Initial and final values for second order circuits

In summary, the capacitor is an open circuit and the inductor is short circuited in the steady state DC circuit. This means that there is no current flow in the circuit, and for a voltmeter placed across the capacitor, the potential of the node connected to the "-" lead will be equal to -20V, as determined by performing a KVL walk. This is because the "-" lead of the voltmeter is used as the reference point for the reading.
  • #1
sugz
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Homework Statement



I am attempting to understand this example shown below:

example.PNG


Homework Equations



During stead state DC, the capacitor is an open circuit and the inductor is short circuited.

The Attempt at a Solution


[/B]
The questions I have are really related to the concepts as I don't seem to understand it well enough even after reading through the chapter many times.
1- How is Vc equal to -20V? Since Vc is an open circuit, how would there be a voltage across it?
2- My understanding is that the inductor current (iL) is zero because there is no voltage source in the circuit during steady state DC. Please confirm if my understanding is correct.
3- I have read through the book many times and I don't seem to understand how this is determined. If someone can provide a detailed explanation, I would really appreciate it!
 
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  • #2
sugz said:
1- How is Vc equal to -20V? Since Vc is an open circuit, how would there be a voltage across it?
2- My understanding is that the inductor current (iL) is zero because there is no voltage source in the circuit during steady state DC. Please confirm if my understanding is correct.

For (1) consider the circuit as drawn here with a voltmeter placed across the "open" capacitor position:

Fig1.png


The voltmeter leads are placed in such a way as to read the capacitor voltage as it is defined on the diagram. Essentially this makes the "-" lead of the capacitor the reference point for the reading. Now note the path marked out in blue: This comprises a single node since it's all connected by conductor. What is the potential of this node with respect to the voltmeter's reference point? So what will the voltmeter read?

For (2) you are correct in this case. The capacitor is "open" so no current can flow from the 20 V source, and the current supply is zero so no current from there. With no sources of current there can be no current in the circuit.
 
  • #3
gneill said:
For (1) consider the circuit as drawn here with a voltmeter placed across the "open" capacitor position:

View attachment 89506

The voltmeter leads are placed in such a way as to read the capacitor voltage as it is defined on the diagram. Essentially this makes the "-" lead of the capacitor the reference point for the reading. Now note the path marked out in blue: This comprises a single node since it's all connected by conductor. What is the potential of this node with respect to the voltmeter's reference point? So what will the voltmeter read?

For (2) you are correct in this case. The capacitor is "open" so no current can flow from the 20 V source, and the current supply is zero so no current from there. With no sources of current there can be no current in the circuit.
gneill said:
For (1) consider the circuit as drawn here with a voltmeter placed across the "open" capacitor position:

View attachment 89506

The voltmeter leads are placed in such a way as to read the capacitor voltage as it is defined on the diagram. Essentially this makes the "-" lead of the capacitor the reference point for the reading. Now note the path marked out in blue: This comprises a single node since it's all connected by conductor. What is the potential of this node with respect to the voltmeter's reference point? So what will the voltmeter read?

For (2) you are correct in this case. The capacitor is "open" so no current can flow from the 20 V source, and the current supply is zero so no current from there. With no sources of current there can be no current in the circuit.
Dear gneill, that was a very clear explanation. From what I understand, the "-" lead is connected to the positive side of the voltage, which would make that point +20V. The "+" lead is connected to the single node, which is at a lower voltage. Therefore, the potential of this node with respect to the voltmeter's reference point is 20V. I was a little confused at where the negative sign comes from? I understand that I am able to get -20V if I perform KVL around the third loop (-10-Vc+40*iL=0, where iL equals 0 and therefore Vc=-10V). Is there another way of doing this?
 
Last edited:
  • #4
sugz said:
Dear gneill, that was a very clear explanation. From what I understand, the "-" lead is connected to the positive side of the voltage, which would make that point +20V.
No, from the point of view of the meter, wherever the "-" lead is is the 0 V reference (for the meter).
The "+" lead is connected to the single node, which is at a lower voltage. Therefore, the potential of this node with respect to the voltmeter's reference point is 20V. I was a little confused at where the negative sign comes from?
Do a "KVL walk" from the meter's "-" lead down through the voltage supply to the blue node. What's the potential of the blue node with respect to the lead?
 
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  • #5
gneill said:
No, from the point of view of the meter, wherever the "-" lead is is the 0 V reference (for the meter).

Do a "KVL walk" from the meter's "-" lead down through the voltage supply to the blue node. What's the potential of the blue node with respect to the lead?
Thank you very much! I understand this now!
 
  • #6
gneill said:
For (1) consider the circuit as drawn here with a voltmeter placed across the "open" capacitor position:

View attachment 89506

The voltmeter leads are placed in such a way as to read the capacitor voltage as it is defined on the diagram. Essentially this makes the "-" lead of the capacitor the reference point for the reading. Now note the path marked out in blue: This comprises a single node since it's all connected by conductor. What is the potential of this node with respect to the voltmeter's reference point? So what will the voltmeter read?

For (2) you are correct in this case. The capacitor is "open" so no current can flow from the 20 V source, and the current supply is zero so no current from there. With no sources of current there can be no current in the circuit.
I am attempting to solve another problem, which carries over the same concepts. The question I have is related to part (a) where the solution suggests "Since it is in series with the +10V source, together they represent a direct short at t=0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor". Can you please explain this? I am confused about the direct short at t=0+.
 

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  • #7
At the instant t = 0+, the capacitor has a potential difference of -10 V. It's in series with the 10 V voltage source. Together they add up to 0 V. For that instant, the two components can be replaced with single zero volt voltage supply. A zero volt voltage supply is, for all intents and purposes, equivalent to a short circuit (a zero resistance path).
 
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  • #8
gneill said:
At the instant t = 0+, the capacitor has a potential difference of -10 V. It's in series with the 10 V voltage source. Together they add up to 0 V. For that instant, the two components can be replaced with single zero volt voltage supply. A zero volt voltage supply is, for all intents and purposes, equivalent to a short circuit (a zero resistance path).
Oh right! That makes sense! Thanks a lot, once again!
 
  • #9
gneill said:
At the instant t = 0+, the capacitor has a potential difference of -10 V. It's in series with the 10 V voltage source. Together they add up to 0 V. For that instant, the two components can be replaced with single zero volt voltage supply. A zero volt voltage supply is, for all intents and purposes, equivalent to a short circuit (a zero resistance path).
Hi,
I know this is a little late but I have been attempting to figure this out myself but had no success. In part (b) of that question, it is indicated that vR=vC+10. However, shouldn't vR=0 since all of the 2A goes to the short circuit (capacitor current)? This would make dvR/dt=0?
 
  • #10
sugz said:
Hi,
I know this is a little late but I have been attempting to figure this out myself but had no success. In part (b) of that question, it is indicated that vR=vC+10. However, shouldn't vR=0 since all of the 2A goes to the short circuit (capacitor current)? This would make dvR/dt=0?
Perhaps we're not looking at the same circuit? The one I see in the problem statement has a 3 A supply. When it turns on at time zero its current will be divided between the 2 Ω and 4 Ω resistors essentially instantaneously since, as you say, the capacitor looks like a short circuit to an instantaneous change. I'd expect that VR (or the point labelled "a" on the circuit shown as figure (b)) to pop up from zero to +4 V. From there I suppose there will be some "ringing" in the LC tank circuit which will also be seen on VR, I guess as a damped sinewave.

Maybe this could be made more obvious if you replaced the current source and 2 Ω resistor with their Thevenin equivalent? You'd then have that node "a" in the middle of a voltage divider.
Fig4.png
 

Related to Initial and final values for second order circuits

1. What are initial values in a second order circuit?

Initial values in a second order circuit refer to the values of the voltage and current at the beginning of a circuit's operation. These values can be determined by looking at the circuit's initial conditions, such as the initial charge on capacitors or the initial current through inductors.

2. What are final values in a second order circuit?

Final values in a second order circuit refer to the steady-state values of the voltage and current after the circuit has reached a stable state. These values can be determined by analyzing the circuit's transfer function and the input signal.

3. How do initial and final values affect a second order circuit?

Initial and final values play an important role in the behavior of a second order circuit. They determine the transient response of the circuit, which is the behavior of the circuit before it reaches a steady state. They also affect the steady-state response, as the final values will be the values of the voltage and current after the transient response has settled.

4. How are initial and final values calculated in a second order circuit?

Initial and final values can be calculated using various methods, such as using circuit analysis techniques like Kirchhoff's laws and Ohm's law, or using differential equations to model the circuit's behavior. It is important to have a thorough understanding of the circuit's components and their properties in order to accurately calculate these values.

5. Why are initial and final values important to know in a second order circuit?

Knowing the initial and final values of a second order circuit is crucial for understanding its behavior and performance. It allows us to predict the circuit's response to different inputs and to design circuits that meet specific requirements. It also helps in troubleshooting and diagnosing issues with the circuit, as any discrepancies between expected and measured initial and final values can indicate potential problems.

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