Infinite Ladder Circuit (Not with Resistors)

[vivekrai]

Homework Statement

Homework Equations

Capacitive Reactance X_c = 1/ωC ; Inductive Reactance X_L = ωL ; Some Phasor Algebra etc.,

The Attempt at a Solution

Well, Basically like we solve the circuit for resistances with DC Supply, I tried to assume the total impedance of the Circuit as Z. Then leaving one block, replace all by an Inductor with impedance Z.

This is where I'm getting stuck!

 

[vivekrai]

Bump! Mentors, If some one is not able to solve in this sub section kindly move it to the appropriate section where some people will answer this.

 

[Redbelly98]

I have moved this to Engineering, Comp Sci, & Technology.

The Attempt at a Solution

Well, Basically like we solve the circuit for resistances with DC Supply, I tried to assume the total impedance of the Circuit as Z.

Yes, that is how I have solved this problem.

Then leaving one block, replace all by an Inductor with impedance Z.

This is where I'm getting stuck!

I don't really understand what you did in this last part. I would focus on figuring out just what Z is first. To do that, realize that the impedance is still going to be Z if you add another inductor + capacitor pair to the left side of the circuit. So you can set up an equation:

Z = Impedance of: (inductor) in series with (parallel combination of capacitor and Z)​

Use the actual impedance expressions for the inductor and capacitor, and solve the equation for Z. The algebra gets a little tedious, but after you get an expression for Z, you can then look at the expression and figure out how to make it consistent with a pure inductance.

 

[ehild]

Look at the dimensions: which version can be angular frequency at all?

An LC circuit without resistors is singular at its resonant frequencies: the impedance can be zero or infinite. Try to build the ladder step by step with ω=1/√(LC). What do you get? Can you apply the "ladder method" ?

If you apply the "ladder method" what do you get for Z? Does the angular frequency C give a meaningful solution?

ehild

 

[gneill]

I think that Redbelly has outlined a valid method, and it seems that it's essentially what vivekrai said he'd tried. But without vivekrai supplying more of his effort I don't see how we can be sure, or how we can do much to help.

I will give one suggestion though. Let s = jω then the impedances of the inductor and capacitors are given by sL and 1/(sC). Use Redbelly's suggestion and find Z such that
$$Z = sL + \frac{Z\,\frac{1}{s C}}{Z + \frac{1}{s C}}$$
The algebra won't be too bad, and you'll end up with a quadratic in Z. From the roots of the equation (quadratic formula) you should be able to spot what s (and hence ω) needs to be in order to leave just an "sL" type term.

 

[vivekrai]

Thanks to everyone for taking interest in solving it, But since its too late here in India now (IST), I'' solve it and tell you people by tomorrow.

 

[vivekrai]

Yes, Got it thanks. The answer turned out to be (C)