Engineering Solve Parallel RLC Circuit Homework, Find ω0 with Expansion

[unscientific]

Homework Statement

Homework Equations

The Attempt at a Solution

I found ω₀ = 1/√LC when I_R 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..

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

I found ω₀ = 1/√LC when I_R 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 ?

 

[unscientific]

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]

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 (I_R)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.

 

[unscientific]

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 (I_R)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]

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]

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.

 

[unscientific]

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]

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.

your tutor is right.