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D Skelton

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Here's a nice little juicy one to get your teeth in to :)

[ElectriciansForums.net] Maths - Inductance and capacitance: Advanced
 
Neglecting my efforts on this here forum. Poor Sintra and his traffic light, I do apologise if he reads this. I've been a little pre-occupired with work at the moment; I had to rewire my first 3036 fuse this week. Given the rarity of such encounters I considered it something of a novelty.

Anyway, once more thanks for the efforts Mr. D. I'll have to retrieve my notes for this one once I've had one of Scaddens apparent pints. Or is it a true pint? I forget...
 
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C'mon Archy, I need some help to inspire these youngsers! :D

Trying to sell maths as 'cool' is like trying to get blood out of a stone!

Anyone gonna have a pop? ANYONE??? :(
 
Selling maths as anything is a tough one. Surely one of the students can make a start, resolving the resistors into a single resistance is an easy first step.
 
Hi mate, spent a fair bit of time with this one. My answer would be that A should be a resistor of value 368.49 Ohms to provide a power factor of 0.90. I think I'm pretty confident with this one, as have been grappling with my calculator to check it. I look forward to hearing from you. Cheers J
 
Just to check I'm going in the right direction, is the power factor initially 0.98 before trying to correct it? As this is what I've calculated so far.
 
No, as in any real life situation you are always trying to achieve unity or at least close to it. Sometimes allowances are made so that you don't tip the other end of the scale and as such you will find a selectable value on all PFCUs. This is a value you can choose to correct the PF to.

In this example, the existing PF is lower than the value I ask for it to be corrected to.

I can't think of any case where you would want to achieve a lower power factor :)
 

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