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

[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

 

[w crane]

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

TRY THESE TWO MEMORY AIDS

1/ "C I V I L" aid.

Capacitance Current leads Capacitance Voltage. and/or lnductance Current lags inductance Voltage

2/
and old Red Indian word: "SOH,CAH,TOA",for phasor dia: ie Sine = Opposite over Hypot
Cosine = Adjacent over Hypot..
Tan =Opposite over Adjcent
see the patten hope it helps.