RLC in Parallel - Calculating instead of Drawing

[hightower]

Morning all,

Last night at college we moved on to RLC circuits in parallel, the type you'd typically find in a fluorescent light to correct power factor.

We drew out some parallelograms to work out the true current usage and power factor after adding a capacitor. The lecturer said that's all we need to know for level 3 - that drawing them out will be sufficient enough to work it all out.

However, I was just wondering if anyone can tell me the way to work it out through calculation? I've fathomed it using trigonometry and it was very accurate, but it felt a bit long winded so wondered if there was a shorter formula/method? I am pretty good at remembering formulas so I think I would rather do it through calculation if possible.

Thanks,

 

[RM11541]

Morning....If your calculator supports complex numbers it might be an option.

 

[Lucien Nunes]

the way to work it out through calculation?

What is the 'it' you want to work out? RLC circuits refer to a very general and broad range of situations and there are many parameters of interest. For power factor correction obviously one is interested in branch currents and their relative phase, and the net pf for a given frequency. For a tuned circuit in a radio one might be more interested in the resonant frequency and Q factor.

By trig methods I assume you are referring to phasors? Actually a good way to visualise what's going on as well as calculate. How good are you with complex numbers?

 

[hightower]

By trig methods I assume you are referring to phasors? Actually a good way to visualise what's going on as well as calculate. How good are you with complex numbers?

Yeah, that's right. Complex numbers? They sound complex, but I'll give anything a try. Like I say, I think I've got a pretty good head for maths but complex numbers might tell me otherwise.

 

[Lucien Nunes]

The idea of complex numbers for AC circuit analysis is to represent both the magnitude and phase of a quantity such as current or voltage with a single number of the form a+jb, instead of the mod-arg form of a phasor. For any given frequency the impedances of passive devices (R, L & C) have simple representations in terms of the pulsatance ω. By using complex equivalents of scalar functions and operators in your working, hey presto the answers contain not only the magnitudes but the phase as well.

E.g. what is the voltage across an impedance of 3-6jΩ when a current of 5+j2A flows through it?
V=IR, hence E=I•Z so we simply find the dot product instead of the scalar product.
E=I•Z
=(5+2j)•(3-6j)
=15+6j-30j+12
=27-24jV

Of course if you need to start or finish with a phase angle of some quantity then you need to do rectangular/polar conversions but one or more of those is usually trivial e.g. applied voltage.

 

[hightower]

Thanks Lucien, I hadn't ignored this, just decided to revisit it when I didn't have a bad head. Today was that day, but needless to say I now have a bad head. I think I'll just stick to drawing it out....

 

[Bobster]

You will cover J notation, (known as I notation when not using it in engineering terms for those that are into pure math) at a higher level, my last apprentices started with this during their HND along with laplace and fourrier transforms. (One to look up if you need a nap)