Engineering Determine Voltage vk(t) across Switch K After Opening

[magnifik]

The circuit below operates with the switch K closed until steady state is reached. At t = 0 the switch K is opened. Determine the voltage v_k(t) across the switch after it is opened. (> 0)

What I attempted to do first was redraw the circuit when the switch is opened:

I am unsure of what the next steps should be. But this is my attempt:
V₀ + V_k + LI₁ = R₁(I₁ - I₂) + Ls(I₁ - I₂)
LI₁ + V₁/s = R₁(I₂ - I₁) + Ls(I₂ - I₁) + R₂I₂ + I₂/Cs

I'm not sure if these equations are correct and do not want to go on if they aren't since everything would be wrong. Thanks in advance.

 

[NascentOxygen]

Perhaps it is intended that K impose the initial conditions on the system, then at t=0 K disconnects the battery from the circuit and leaves the circuit to oscillate until its energy is dissipated? The currents in the two branches therefore become equal; the current now being a loop current.

 

[magnifik]

Perhaps it is intended that K impose the initial conditions on the system, then at t=0 K disconnects the battery from the circuit and leaves the circuit to oscillate until its energy is dissipated? The currents in the two branches therefore become equal; the current now being a loop current.

so i(0) = V₀/R₁?

 

[gneill]

so i(0) = V₀/R₁?

Yes, that would be the initial current in the "new" loop -- the inductor will "insist" that it be so!

But the problem is going to be that this is a serial RLC circuit. As such it can have three distinct solution forms depending upon the particular values of the components; The circuit can be underdamped, overdamped, or critically damped.

A straight inverse Laplace transform of the circuit's Laplace domain equation for the voltage at the top node (top horizontal rail), containing all the individual circuit component values, will be a real "dog's breakfast"! Loads of sine and cosine terms, exponentials, and so on.

It might be worthwhile trying to batter it into a more familiar form by converting the terms that represent the natural frequency ω_o, the damping ratio \alpha and any time constants \tau into those symbols.

 

[magnifik]

Yes, that would be the initial current in the "new" loop -- the inductor will "insist" that it be so!

But the problem is going to be that this is a serial RLC circuit. As such it can have three distinct solution forms depending upon the particular values of the components; The circuit can be underdamped, overdamped, or critically damped.

A straight inverse Laplace transform of the circuit's Laplace domain equation for the voltage at the top node (top horizontal rail), containing all the individual circuit component values, will be a real "dog's breakfast"! Loads of sine and cosine terms, exponentials, and so on.

It might be worthwhile trying to batter it into a more familiar form by converting the terms that represent the natural frequency ω_o, the damping ratio \alpha and any time constants \tau into those symbols.

unfortunately we have to do Laplace transform and then convert back to the time domain. i just realized that shouldn't it simplify to the circuit below with the switch open?

Then I can do I(s) = V(s)/Z(s) where V(s) is the sum of the transform voltages and Z(s) is the sum of the transform impedances so...
I(s) = [LI₁ - V₁/s]/[Ls + R₁ + R₂ + 1/Cs]

err I suppose I should do V(s) = I(s)/Y(s)

 

[gneill]

Sure, you can do that (Be careful about the polarity of the initial voltage source associated with the initial inductor current. Its polarity should be in the same direction as the initial current).

Once you have the current in the time domain, presumably you can use it to work out an expression for the voltage at the top. It may involve some calculus.

 

[magnifik]

Sure, you can do that (Be careful about the polarity of the initial voltage source associated with the initial inductor current. Its polarity should be in the same direction as the initial current).

Once you have the current in the time domain, presumably you can use it to work out an expression for the voltage at the top. It may involve some calculus.

is my polarity incorrect? I know usually it's the other way around, but in this case I₁ is defined as going up so shouldn't the voltage source be - to + ?

 

[gneill]

is my polarity incorrect? I know usually it's the other way around, but in this case I₁ is defined as going up so shouldn't the voltage source be - to + ?

Disregard the labelled current and determine the actual direction of the initial current flow. The true polarity of the model source will be determined by that.

Otherwise, calculate the value of your I1 according to the label and you'll find it to be negative. Again, set your model voltage source accordingly.

 

[magnifik]

So after combining impedances I have

Then I did a source transformation

is what i have done so far correct?? can i now apply nodal analysis to get Vk?

 

[gneill]

Looks okay. Sure, you can do nodal analysis, or you could continue to simplify the circuit; the current sources and impedances are in parallel...

 

[magnifik]

Looks okay. Sure, you can do nodal analysis, or you could continue to simplify the circuit; the current sources and impedances are in parallel...

awesome. thanks!