Discussion in 'Homework Help' started by peeka, Apr 25, 2009.

1. ### peeka Thread Starter Member

Apr 25, 2009
21
0
a 120 volt 60 hz circuit consists of coil that draws 4.8 amps and dissipates 205 watts,
if a 230 microfarads capacitor is connected in series with the coil the new ciruict current would b ?

the answer says 8.1 but i keep getin 4.3 or 4.8.

btw this is no homework, this is study and im trying to clarify from wat i did bout 2 years ago since ive completed my apprenciship in australia and now converting it over here in canada so its really hard no one to help me so yea. hopefully someone answers me soon coz this is really pissing me off.

2. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
Did you account for the inductance of the original coil? (Yes, you are given enough information to figure that out! Hint: dissipation only accounts for the RESISTIVE part of the original impedance.)

Let me know if that helps.

Eric

3. ### peeka Thread Starter Member

Apr 25, 2009
21
0
i havnt tried that question but i need some clarifycation with this formula that canada has, active current = total current * power factor.

now the thing i cant wrap my head around is i use the triangle to do everything neally.pythagoras one, anyways ive neva seen this formula and im jsut wondering wat else is called the active current ? coz the total current is the total line current which is obvous and the pf is powerfactor simple, but the active current is that the phase current ? for me would b Ip. and total current would b IL.

thanx for the really early reply wasnt expecting it but i will try that question again the dissapation (watts) or power, ive taken into account sorta
ill tell u wat ive done it mite help u lol.
r= v/i , v = voltage. that , 120/4.8 = 25
then the capacitor xc = 106/ 2pie x 60 x 230 = 11.53
i found impedence of that use the triangle Z = 27.53.
then I = v/z which is 120 / 27.53 = 4.35 n thats the answer i get. and ideas lol

most of the math questions i cant wrap my head around it but theory is easy dont ask me why lol.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The solution can go as follows:-

1. Find the coil resistance,
Rcoil= Power/(I^2) = 205/(4.8^2) = 8.9 Ω

2. Find the coil reactance,
Xcoil = Vx/I
Xcoil = (√(120^2 - Vr^2))/4.8

where Vr = volt drop across coil resistance and

Vr = Rcoil * I = 8.9x4.8 = 42.71 V

Hence Xcoil = (√(120^2-42.71^2))/4.8 = 23.36 Ω

3. With 230uF capacitance in series - capacitive reactance
Xc = 1/(2*Pi*60*C)
Xc = 11.53 Ω

4.Total reactance,
Xt = Xcoil - Xc
Xt = 23.36-11.53
Xt = 11.83 Ω

5.Total series impedance,
Zt = √(Rcoil^2+Xt^2)
Zt = √(8.9^2 + 11.83^2)
Zt = 14.8 Ω

6. New current = 120/Zt = 120/14.8 = 8.11 Amp

5. ### peeka Thread Starter Member

Apr 25, 2009
21
0
jeez dont i feel like an idiot

oh well live n learn i spose thank you t_n_k appreciate the help and that lovely explanation very easy to follow ill let u know how i go thanks

6. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
No need to fool feelish at all! This was actually a rather tricky problem (though a very good one!)

There are two things to watch out for....problems that give you more information than you need....these can lead you down the garden path.....and those that don't SEEM to give you enough information (as in this case).

Another approach you can use here is to first figure out the Real and Apparent power, and the phase angle. Then you can work backwards and see if you get the same answer! Quite a few more algebraic gymnastics involved, but it does lead to some great insights.

Eric

7. ### peeka Thread Starter Member

Apr 25, 2009
21
0
okay i understand real and apparent power and phase angle coz thats easy for me to understand. but that xcoil is funky neva seen that b4 but oh well like i said ill go thru the examples u guys gave me and the hints and ill work it out it looks like t_n_k used the reactive capacitance formula which i know but that xcoil is weird neva seen that.
well ill let u know how i go with this one ill try probs today at work thank u so much for the help

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I was a little cryptic in my derivation of Xcoil. It's never easy knowing how much info to write in a "succinct" response.

Perhaps some further explanation might help.

The coil may be viewed as a series combination of its resistance and inductive reactance - the latter I called Xcoil.

So Vcoil = √(Vr^2+Vx^2) where Vr is the resistive voltage drop and Vx is the reactive voltage drop. Of course these two voltages are out of phase & hence the pythagorean summation. I haven't concerned myself with phase angles - just resultant magnitudes.

BTW - I had in mind your earlier statement that you used Pythagoras for your calculations so I had tried to stick with that approach.

So rearranging the previous equation I get

Vx = √(Vcoil^2-Vr^2) - Vcoil and Vr are both known

and since we were given the current I we can find the coil reactance using

Xcoil = Vx/I

Hope that clarifies the mystery of Xcoil....

Knowing Xcoil and Xc (capacitive reactance) I was then able to calculate the new series reactance (i.e. the difference between the two) when the capacitor was added to the mix. From there I went on to calculate the new total series impedance (magnitude only) which includes the original resistance term.

And so on ...

9. ### peeka Thread Starter Member

Apr 25, 2009
21
0
where u said
"since we were given the current I we can find the coil reactance using"
Xcoil = Vx/I

thats just the resistance of the coil ? am i right ?

and where u say Vx is the reactive voltage drop ? is that the capacitors voltage drop wit the resistance of 11.53 ? arghhh damit cant understand anything anymore

thers a part , part 4. total reactance
xt = xcoil - xc (now where and how did u get that formula or knowledge to do that ?)

as u know im not the smartest but im very stuborn to get something perfect which i spose is very good. well many thanks to ur replys they are helping me alot im taking off friday to study all day coz im hoping to sit the exam in june.so if u got msn would b nice to talk to u bout that stuff always fun. djpeeka@hotmail.com

cya

10. ### peeka Thread Starter Member

Apr 25, 2009
21
0
im also a lil hazy on the "coil reactance"

like wehre did u find that formula ?

xcoil = (root(120^2 - Vr^2))/4.8

or did u make this formula up from ohms law v=ir and something ?

is there another way around this question using the triangle pythagoras method ?

where Z is hypotenuse
Xl or Xc is opposite

11. ### peeka Thread Starter Member

Apr 25, 2009
21
0
another question how do u add phasor angles god im sooo fkn pissd cant even do that

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
To read about the difference between resistance and reactance may I suggest you read some of the sections in VOL.II - Ac on this site. This is general link above but the following link discusses series LR circuits.

When you "add" inductors & capacitors in series to obtain an equivalent reactance, then you keep in mind that their individual reactance terms need to be added with regard to their sign. This has to do with the complex algebra that is used to analyze AC circuits. For the simple series case in your problem consider the inductive reactance as being positive sign and the capacitive reactance as being negative sign. To fully appreciate this you need to review complex numbers - if indeed you have used them in the past. If not then you'll need to get a handle on there use in AC circuit analysis.

While it's possible to analyze everything just using pythagoras triangles, more complex problems need more powerful techniques to solve them efficiently.

I understand your frustration - unfortunately it's usually a matter of time and hard work required for this stuff to sink in. If you are doing this "solo" it's much harder still - are there any refresher classes or courses you might be able to attend? Anyway - read through the notes on this site as suggested.

13. ### peeka Thread Starter Member

Apr 25, 2009
21
0
and yes im working on this solo but im going to call this tutor today who lives around my area. so hopefully he can help me

14. ### peeka Thread Starter Member

Apr 25, 2009
21
0
okay ive found a close answer to it,

i got 8.24 amps.

now the way i did it was from that example u passed me.i found the resistance of the full resistor R = p/i2
8.89
and found Xc which is 11.53,
so root(11.53^2 + 8.89^2 )
gives me 14.55 ohms
so then thats the full resistance of it, or reactance wateva u wana call it
so then I = V/Z which is 120/14.55 gives me 8.24 amps.
now is this correct coz i dont see how and why u would want to find the coil reactance of the resister.coz 1. i didnt see that in the example. and 2 neva seen that b4 so please let me know okay im dieing to know

15. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295

col reactance (XlL)= 2 pi F L

Eric

16. ### peeka Thread Starter Member

Apr 25, 2009
21
0
im having trouble once again with this question.

a 240 volt 60 hz , has a heater and a singlephase motor connect in parallel.motor uses 1300 watts power at pf .4 and a heater dissapates 600 watts.wats the total line currents (its spose to b 14.72A)

now i know i gota do something similar to the other question.i just cant rack my brain around it at this point hence why im asking and yes ive called the tutor guy up got answering message hopefulyl im not anoying u guys .
most of my answers for this question i get around 13.7 or 21a so i know imdoing something rong

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The power factor tells you that not all of the current is going towards producing real power losses (= output torque) in the motor.

Remember that Real Power = apparent power * pf

So for the motor first ...

Real power = 1300W
Apparent power = 1300/0.4 = 3250 VA

Using pythagoras

Apparent power = √((real power)^2+(reactive power)^2)

Reactive power = √((apparent power)^2-(real power)^2)

Reactive power = √(3250^2 - 1300^2) = 2979 VA

For the heater it's OK to assume the pf = 1 (since you were not told otherwise)

Real power = 600 W
Apparent power = 0

For the combined system

Total Real Power = 1300 + 600 W = 1900W
Total Reactive Power = 2979 + 0 = 2979 VA

Total Apparent Power = √(1900^2+2979^2) = 3533 VA

The Line current = Total Apparent Power / Vsupply = 3533 / 240 = 14.72A

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783

Real Power = 600 W
Apparent Power = 600 VA
Reactive Power = 0

19. ### peeka Thread Starter Member

Apr 25, 2009
21
0
god that was an easy question lol once u know how and wat to do with it ,it really becomes easy.maybe im just stressing to much and therefor i get jumbled up in everything.

well thanx again for explaining t_n_k appreciate it.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
No problem peeka.

Get out and enjoy some fresh air away from the books - it's the weekend!