Question about re-drawing circuit

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In summary: It's okay, I understand what you are saying. I was just trying to see if there was a way to simplify the equation and I didn't think there was. I will try to rephrase it that way.
  • #1
Master1022
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Homework Statement
Find an expression for the current out as a function of the difference between the input and output voltage when the reference voltage is the same as the output voltage
Relevant Equations
V = IR
Hi,

I was attempting this problem where the aim is to find the output current as a function of the voltage difference ## P_{in} - P_{out} ##. I just have two quick questions about this circuit:
1. If the 'reference voltage' (which I assume is how the ground voltage is referred to) is the same as the output voltage, am I correct in thinking that I can ignore the last capacitor ## C_v##?
2. Am I allowed to re-draw the following circuit such that the bottom connection wire is removed? It doesn't make any difference to my analysis, but just wondering?
3. Does my attempt below look like it is on the right tracks?
Screen Shot 2021-03-20 at 8.52.49 AM.png


re-drawn as:
Screen Shot 2021-03-20 at 8.52.14 AM.png


Attempt:
Making the above assumptions, we could use KVL to get an equation for the first junction, which we can denote as ## P_x ## (in ##s##-domain):
[tex] \frac{P_{in} - P_{x}}{R_a + I_a} + \frac{P_{ref} - P_{x}}{\frac{1}{s C_a}} + \frac{P_{out} - P_x}{(R_c + R_v) + s I_v} = 0 [/tex]

and an equation for the output junction as:
[tex] \frac{P_x - P_{out}}{(R_c + R_v) + s I_v} + \frac{P_{ref} - P_{out}}{\frac{1}{s C_v}} = i_{out} [/tex]
which then simplifies to: (if ## P_{ref} = P_{out} ##)
[tex] \frac{P_x - P_{out}}{(R_c + R_v) + s I_v} = i_{out} [/tex]

which I can then go on to solve. However, I cannot easily see how this will take the form of an LPF. When re-drawing the circuit above, it looks as if it turned into two potential dividers, each of which had a transfer function of the form:
[tex] \frac{1}{1 + sRC + s^2 LC} [/tex]
When cascaded without making any assumptions about the reference voltage, it looks as if the complete circuit would form a fourth order LPF... I can't seem to reconcile these two viewpoints.

Any help is greatly appreciated.
 
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  • #2
You can draw the reference as you did in the second picture okay.

The letters are used in an unusual way here. I mostly see ##V## used instead of ##P## because ##P## is usually used for power; inductors usually use ##L## because ##I## is typically used for current.

The first equation you accidentally dropped the ##s## in front of the inductor A. In the second equation is there something else connected after what you called ##P_{out}##? I don't think so. If there isn't, then you should be saying the sum of the currents is equal to zero instead of ##i_{out}##.

I don't understand the last part why you would say ##P_{ref} = P_{out}##. Are you meaning to short the output to the ground reference? That's very unusual. I don't think that's right. I think you were meaning to say that ##P_{ref} = 0##. I would recommend trying that unless you could clarify on what you were intending to do?
 
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  • #3
Thank you very much for your reply @Joshy ! Apologies, I should have added more context to the post. This question is within the general topic of haemodynamics and thus some of the naming conventions are different. However, I didn't include the context in fear that it may just overcomplicated a question which is, at its core, just a circuit analysis question.

Joshy said:
You can draw the reference as you did in the second picture okay.
Okay thank you.

Joshy said:
The letters are used in an unusual way here. I mostly see ##V## used instead of ##P## because ##P## is usually used for power; inductors usually use ##L## because ##I## is typically used for current.
Indeed, so in this case the ## P ## is actually "pressure", but this question actually uses it in an analogous way to voltage (i.e. the pressure is the driver and the blood flow is the flux). With the inductance, I am not sure why they change the convention, so I do agree (especially as this is going to be analyzed with circuit theory concepts).

Joshy said:
The first equation you accidentally dropped the ##s## in front of the inductor A.
Yes, thanks for spotting that. Apologies for the typo!

Joshy said:
In the second equation is there something else connected after what you called ##P_{out}##? I don't think so. If there isn't, then you should be saying the sum of the currents is equal to zero instead of ##i_{out}##.
I was thinking the same thing and I am not sure whether there is anything else afterwards. The reason I did that is because of the following (the only example of this content in the lectures) shown in the image below.

Screen Shot 2021-03-22 at 11.37.12 AM.png


It also uses ## q_{out} ## instead of 0 for some reason. I will keep thinking to see why that is the case.

Joshy said:
I don't understand the last part why you would say ##P_{ref} = P_{out}##. Are you meaning to short the output to the ground reference? That's very unusual. I don't think that's right. I think you were meaning to say that ##P_{ref} = 0##. I would recommend trying that unless you could clarify on what you were intending to do?
Yes, I do agree. However, that is what I interpreted the wording of the question as saying. The wording of the question is:
"Find an expression for the current out as a function of the difference between the input and output voltage when the reference voltage is the same as the output voltage"
(Note: I changed the words pressure for voltage and blood flow for current, but wording is otherwise the same)
Perhaps there is a typo in the sheet and I ought to ask the teacher.
 

Related to Question about re-drawing circuit

1. How do I re-draw a circuit?

To re-draw a circuit, you will need to have a clear understanding of the original circuit and its components. Begin by drawing a rough sketch of the circuit, labeling each component and its connections. Then, use a software program or a ruler and pencil to create a neat and accurate drawing of the circuit.

2. Why would I need to re-draw a circuit?

There are several reasons why you may need to re-draw a circuit. It could be to make modifications or improvements to the original design, to troubleshoot and identify any issues with the circuit, or to create a more professional and organized representation for a project or presentation.

3. What tools do I need to re-draw a circuit?

You will need a basic understanding of circuit components and their symbols, as well as a software program or paper and pencil. Some helpful tools for creating accurate circuit drawings include a ruler, protractor, and a compass. You may also need a multimeter to test the circuit and ensure its accuracy.

4. How do I ensure the accuracy of my re-drawn circuit?

To ensure the accuracy of your re-drawn circuit, it is important to double-check all connections and component values. You can also use a multimeter to test the circuit and make sure it functions as intended. Additionally, referencing the original circuit diagram and consulting with other experts can help identify any errors or discrepancies.

5. Are there any tips for re-drawing a circuit efficiently?

To re-draw a circuit efficiently, it is helpful to have a clear plan and understanding of the circuit beforehand. Start with a rough sketch and then use a software program to create a neat and organized drawing. It is also important to label all components and connections accurately and consistently. Additionally, using templates or pre-made symbols can save time and ensure accuracy.

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