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Author Topic: A.c circuit calculation  (Read 2563 times)

Offline Chikis

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A.c circuit calculation
« on: 10/11/2012 19:52:33 »
A source of e.m.f 240V and frequency 50 HZ is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V.

Calculate the:
(i) potential difference across the capacitor,
(ii)capacitance of the capacitor,
(iii) inductance of the inductor

 I got the answer to  (i)
as 245V

that of (ii) I got to this extent C= 1/2450(pi)
= 1/7700
= 0.000129870
aprox. 0.0001299
= 1299*10^-4 uf

my problem now is that when I checked my answer booket to see wether the answer I got for (ii) was correct, I was highly disapointed to find the answer as 130 uf instead of 1299*10^-4 uf.
What really happened?


 

Offline syhprum

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Re: A.c circuit calculation
« Reply #1 on: 10/11/2012 23:06:07 »
This question does not fall into the group of Physics, Astronomy & Cosmology may I suggest it be moved to Technology or Geek speak.
 

Offline RD

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Re: A.c circuit calculation
« Reply #2 on: 11/11/2012 00:42:45 »
I got to this extent C= 1/2450(pi)
= 1/7700
= 0.000129870
aprox. 0.0001299
= 1299*10^-4 uf

1/(2450*pi) is 1.299*10^-4  [to 4 sig figs]

which is the same as 129.9*10^-6

which is 129.9 μF  (μ is 10^-6)

[ you said correct answer given is 130μF , so the disagreement is less than a tenth of one percent ]
« Last Edit: 11/11/2012 00:55:07 by RD »
 

Offline syhprum

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Re: A.c circuit calculation
« Reply #3 on: 11/11/2012 08:52:29 »
There seems to be some simple error in the arithmetic 1/7700 F is 130 micro farad
 

Offline syhprum

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Re: A.c circuit calculation
« Reply #4 on: 11/11/2012 14:15:09 »
It is probably the way the question is written but I cannot see how the voltage of 245 volts appears across the capacitor, we are told that 140 volts are dropped by the resistor so the circuit now boils down to a 100 volt supply feeding an inductor and capacitor in series with 50 volts across the inductor it would seem that 50 volts must also appear across the capacitor.
I take it the voltages are measured relative to some common point and not across the components as stated.
I know all about i and -i but as regards voltage as the question is posed they are not relevant.
« Last Edit: 11/11/2012 16:33:35 by syhprum »
 

Offline Bored chemist

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Re: A.c circuit calculation
« Reply #5 on: 11/11/2012 15:05:29 »
It would seem that way, but that's not what happens.
You can get voltages across the inductor or capacitor in this sort of circuit which are larger than the supply voltage.
 

Offline syhprum

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Re: A.c circuit calculation
« Reply #6 on: 11/11/2012 16:26:21 »
I withdraw what I said previously about understanding i and -i I have drawn some right angled triangles and applied Pythagoras and the penny has dropped.
when I calculate the voltage across the capacitor I find it is in fact 244.936 volts which might account for the small error that arises assuming it is 245 volts.
After a little battle with brackets I coaxed it into Mathematica and got the right answer
         10/((((240^2 - 140^2)^.5) + 50)10^-6 *2*50*Pi) = 129.956 μF
« Last Edit: 12/11/2012 13:54:31 by syhprum »
 

Offline techmind

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Re: A.c circuit calculation
« Reply #7 on: 11/11/2012 21:42:59 »
This question does not fall into the group of Physics, Astronomy & Cosmology may I suggest it be moved to Technology or Geek speak.
While the subject is arguably electronic engineering, basic Ohm's Law and with complex impedances will fall within the scope of many advanced physics courses - therefore I'm happy for it to stay. The "Tech" or "Geek speak" tends to be much more applied technology or software anyway.

Without actually having tried this specific problem (but I've been getting my head around a somewhat more complicated one in the past 2 weeks at work) it should work using
Z (resistor) = R,   Z (inductor) = jwL,     Z (capacitor) = 1/jwC = -j/wC
and V = I.Z
and I is the same throughout the circuit as everything is in series in this example.

When the maths gives "complex" voltages across the inductor or capacitor, you need to take the magnitude (i.e. sqrt(real^2 + imag^2)) to get the voltage you'd measure with a physical voltmeter.

And inductor/capacitor circuits can certainly give you much higher voltages than you put in, e.g. a sonoluminescence experiment I built years ago generated about 700V on the capacitive piezo-transducers from only 30V input.
 

Offline syhprum

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Re: A.c circuit calculation
« Reply #8 on: 11/11/2012 23:16:00 »
I am pleased for the opportunity blow the cobwebs from my mind to tackle problems that I learnt to solve seventy years ago, but advanced physics surely this is rather elementary stuff advanced physics is QD and general relativity.
I certainly do not consider myself to have any understanding of advanced physics and only learnt how to solve these problems to build crystal radio sets.
« Last Edit: 11/11/2012 23:20:46 by syhprum »
 

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Re: A.c circuit calculation
« Reply #8 on: 11/11/2012 23:16:00 »

 

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