Determine Voltage vk(t) across Switch K After Opening

In summary, the circuit discussed operates with a closed switch K until steady state is reached. When the switch is opened at t=0, the voltage Vk(t) across the switch can be determined using nodal analysis or simplifying the circuit further. The circuit can have three distinct solution forms depending on the component values and the voltage expression may involve calculus.
  • #1
magnifik
360
0
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 vk(t) across the switch after it is opened. (> 0)
i3vts2.png



What I attempted to do first was redraw the circuit when the switch is opened:
r6ztar.png


I am unsure of what the next steps should be. But this is my attempt:
V0 + Vk + 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.
 
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  • #2
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.
 
  • #3
NascentOxygen said:
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?
 
  • #4
magnifik said:
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 ωo, the damping ratio [itex]\alpha[/itex] and any time constants [itex]\tau[/itex] into those symbols.
 
  • #5
gneill said:
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 ωo, the damping ratio [itex]\alpha[/itex] and any time constants [itex]\tau[/itex] 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?
vwxzx4.png


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)
 
  • #6
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.
 
  • #7
gneill said:
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 + ?
 
  • #8
magnifik said:
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. :smile:
 
  • #9
So after combining impedances I have
2z86iz9.png


Then I did a source transformation
smzwp3.png


is what i have done so far correct?? can i now apply nodal analysis to get Vk?
 
  • #10
Looks okay. Sure, you can do nodal analysis, or you could continue to simplify the circuit; the current sources and impedances are in parallel...
 
  • #11
gneill said:
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!
 

FAQ: Determine Voltage vk(t) across Switch K After Opening

What is the purpose of determining the voltage across switch K after opening?

The purpose of determining the voltage across switch K after opening is to understand and analyze the change in voltage in a circuit when a switch is opened. This can help in troubleshooting any issues in the circuit and also in designing more efficient and safe circuits.

How do you determine the voltage across switch K after opening?

The voltage across switch K after opening can be determined by using Ohm's law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). By measuring the current and resistance in the circuit, the voltage across switch K can be calculated.

Why is it important to open switch K before determining the voltage?

Opening switch K before determining the voltage is important because it allows for a break in the circuit, which can help in isolating the specific section of the circuit that is affected by the switch. This also prevents any potential damage to the circuit or the measuring equipment.

What factors can affect the voltage across switch K after opening?

The voltage across switch K after opening can be affected by various factors such as the type and size of the circuit, the resistance of the switch and other components in the circuit, the amount of current flowing through the circuit, and any external factors such as temperature or humidity.

How can the voltage across switch K after opening be used in practical applications?

The voltage across switch K after opening can be used in various practical applications such as designing and troubleshooting electronic circuits, measuring power consumption, and determining the efficiency of a circuit. It can also be used in the development of new technologies and devices that require precise voltage control.

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