Power dissipation formulas?

[MFeely88]

Hi all, new here. I'm an apprentice near the end of my time, and taking my resit for the H+S and Electrical Principles exam 2330-301 either tomorrow or next Monday. I've got some practice questions, and I'm stuck on a few of them. I've got the answers so I don't need them, I just need to know what calculations to do as I'm not getting the right answers.

If a pure inductor of reactance 100 ohms is connected across a single phase 230 V a.c supply, the power dissipated in the circuit will be:a) 0 W B) 529 W c) 2.32 W d) 23 kW

The calculations I did landed me with the answer 529 W, but the answer sheet says the answer is 0 W, so I'm stumped. Any advice on power dissipation formulas and the like would be greatly appreciated.

Thanks, Martyn.

 

[Andy™]

a pure inductive, or pure capacitance load does not dissipate any power

 

[MFeely88]

I see. Thanks a lot, I was doing 230/100 = 2.3, then 2.3x230 = 529. If it wasn't a pure inductor, would this have been correct?

 

[Andy™]

if it was resistance, then yes

 

[MFeely88]

Ah, so which calculations should I be using for reactance? Thanks again mate.

 

[Robin Spark]

I am with Andy on this one and I think its answer a O watts. Oops sorry just realised you know this anywayBlushing

As for calculating power you could use power (watts)= volts squared/ Resistance

or

power(watts) = volts *amps

As for reactance then

inductive reactance Xl=2*pie*f*L

where pie =3.142 and f = frequency and L = inductance of circuit

 

[green-hornet]

Try this link

http://en.wikiversity.org/wiki/Reactance

 

[MFeely88]

I am with Andy on this one and I think its answer a O watts. Oops sorry just realised you know this anywayBlushingAs for calculating power you could use power (watts)= volts squared/ Resistance

or

power(watts) = volts *amps

As for reactance then

inductive reactance Xl=2*pie*f*L

where pie =3.142 and f = frequency and L = inductance of circuit

Thanks for the formula, I've got that one written down and have actually memorised it already, so it's nice to actually have piece of mind that I had the right one.

And thanks a lot for the link green-hornet, I'll check it out.

 

[sonasardar]

ohms law from P is derived from volts x current or v2/r

 

[sonasardar]

the answer is 23kw

power - volts x resistence

- 230 x 100

- 23kw

 

[a1]

Anyone care to update this page for me?

http://www.electricianforum.co.uk/electricalformulas.html

 

[a1]

the answer is 23kwpower - volts x resistence

- 230 x 100

- 23kw

The answer is 0w, I had this actual question in one of my exams.

 

[a1]

OP - Don't forget Swindon-Massive - Kevin has a great resource place, there.

 

[MFeely88]

OP - Don't forget Swindon-Massive - Kevin has a great resource place, there.

I've just Googled that, thanks for the tip. I've got one more formula that I'm stuck on here though and I've searched everywhere in my paperwork and Google but I can't seem to find anything.

An electric machine has a power output of 1500 Watts. The input current is 4A and the applied voltage is 400V. The efficiency of the machine will be:a) 50% B) 75% c) 87.55% d) 93.75%

I'm seriously stumped by this. I haven't yet looked at the answer as I want to work this one out and then check the answer, I just can't seem to remember (or find) a formula to help solve the equation. Any ideas?

 

[a1]

I've just Googled that, thanks for the tip. I've got one more formula that I'm stuck on here though and I've searched everywhere in my paperwork and Google but I can't seem to find anything.I'm seriously stumped by this. I haven't yet looked at the answer as I want to work this one out and then check the answer, I just can't seem to remember (or find) a formula to help solve the equation. Any ideas?

The formula to that one is on the link that I gave you.

http://www.electricianforum.co.uk/electricalformulas.html

 

[MFeely88]

The formula to that one is on the link that I gave you.http://www.electricianforum.co.uk/electricalformulas.html

I did see the efficiency formulas, but I don't know how the hell to get my power input worked out.

I'M AN IDIOT hahaha.

I've just worked it out using IxV (My input power), then dividing the 1500 Watts by the IxV, which gives me 0.9375. Multiply by 100 and bingo! 93.75%.

Thanks a lot Admin, appreciate all the help. I'm not as nervous about the exam now as I was the first time around. Now all I need to learn about motors, transformers and DC systems, piece of cake! (I wish...)