# Equivalent Impedance

Discussion in 'Homework Help' started by hitmen, Jan 6, 2010.

1. ### hitmen Thread Starter Active Member

Sep 21, 2008
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A residental apartment has
8 lamp of each of resistance 960ohm
3 ceiling fan each of impedance 345< 53.13 degrees (phasor)
2 air-con each of impedance of (48 + 36j)

Q is to determine the equivalent impedance.

I got the WRONG answer after I have separated everything into real and imaginery part and added many together. why?

Z real = (8*960)+ 3(345 cos 53.13) + 2*48
Z imaginery = 3(345 cos 53.13) + 2*36
I put them together and express them in phasor form.
Why am I wrong?

2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,621
468
Because the loads are in parallel, not in series.

3. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
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May I ask: How do you know that it is parallel?
How am I supposed to solve the problem then?

4. ### thyristor Active Member

Dec 27, 2009
94
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Well if they were all in series, you would go to switch on a lamp and instead of one lamp, you would get 8 lamps, 3 fans and 2 air-cons coming on simultaneously!!!!!!

Now do you understand why they must be in parallel?

The algebra needed to work out the answer is much simpler than for the poor guy who had to think up the question and make the complex number answers simple, which he has done in spades.

You should see that you now have 3 complex impedances in parallel made up of:

a) 8 (resistive) lamps in parallel

b) 3 fans in parallel (x + jy)

c) 2 air-cons in parallel (p + jq)

The maths is actually very simple owing to the numbers involved.

5. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
0
The maths is super complicated!

For 8 lamps, Z = 960^8 / (960*8) = 9.393093476*10^19
For 3 fans Z = (345<53.13)^3 / 3*345<53.13 = 39675<106.26
For 2 air-con Z = 1008+j3456 / 96+72j = 30<36.86989

And I have no idea how to combines all these impedances that are parallel!

Thanks! What is next?

Last edited: Jan 7, 2010
6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,621
468
Doesn't your text show the formula to calculate the equivalent of a number of paralleled impedances?

7. ### thyristor Active Member

Dec 27, 2009
94
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You may have a fundamental gap in your knowledge here.......

if you simply do a sanity check, you should realise that resistors in parallel cannot produce an answer for the overall resistance that is larger than any of the individual resistors !!!!!! Ergo, your calculation must be wrong.

eg: 8 identical resistances (R) in parallel give a combined resistance of R/8 so the answer for the lamps is 960/8 = 120 ohms. Parallel resistors are only PRODUCT/SUM if there are just two resistors involved

The other calculations are easy too...........

8. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
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I have been thought only the product/sum method.
But if all the resistors are parallel, wouldnt using product/sum on them simultaneously be the same

9. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
0
ok ok. I get it.
Ultimately, I get (1/R) = (1/120) + (1/ (69+92j)) + (1 / (24+18j))

Then it starts to get messy here, am i correct?
Then I convert it into phasor form and solve it.

10. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
0
Thanks for pointing out the gap in my knowledge