Finding Instantaneous UL and iL After Commutation for Transients (LC) Physics

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In summary: Closed. So these are the initial conditions for after the switch is closed.You should be able to calculate the steady state value of the voltage on the capacitor, and the steady state current flowing through the inductor before the switch is closed.These are going to serve as initial conditions for after the switch is closed.In summary, before the switch is closed in the given circuit, the capacitor will have a charged constant voltage and no current will flow through it, while the inductor will have a constant current and no voltage drop. After the switch is closed, the circuit will be modified and the capacitor will still retain its initial voltage, while the inductor will still have its initial current. These initial conditions will serve as starting points for the
  • #1
builder_user
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Homework Statement


Find instanteous UL, iL after commutation
Type of commutation - switch key

Homework Equations





The Attempt at a Solution


J=4*K1
L1=1mH*K2
C1=10mkF8*K2
R1=4*K3
R2=4*K3
R3=2*K3
K1=1.2
K2=0.7
K3=1

How to make it?
 

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  • #2


before commutation
Uc1(0_)=Uc1(0+)=0
UL1(0_UL1(0+)=0?
 
  • #3


MisterX said:
I assume you should assume that the circuit is in steady state before the switch is closed.

If that is the case, no current will flow through the capacitor, and the inductor will be like a short circuit.
It means that all circuit will have only resistors?Ok.
After I'll find U on the capacitor and current through the inductor what to do next?
 
  • #4


I assume you should assume that the circuit is in steady state before the switch is closed.

If that is the case, no current will flow through the capacitor, and the inductor will be like a short circuit.

Solve that resistor network and get the voltage on the capacitor and the current through the inductor. Do these quantities change instantaneously? You should be able to get the instantaneous values at the moment of switch closure this way.

You could use the Laplace transform of the circuit after the switch is closed to get the transients.
 
  • #5


MisterX said:
You could use the Laplace transform of the circuit after the switch is closed to get the transients.

How?
 
  • #7


builder_user said:

The Attempt at a Solution


J=4*K1
L1=1mH*K2
C1=10mkF8*K2
:
:
Is that 8 a typo, or is it really 8*K2 ? These problems with the "K" constants seem a bit odd. Why not just specify the component values and be done with it? Also, are we to take the resistance values to have implied Ohms units? (the capacitors and inductors have specified millihenries and microfarads). The current, too, lacks explicit specification for units. Do we assume Amps and not milliamps?
 
  • #8
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  • #9


gneill said:
Is that 8 a typo, or is it really 8*K2 ? These problems with the "K" constants seem a bit odd. Why not just specify the component values and be done with it? Also, are we to take the resistance values to have implied Ohms units? (the capacitors and inductors have specified millihenries and microfarads). The current, too, lacks explicit specification for units. Do we assume Amps and not milliamps?

10mkF*K2

Before commutation and after inductor Ul(0).But what's happen with capacitor?After commutation it seems that there is no current in his circuit.
 
  • #10


It's the last task.
What's happen with capacitor?

I need to find Thevenin equivavalent before and after commutation right?
 
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  • #11


builder_user said:
It's the last task.
What's happen with capacitor?

I need to find Thevenin equivavalent before and after commutation right?

The capacitor will have some initial voltage on it when the switch is closed. The inductor will be carrying some initial current.

Are you required to find U(t) and i(t) for the inductor in the problem, or just the initial values after the switch closes? Otherwise, since there's both L and C in the circuit there will probably be some oscillation after the switch closes; you might have to deal with damped oscillation (second order differential equation rather than first order).

You can probably skip finding the Thevenin equivalent for the 'before' circuit if you can determine the voltage on the capacitor and current through the inductor before the switch is closed.
 
  • #12


gneill said:
The capacitor will have some initial voltage on it when the switch is closed. The inductor will be carrying some initial current.

Are you required to find U(t) and i(t) for the inductor in the problem, or just the initial values after the switch closes? Otherwise, since there's both L and C in the circuit there will probably be some oscillation after the switch closes; you might have to deal with damped oscillation (second order differential equation rather than first order).

You can probably skip finding the Thevenin equivalent for the 'before' circuit if you can determine the voltage on the capacitor and current through the inductor before the switch is closed.

The same task as in previous topic.But for circuits with capacitor and inductor.Before "switch is closed" in the circuit will be capacitor and inductor?Or only capacitor?
 
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  • #13


Before the switch is closed:

The capacitor will be charged to some constant value, it will be passing no current.
The inductor will be passing some constant current, it will have no voltage drop.

After the switch is closed:

The switch will short circuit R1, so R1 "disappears". This leaves R2, C1 and R3+L1 in parallel with the current source. C1 has its initial voltage from before, and L1 has its initial current from before.

So the question is, before the switch is closed, what is the voltage on C1 and the current through L1?
 
  • #14


The scheme for "before"?I forgot delete program' values...(
 

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  • #15


After?
 

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  • #16


The "after" circuit looks fine. Can you tell the capacitor voltage and inductor current from the first circuit?
 
  • #17


gneill said:
The "after" circuit looks fine. Can you tell the capacitor voltage and inductor current from the first circuit?

I...know it?Or I need to find it?
 
  • #18


builder_user said:
I...know it?Or I need to find it?

You should be able to calculate the steady state value of the voltage on the capacitor, and the steady state current flowing through the inductor before the switch is closed.

These are going to serve as initial conditions for after the switch is closed.
 
  • #19


gneill said:
You should be able to calculate the steady state value of the voltage on the capacitor, and the steady state current flowing through the inductor before the switch is closed.

Voltage on capacitor can be the same as voltage on R2
 
  • #20


Yes...
 
  • #21


gneill said:
Yes...

So R2 parrallel C and R1&&R2 parallel to J and to R&&L?
And I can delete C from the circuit?
 
  • #22


I'm not sure what your "&&" notation implies. But if you're trying to calculate the voltage on the capacitor, then notice that at steady state the capacitor "looks like" an open circuit and the inductor "looks like" a short circuit. So the current supply J will be split between the R1+R2 path and the R3 path. So it's a current divider situation.

i2 = J*R3/(R1 + R2 + R3)
 
  • #23


Before..?
 

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  • #24


builder_user said:
Before..?

Yes. That's all the paths that are carrying current and making voltage drops.
 
  • #25


gneill said:
i2 = J*R3/(R1 + R2 + R3)

But R1&R2 have parallel connection with R3?
 
  • #26


That's right. It's a current divider.

Draw a circuit with a current supply I driving two parallel resistors, Ra and Rb. How does the current divide between the Ra path and the Rb path?

The parallel resistance is Ra*Rb/(Ra+Rb). So the voltage across the pair is V = I*Ra*Rb/(Ra+Rb). Thus, the current through, say, Ra, is V/Ra, which is I*Rb/(Ra+Rb).

Now go back to your circuit. One path has resistance R1+R2, while the other has resistance R3.
 
  • #27


gneill said:
That's right. It's a current divider.

Draw a circuit with a current supply I driving two parallel resistors, Ra and Rb. How does the current divide between the Ra path and the Rb path?

The parallel resistance is Ra*Rb/(Ra+Rb). So the voltage across the pair is V = I*Ra*Rb/(Ra+Rb). Thus, the current through, say, Ra, is V/Ra, which is I*Rb/(Ra+Rb).

Now go back to your circuit. One path has resistance R1+R2, while the other has resistance R3.

A...I understand.I thought that it was voltage across all circuit.But it is voltage on R1+R2.I can find current by dividing U/R1+R2. Then I can find voltage on R2 and voltage on capacitor right?
 
  • #28


I found UR2.what's next?
 
  • #29


Find the current in the inductor. You can use the same voltage divider method; it's the current through R3 this time.
 
  • #30


Done.What's next?
 
  • #31


Before proceeding, let's make an educated guess at what the voltage across the inductor is going to look like when the switch closes. That way we'll know where we're headed.

When the switch closes you'll have the current supply J = 12A in parallel with R2 and C1 and the branch R3&&L1 (using your && notation). The capacitor has an initial voltage of UC = UR2 = 9.6V (correct?). The inductor is carrying initial current iL = 9.6A (correct?).

The instant the switch closes, the capacitor will want to maintain its voltage and the inductor will want to maintain its current. So the top of R3 will be at potential UC = 9.6V. The current through R3, however, will still be iL = 9.6A, so the voltage drop on R3 is going to be 19.2V. That means the voltage across L1 must shoot up to 9.6V (- on top, + on bottom of L1) in order to satisfy Kircchoff, because UR3 + UL1 = UC at that instant.

Eventually the voltage across L1 must return to zero in a new steady state. So there will be an exponential decay (with time constant to be determined) from 9.6V down to zero. Superimposed on that will be whatever oscillations, if any, that result from the interaction of L and C. But the "big picture" is that UL1 will shoot up to 9.6V and decay.

Does all that make sense?
 
  • #32


gneill said:
Before proceeding, let's make an educated guess at what the voltage across the inductor is going to look like when the switch closes. That way we'll know where we're headed.

When the switch closes you'll have the current supply J = 12A in parallel with R2 and C1 and the branch R3&&L1 (using your && notation). The capacitor has an initial voltage of UC = UR2 = 9.6V (correct?). The inductor is carrying initial current iL = 9.6A (correct?).

The instant the switch closes, the capacitor will want to maintain its voltage and the inductor will want to maintain its current. So the top of R3 will be at potential UC = 9.6V. The current through R3, however, will still be iL = 9.6A, so the voltage drop on R3 is going to be 19.2V. That means the voltage across L1 must shoot up to 9.6V (- on top, + on bottom of L1) in order to satisfy Kircchoff, because UR3 + UL1 = UC at that instant.

Eventually the voltage across L1 must return to zero in a new steady state. So there will be an exponential decay (with time constant to be determined) from 9.6V down to zero. Superimposed on that will be whatever oscillations, if any, that result from the interaction of L and C. But the "big picture" is that UL1 will shoot up to 9.6V and decay.

Does all that make sense?

Why 12?J=4*K1=4*1.2=4.8
 
  • #33


Isn't J = 4*K1 = 4*3 = 12? Or are the units something other than amps?
 
  • #34


gneill said:
Isn't J = 4*K1 = 4*3 = 12? Or are the units something other than amps?

But K1=1.2
 
  • #35


Ah! My mistake. Very sorry. These K's will be the end of me...

So, make that:

iL = 3.84A
UC = 3.84V
UR3 = 7.68V
UL = 3.84V (+ on bottom end)

The general picture still stands, though, with these modified values. Do your calculated values agree, and does the argument I presented make sense?
 
<h2>1. What is the purpose of finding instantaneous UL and iL after commutation for transients?</h2><p>The purpose of finding instantaneous UL and iL after commutation for transients is to analyze the behavior of an electrical circuit during transient events, which are sudden changes in voltage or current. This information can help engineers design more efficient and reliable circuits.</p><h2>2. How is instantaneous UL and iL calculated?</h2><p>Instantaneous UL and iL are calculated using the equations UL = L(diL/dt) and iL = (1/L)∫Vdt, where L is the inductance of the circuit, diL/dt is the rate of change of current, and V is the voltage across the inductor.</p><h2>3. What is the significance of commutation in this process?</h2><p>Commutation refers to the process of switching the direction of current flow in an electrical circuit. In the context of finding instantaneous UL and iL, commutation is important because it allows us to analyze the behavior of the circuit during transient events, which occur when there is a sudden change in current direction.</p><h2>4. What factors can affect the accuracy of finding instantaneous UL and iL?</h2><p>The accuracy of finding instantaneous UL and iL can be affected by several factors, including the quality of the circuit components, the precision of the measurement equipment, and the complexity of the circuit. Additionally, external factors such as electromagnetic interference can also impact the accuracy of the results.</p><h2>5. How is this information used in real-world applications?</h2><p>The information obtained from finding instantaneous UL and iL after commutation for transients can be used in various real-world applications, such as designing power electronics for electric vehicles, optimizing power distribution systems, and improving the efficiency of renewable energy systems. This data can also be used for troubleshooting and diagnosing issues in electrical circuits.</p>

1. What is the purpose of finding instantaneous UL and iL after commutation for transients?

The purpose of finding instantaneous UL and iL after commutation for transients is to analyze the behavior of an electrical circuit during transient events, which are sudden changes in voltage or current. This information can help engineers design more efficient and reliable circuits.

2. How is instantaneous UL and iL calculated?

Instantaneous UL and iL are calculated using the equations UL = L(diL/dt) and iL = (1/L)∫Vdt, where L is the inductance of the circuit, diL/dt is the rate of change of current, and V is the voltage across the inductor.

3. What is the significance of commutation in this process?

Commutation refers to the process of switching the direction of current flow in an electrical circuit. In the context of finding instantaneous UL and iL, commutation is important because it allows us to analyze the behavior of the circuit during transient events, which occur when there is a sudden change in current direction.

4. What factors can affect the accuracy of finding instantaneous UL and iL?

The accuracy of finding instantaneous UL and iL can be affected by several factors, including the quality of the circuit components, the precision of the measurement equipment, and the complexity of the circuit. Additionally, external factors such as electromagnetic interference can also impact the accuracy of the results.

5. How is this information used in real-world applications?

The information obtained from finding instantaneous UL and iL after commutation for transients can be used in various real-world applications, such as designing power electronics for electric vehicles, optimizing power distribution systems, and improving the efficiency of renewable energy systems. This data can also be used for troubleshooting and diagnosing issues in electrical circuits.

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