# Parallel RLC Circuit

• Engineering

## Homework Statement ## The Attempt at a Solution

I found ω0 = 1/√LC when IR is at its maximum. (Purely resistive)

I have a feeling that the last part requires an approximation and then an expansion which gives 2 values of ω.

But the thing is, (1 + x)n ≈ 1 + nx + .....

only when x << 1

In this case it's L/RC << 1 which doesn't really fit.. Related Engineering and Comp Sci Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement ## The Attempt at a Solution

I found ω0 = 1/√LC when IR is at its maximum. (Purely resistive)

I have a feeling that the last part requires an approximation and then an expansion which gives 2 values of ω.

But the thing is, (1 + x)n ≈ 1 + nx + .....

only when x << 1

In this case it's L/RC << 1 which doesn't really fit..

[ IMG]http://i44.tinypic.com/530jdy.png[/PLAIN]
Well, what did you get for ω1 and ω2 ?

Well, what did you get for ω1 and ω2 ?
I can't do the expansion, because I can't find the small approximations in the square root!

SammyS
Staff Emeritus
Homework Helper
Gold Member
I can't do the expansion, because I can't find the small approximations in the square root!
You can find ω+ and ω- without an approximation.

However, I do see that the instructions do say to find approximate values assuming RC/L >> 1 .

Also notice that those values of ω are for (IR)2 = (I0)2/2 , so the square root goes away.

If they had said R2C/L >> 1, then I could see that helping.

Sorry. I can't be of more help.

You can find ω+ and ω- without an approximation.

However, I do see that the instructions do say to find approximate values assuming RC/L >> 1 .

Also notice that those values of ω are for (IR)2 = (I0)2/2 , so the square root goes away.

If they had said R2C/L >> 1, then I could see that helping.

Sorry. I can't be of more help.
Hmmm it's alright, let's leave this question up for other takers.

NascentOxygen
Staff Emeritus
Given RC/L ≫1, then for R>1Ω it follows that 4R2C/L ≫1.

Taking R>1Ω seems very reasonable for practical tuned circuits, I'd say.

rude man
Homework Helper
Gold Member
First: it's easier to use admittance and conductance rather than impedance and resistance when dealing with components connected in parallel. So for example for your circuit Y = G + jwC - j/wL where G = 1/R and Y = 1/Z.

Then: What is V as a function of w, where V is the voltage across your network? V is complex, but take |V|^2 which isn't.

Now, using |V|^2, what is I thru R? And I^2 thru R?
Then form I0^2/I^2 where I = Y*V. The V's cancel, giving you what for this ratio?

More when you get to this point.

First: it's easier to use admittance and conductance rather than impedance and resistance when dealing with components connected in parallel. So for example for your circuit Y = G + jwC - j/wL where G = 1/R and Y = 1/Z.

Then: What is V as a function of w, where V is the voltage across your network? V is complex, but take |V|^2 which isn't.

Now, using |V|^2, what is I thru R? And I^2 thru R?
Then form I0^2/I^2 where I = Y*V. The V's cancel, giving you what for this ratio?

More when you get to this point.
I have discussed this problem with my tutor, apparently the [RωC - R/(ωL)]2 = 1

which literally stems from the definition I = I0/√(2)
Then we get two values of ω.

rude man
Homework Helper
Gold Member
I have discussed this problem with my tutor, apparently the [RωC - R/(ωL)]2 = 1

which literally stems from the definition I = I0/√(2)
Then we get two values of ω.