Capacitive or Inductive: R//L//C

[ElectroNewby]

Hello, folks! I need a little hand to get an understanding in this one.

Capacitive or Inductive: configuration is R//L//C with 5VAC 12khz, R=100Ω, L=15mH and C=0.022µF

I know that in series it would be easy... Xtotal would be Xl-Xc or the inverse. However, with this parallel config, I'm hurting a rock. Another thing also, is that in my notes, when we have // circuits, they gave us the Xl and Xc values, but here I need to find them.

I cannot find the formula for Xc in parallel.
It's Xc=1/(2*pi*f*C) in series and Xl=2*pi*f*L in serie, right ?

Thank you in advance once more !

 

[Ratchit]

ElectroNewby,

Inductors and capacitors have the same reactance in parallel as they do in a series configuration. You asked about a formula, but what are you trying to solve in the problem? It appears you are trying to find the answer by rote plug and chug instead of actually knowing what is happening. Ratch

 

[MrAl]

Hi,

Im not sure either what you are asking for as xC does not change depending on how you use it.
Did you mean you wanted to know the impedance of the RC circuit, or the total impedance of RLC in parallel?
Perhaps you can post a circuit diagram too.

 

[ElectroNewby]

I need to know if the circuit if more L or more C

 

[MrAl]

Hi again,

Oh you mean you need to calculate the current through the circuit?

 

[birdman0_o]

You are forgetting the ever important j's. Both Xc and Xl are imaginary. You need to be able to do some complex algebra to solve for the parallel impedance.

 

[RCinFLA]

The simplest way to do parallel, or parallel/series combination of reactance is to get familar with polar-rectangular conversion.

Parallel is inverse of the sum of the invererses. For example @ 12KHz, L=+j1131, C = -j603, R = 100

If all three are in parallel the reactance is 1/(1/R + 1/XL + 1/Xc). You must do this keeping complex numbers.

Xtotal = 1/(1/100 + 1/+j1131 + 1/-j603) = 1(0.01 + (-J8.84e-4 +j1.658e-3)) = 1/(0.01 + j7.742e-4)

Convert denominator to Polar: Rect(0.01 + j7.742e-4) = Polar (1.003e-2 at angle of +4.427 degrees)

Invert polar denominator: (invert magnitude and change polarity of angle) -> 1/1.003e-2 at angle of -4.427 degrees = 99.7 at angle of -4.427 degs. The polar form makes inversion easy.

Convert back to Rectangular format , final answer -> 99.4 - j7.696 ohms

At 12 kHz this equals a 99.4 ohm resistor in series with a 1.72 uF cap.

This assumes you have a calculator with Polar to Rectangular and Rectangular to Polar conversion. If you don't have one, get one.

 

[Ratchit]

ElectroNewby,

I need to know if the circuit if more L or more C

OK, that is easy. The inductive reactance is 1131 ohms and the capacitive reactance is 603 ohms. With a constant frequency voltage existing in parallel across both components, more current will exist in the capacitor than in the inductor. Therefore, the circuit will be capacitive. You only have to know the reactances to answer your question.

Ratch

 

[ElectroNewby]

Yes actually, Ratchit answered exactly what I needed. In // more resistance in Xl than Xc will result in a current going more in the Xc.
My bad, it was very simple.

thanks a lot !

 

[Ratchit]

ElectroNewby,

In // more resistance in Xl than Xc will result in a current going more in the Xc.

Now my pedantic compulsion kicks in. Be careful referring to a reactance as a resistance. They both restrict current and have the same units (ohms), but each does it a different way. For the problem given, a decrease of current in the inductor branch will not change the current in the capacitor branch. Only the current proportions change.

Ratch

 

[ElectroNewby]

Yes I know, sorry . I've learned about the active, reactive and apparent power this weekend. This science rocks