Help please!!

[Lainge]

Hi all,

Im having trouble with a calculation for college & keep coming back with the same answer which looks wrong. Please could someone give me some help with this one?

A resistor of 4ohms is connected in series with an inductor of 3ohms. The supply voltage is 200v 50hz; what is the power dissipated in the circuit?

 

[ADS]

Hi all,Im having trouble with a calculation for college & keep coming back with the same answer which looks wrong. Please could someone give me some help with this one?

A resistor of 4ohms is connected in series with an inductor of 3ohms. The supply voltage is 200v 50hz; what is the power dissipated in the circuit?

You need to find the Impedance of the circuit first.

Then you need the 'Power' formula - the one using 'voltage'.

Done correctly, you should get 800 watts.

As Canoeboy says - let's see your calculations, then.

 

[Lainge]

Ok so what Im doing is to get impedance I need the square root of 4squared + 3 squared = z 5ohms

Then I use I = v/z instead of R 200/5= 40A

Then as P= I*V 40A * 200= 8kw ?

 

[GeoffreyBubbles]

Ok so what Im doing is to get impedance I need the square root of 4squared + 3 squared = z 5ohmsThen I use I = v/z instead of R 200/5= 40A

Then as P= I*V 40A * 200= 8kw ?

I get the same!!

Z= Root sq 4 Sq + 3 Sq = 5 ohms

Then,

P= V Sq/R so: 200 Sq/5 = 8000 W

 

[ADS]

Sorry

I did the 200 x 200 in my head - left a nought off ha ha

It is 8000, not 800.

Lainge, use P = V squared / R instead

 

[meerkat]

Hi Lainge

the formula for single phase power dissipated is

V x I x PF

I = V/R or as in this case Z

PF = R/Z

Answer = 6400Watts

V squared divided by R is the total power, not the power dissipated

if you divide your answer (8000) by 100 and times it by 80 you will get (6400) this is a good way of checking your answer

80 represents the 0.8 PF in other words 80% of (8000) = 6400 Watts

converesly 8000 divided by 100 and x by 20 = 1600

6400 + 1600 = (8000) Watts total

I hope this helps (MEERKAT)

 

[Larnacaman]

It should do meerkat, as your calculations and summary are spot on .... The others have neglected what was actually asked for in the question. Where there is inductance there will always be a PF content.... and the R (4ohms) divided by Z (5 ohms) = 0.8 (80%)

 

[ADS]

.. The others have neglected what was actually asked for in the question. Where there is inductance there will always be a PF content.... and the R (4ohms) divided by Z (5 ohms) = 0.8 (80%)

Not necessarily - 'dissipated' is just a fancy word for 'used'.

The question didn't stipulate whether it required 'apparent' power usage of the circuit, (the one you pay for), or the 'real' power usage,(what's actually used)

The problem with our answer was that if we're going with 'apparent' power, then the answer should be in KVA - so 8 KVA

'Real' power would require correction and, correctly, would be 6.4 KW

But, in the circumstances, neither answer could be deemed wrong.

 

[Larnacaman]

You have to remember here, that this was a collage question.

I don't know what the emphasis is these days, on exam/test questions at college level, but in my training and collage days, there was often an awareness element to such questions. Marks were often attributed to each part of a question.

So the fact that an inductor was part of this circuit changed it from a simple resistance equation, to one that needed an added element to arrive at the desired answer.

Your right that ''dissipated'' is another word for used, but again in an exam environment it can also mean much more than that!!! Words are chosen very carefully in Exam papers, to be fair to both those taking the exam and those marking the exam paper.

In essence, what i'm trying to say is, those that can show there calculation in this case to be 8KW, will receive one mark, those that identified a need for PF correction will receive another.

To this day, i always read equations carefully, i was caught out too many times in my training days, Not too!!! ...lol!!!

 

[ADS]

You have to remember here, that this was a collage question. I don't know what the emphasis is these days, on exam/test questions at college level, but in my training and collage days, there was often an awareness element to such questions. Marks were often attributed to each part of a question.

So the fact that an inductor was part of this circuit changed it from a simple resistance equation, to one that needed an added element to arrive at the desired answer.

Your right that ''dissipated'' is another word for used, but again in an exam environment it can also mean much more than that!!! Words are chosen very carefully in Exam papers, to be fair to both those taking the exam and those marking the exam paper.

In essence, what i'm trying to say is, those that can show there calculation in this case to be 8KW, will receive one mark, those that identified a need for PF correction will receive another.

To this day, i always read equations carefully, i was caught out too many times in my training days, Not too!!! ...lol!!!

Totally agree.

The other problem, of course, is how ambiguous some questions can be - especially when you're talking about City and Guilds.

I found this to be a constant problem in the courses I've taken - even the 2391 - City and Guilds examiners seem to have a problem with just asking a straight question.