the true power is 100kW and the reactive power is 75Kvar that would be handy.

You should be working these out for yourself now enough people have explained it - have a look back at some of the replies from NovusSparkus, JUD and others. Not wishing to be a pain in the ---, but in your exam you will be there by yourself. This example is very similar to others you have posted.

 

[Phil]

Re: Cant remember studying this but if snyone can help me out that that would be swee

if anyone knows why the kva of a load is 125 Kva given that the true power is 100kW and the reactive power is 75Kvar that would be handy.

i will do the exam for 50 quid if you want plus travelling

 

[lofty84]

Re: Cant remember studying this but if anyone can help me out that that would be swee

If you was local id take you up on it


A transformer has a primary winding with 1100 turns and 380 secondary turns. It is connected to a 230v supply. What will the secondary voltage be?

 

[JUD]

Re: Cant remember studying this but if anyone can help me out that that would be swee

If you was local id take you up on it


A transformer has a primary winding with 1100 turns and 380 secondary turns. It is connected to a 230v supply. What will the secondary voltage be?

380 ÷ 1100 x 230 = 79.5V

or

230 ÷ (1100 ÷ 380) = 79.5V

 

[lofty84]

Re: Cant remember studying this but if anyone can help me out that that would be swee

ive tried to work this out one out but dont seem able to

a capacitor of capacitive reactance 5 ohms an inductor of 8 ohms inductive reactance and a resistor of 4 ohms are connected in series to a 230v 50 hz supply. the total impedance of the circuit will be

exams in half hour

 

[Sintra]

Re: Cant remember studying this but if anyone can help me out that that would be swee

ive tried to work this out one out but dont seem able to

a capacitor of capacitive reactance 5 ohms an inductor of 8 ohms inductive reactance and a resistor of 4 ohms are connected in series to a 230v 50 hz supply. the total impedance of the circuit will be

exams in half hour

Z= SqRt(R x R + XL x XL)

Inductive reactance and capacitive reactance cancel each other out so you are left with 3 ohms inductive. Therefore.

Z= SqRt(4 x 4 + 3 x 3)

Z= SqRt(16 + 9)

Z= SqRt(25)

Z= 5 ohms

 

[Adam W]

Re: Cant remember studying this but if anyone can help me out that that would be swee

ive tried to work this out one out but dont seem able to

a capacitor of capacitive reactance 5 ohms an inductor of 8 ohms inductive reactance and a resistor of 4 ohms are connected in series to a 230v 50 hz supply. the total impedance of the circuit will be

exams in half hour

You don't even have to find XL and XC so it's easy:

Z = √R²+(XL~XC)²
Z = √4²+3²
Z = 5Ω

 

[Adam W]

Re: Cant remember studying this but if anyone can help me out that that would be swee

Well same answer as Sintra, just remember:

Formula
Figures in the formula
Answer.

... Not that you'll get anything that dificult in the exam.
Good luck!

 

[lofty84]

Re: Cant remember studying this but if anyone can help me out that that would be swee

Cant beleive it as I got a distinction. Thanks to everybody who helped me you were superb.

 

[JUD]

Re: Cant remember studying this but if anyone can help me out that that would be swee

Maybe if you tell us what you're not getting we could help you.

Here are some of my notes:

Capacitive Reactance Current leads

Xc = 1 ÷ (2 x π x f x C)

f = 1 ÷ (2 x π x C x Xc)

C = 1 ÷ (2 x π x f x Xc)

Xc = Capacitive Reactance (Ω), f = Frequency (Hz), C = Capacitance (F)

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

Inductive Reactance Current lags

XL = 2 x π x f x L

XL = Inductive Reactance (Ω), f = Frequency (Hz), L = Inductance (H)

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

Total Reactance

X = XL ~ Xc

X = Total Reactance (Ω), XL = Inductive Reactance (Ω), Xc = Capacitive Reactance (Ω)

Subtract the lower value out of your Capacitive and Inductive Reactances from the higher value.

In the case of the above question that would be 8 - 5 = 3Ω Inductive Reactance.

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

Impedance

Z = √(R² + X²)

R = √(Z² - X²)

X = √(Z² - R²)

Z = Impedance (Ω), R = Resistance (Ω), X = Reactance (Ω)

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

So you now have your values for resistance (R) and reactance (X).

To work out your impedance (Z), the formula is:

Z = √(R² + X²)
Z = √(4² + 3²)
Z = √(16 + 9)
Z = √(25)
Z = 5Ω