Maths - Inductance and capacitance: Advanced

[jimmyray]

I started with the resistors, working out both parallel sides, then combining them in series to get a total resistance of 31.21 Ohms. Then calculated overall capacitance by working out the two in series, followed by a parallel calculation including the one on its own, with the result of 100 micro farads. Calculated Inductive reactance and capacitive reactance, with the results as: Xl = 10.68 Ohms and Xc = 3.75 Ohms. Impedance came out at 31.97 Ohms, using the Resistance and Reactance results. I may be using the wrong formula for power factor, as I was using Pf= R/Z. Apologies for essay response, and thanks again for your help.

 

[D Skelton]

Ok, somewhere you've gone wrong calculating Xc. Try that part again. I'm guessing you calculated the overall capacitance carrying out the parallel calculation on the series capacitors and the opposite on the two parallel capacitance values as you have the correct value of overall capacitance?

 

[jimmyray]

Will come back to the calculations tomorrow, as my brain is fried right now! That's exactly what I did for the capacitance.

 

[D Skelton]

No worries, just remember; Xc = 1/(2 x π x f x l)

 

[jimmyray]

I've recalculated the Xc = 18.724 = 18.72Ohms. To work out Impedance, I'm using the formula Zt = sqrt(Rsq + (Xl - Xc)sqd) (please excuse the awful formula layout!) basically the square root of ((resistance squared plus (Xl - Xc)squared)). Any help appreciated.

 

[D Skelton]

Ok, what you're trying to do there is work out impedance using resistive and reactive values in parallell configuration. The way you are doing it won't work and to be honest, you don't even need to know the impedance to work out this calculation.

Instead, what you need to concentrate on doing now is working out your resistive current; I[SUB]R[/SUB], then you need to work out your reactive currents; I[SUB]Xc[/SUB] and I[SUB]Xl[/SUB]. This is simple ohms law and is the value of current as a product of the resistive element of the circuit, the capacitive element of the circuit and the inductive element of the circuit.

Your power factor is then equal to: I[SUB]R[/SUB] / I[SUB]T[/SUB].

I[SUB]T[/SUB] being equal to; sqrt(I[SUB]R[/SUB][SUP]2[/SUP] + (I[SUB]Xl[/SUB]- I[SUB]Xc[/SUB])[SUP]2[/SUP])

See how you get on with that, you should be able to find the power factor. Once that is done I can help you through how we can 'scrub' what inductance we dont need with a little capacitance

 

[Rockingit]

This is quite good fun all round, making me remember college stuff I haven't used for 20+ years!! Well done chaps.

 

[D Skelton]

No worries, just remember; Xc = 1/(2 x π x f x l)

Sorry, I've just realised that the above formula in post #19 should read; Xc = 1/(2 x π x f x c) instead.

 

[jimmyray]

Apologies for delay, had a hectic weekend! Been looking at this for a while, and from my calculations the circuit to begin with has a power factor of 0.62 Inductive. In order to get the power factor up to 0.90, a capacitor of 46uF(micro Farads) should be placed in parallel with the other capacitors. Please say this is somewhere close, as my brain has started to go round in circles with all those formulas!! I used the Current formulas. Cheers. J

 

[D Skelton]

Bang on the nose mate!

Without wanting to sound patronising at all, you've got a promising career ahead of you if as a student who hasn't yet covered this topic you're already coming out with answers like that!

Top stuff!

 

[jimmyray]

Got there in the end! (Got about 5 sheets of A4 filling with workings!) Thanks mate, really appreciate all your help.

 

[D Skelton]

No worries mate, it's a pleasure. Anything else you know where I am

 

[mr7526]

It's a long time since I've covered power factor correction, but I've worked out the power factor. I'd appreciate someone looking over my work:

The schematic simplifies to being a 41μF capacitor + 20mH inductor + 31Ω in parallel with a 1060V supply @ 170π rad / s.

1 / r = 1 / r1 + 1 / r2 + ... + 1 / rn

= 170π * 41E-6 * j + 1 / 170π * 20E-3 * j + 1 / 31.2

r = 5.19 + 11.6j Ω

By Ohm's law, I = V / R = 1060 / 5.19 + 11.6j = 83.4 /_ - 65.9°

As power factor is the angle between the voltage and current, the power factor is -65.9 lagging.

Is that along the right lines?

 

[telectrix]

40 years ago, i'd have done this but several thousand beers and countless packets of smokes, the brain hurts adding up a bar bill. LOL. :rofl:

 

[D Skelton]

It's a long time since I've covered power factor correction, but I've worked out the power factor. I'd appreciate someone looking over my work:

The schematic simplifies to being a 41μF capacitor + 20mH inductor + 31Ω in parallel with a 1060V supply @ 170π rad / s.

Your capacitance is out. Remember that working out capacitance in series and parallel is the exact opposite of working out resistance.

1 / r = 1 / r1 + 1 / r2 + ... + 1 / rn

= 170π * 41E-6 * j + 1 / 170π * 20E-3 * j + 1 / 31.2

Can't work out exactly what you're trying to do here and why? X[SUB]c[/SUB] + 1/X[SUB]l[/SUB] + 1/R? I might be wrong but I think you may be considering a linear equation for a non linear circuit using the resistive and reactive values? What you need to do is use the values of resistive and reactive currents in a non linear equation such as sqrt(I[SUB]R[/SUB][SUP]2[/SUP] + (I[SUB]Xl[/SUB]- I[SUB]Xc[/SUB])[SUP]2[/SUP]). This gives you a value to use in working out PF.

r = 5.19 + 11.6j Ω

What are these values?

By Ohm's law, I = V / R = 1060 / 5.19 + 11.6j = 83.4 /_ - 65.9°

This needs simplifying, you're over thinking things. The power factor is I / I[SUB]T[/SUB] it's that simple.

As power factor is the angle between the voltage and current, the power factor is -65.9 lagging.

The power factor isn't the angle, it's the cosine of the angle. The existing angle can be worked out as being 51.7°, the cosine of which is 0.62. That is your existing power factor.

Is that along the right lines?

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