# Electrical analog and impedance

• Shackleford
In summary: It should be kx_1^2, not kx_1.You're on the right track. Take your friend's hint, replace \ddot{q} with \frac{di}{dt} and \dot{q} with i; you will see it clearer. And yes, each term corresponds to voltage across some electrical device.P.S.: I saw the term corresponding to kx_1 is a bit incorrect. It should be kx_1^2, not kx_1.
Shackleford
I'm a bit lost on this one. Also, what formula do I use for impedance?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19202315.jpg?t=1284947003

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19202817.jpg?t=1284947014

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What are the differential equations for m1 and m2? Notice that now the damping force on m1 is $$-b_1(\dot{x}_1-\dot{x}_2)$$ (that is, it depends on the RELATIVE velocity between the masses).
After write down the 2 equations, relate it to an electrical circuit by using electromechanics analogy:
q ~ x
i ~ v (or dx/dt)
k ~ 1/C
L ~ m
b ~ R
Fcos(wt) ~ Vcos(wt)

The impedance is calculated from the circuit built

hikaru1221 said:
What are the differential equations for m1 and m2? Notice that now the damping force on m1 is $$-b_1(\dot{x}_1-\dot{x}_2)$$ (that is, it depends on the RELATIVE velocity between the masses).
After write down the 2 equations, relate it to an electrical circuit by using electromechanics analogy:
q ~ x
i ~ v (or dx/dt)
k ~ 1/C
L ~ m
b ~ R
Fcos(wt) ~ Vcos(wt)

The impedance is calculated from the circuit built

That makes sense. The problem doesn't imply they move with the same velocity. That's why I was adding the masses.

Actually, the damping force on m2 is $$-b_2(\dot{x}_2)$$. I'll write out the DEs tomorrow. I'm tired. Thanks for the help.

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Are my equations correct? If so, do I then start substituting the electrical analogs? If so and assuming I get that correct, I'm not sure what to do after that. I suppose I could find my electrical solutions and then plug those into the impedance formula (whatever that is).

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-22220323.jpg?t=1285211103

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It's not correct, because you forgot the elastic force F=-kx
Well, the next thing to do is to convert these into an equivalent circuit Just replace F with V, k with 1/C, etc, and what you get is 2 voltage equations. You will then know what to do.

hikaru1221 said:
It's not correct, because you forgot the elastic force F=-kx
Well, the next thing to do is to convert these into an equivalent circuit Just replace F with V, k with 1/C, etc, and what you get is 2 voltage equations. You will then know what to do.

Crap. I knew I was forgetting something. I was pretty tired last night. Let me re-write this. It's absolutely important that I firstly understand the situation physically then construe it mathematically.

However, I have a question. The mass m1 has the spring restoring force, but the mass m2 does not. M2 has the frictional force from the floor based on its velocity and then again the force of friction from m1.

The mass m1 has the spring restoring force, but the mass m2 does not. M2 has the frictional force from the floor based on its velocity and then again the force of friction from m1.

Yes, you're right
But... is that a question?

hikaru1221 said:
Yes, you're right
But... is that a question?

Haha! Oops.

Okay, I'll give you a question now. Is this correct?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-23191532.jpg?t=1285288944

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Your equations are correct.

hikaru1221 said:
Your equations are correct.

Hey, that's great. Now, I can attempt my electrical analogs. However, should I write it in terms of I or q? And, isn't impedance equal to the square root of the sum of the squares of RL and RC. I vaguely remember the equation.

Just do as I suggested earlier. Convert all the mechanical quantities into the equivalent electrical quantities, and you have 2 voltage equations. Each term of the voltage equations corresponds to a device (L, R, C, source). From the equation, you can draw the equivalent circuit and then from the circuit, calculate the impedance.

hikaru1221 said:
Just do as I suggested earlier. Convert all the mechanical quantities into the equivalent electrical quantities, and you have 2 voltage equations. Each term of the voltage equations corresponds to a device (L, R, C, source). From the equation, you can draw the equivalent circuit and then from the circuit, calculate the impedance.

Here are my equations. One of my classmates was saying something about the delta-q being the current through the R1 resistor or something like that. Each of these terms has to be voltage.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25172307.jpg?t=1285453923

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You're on the right track. Take your friend's hint, replace $$\ddot{q}$$ with $$\frac{di}{dt}$$ and $$\dot{q}$$ with $$i$$; you will see it clearer. And yes, each term corresponds to voltage across some electrical device.

P.S.: I saw the term corresponding to $$kx_1$$ is a bit incorrect.

hikaru1221 said:
You're on the right track. Take your friend's hint, replace $$\ddot{q}$$ with $$\frac{di}{dt}$$ and $$\dot{q}$$ with $$i$$; you will see it clearer. And yes, each term corresponds to voltage across some electrical device.

P.S.: I saw the term corresponding to $$kx_1$$ is a bit incorrect.

Ah. You're right. I accidentally put the dot over the q. Let me see what I can come up with. I took Physics II two years ago. That's the last time I really dealt with impedance, circuits, and so forth.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25233241.jpg?t=1285475708

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The last term of 2nd equation is a bit incorrect. Anyway, from here, can you build a corresponding circuit?

hikaru1221 said:
The last term of 2nd equation is a bit incorrect. Anyway, from here, can you build a corresponding circuit?

Crap. I put the dot somewhere but not where it belonged!

The circuit has two inductors, two resistors, and on capacitor. Does it have a parallel branch?

Yes. That's why I told you to replace q' with i There should be some parallel branches. Hint: You can rewrite the 2nd equation as $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$ so that you can see a common term of both equations. That shows something about the parallel branches.

hikaru1221 said:
Yes. That's why I told you to replace q' with i There should be some parallel branches. Hint: You can rewrite the 2nd equation as $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$ so that you can see a common term of both equations. That shows something about the parallel branches.

Unfortunately, I don't have a damn idea what to do or how to calculate the impedance. I suppose I could insert the common term into equation one, but I don't see how that helps me. I guess I first need to know how I can calculate the impedance then try to see how my equations could get me there.

Well you must first build the equivalent circuit There is no way around I suppose.
You don't have to substitute the common term into the 1st equation. Let me explain the circuit for you.
_ From the 1st equation: $$L_1\frac{di_1}{dt}+R_1(i_1-i_2)+\frac{q_1}{C}=V$$

That is, there is a source V, connected in series with an inductor L1, a resistor R1, a capacitor C. If you look at the current accompanying each term, you can see that L1 is accompanied by I1 and C is accompanied by Q1 which is directly related to I1. However, R1 is accompanied by (I1 - I2), i.e. the current through R1 is only a part of I1. Therefore there must be something connected in parallel to R1.

_ Indeed, the 2nd equation shows so: $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$

We can see that, sharing the same voltage as R1, there is a branch consisting of L2 and R2, both of which are accompanied by I2.

So now can you draw the equivalent circuit? From the circuit, you can easily find the impedance.

That makes sense.

This is probably wrong, but it kind of makes sense since this in this electrical analog the voltage source would provide AC.

Edited for embarrassment.

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Sorry, it's really wrong
Remember that the branch (L2 and R2) is parallel to R1. Besides the source has 2 terminals.

hikaru1221 said:
Sorry, it's really wrong
Remember that the branch (L2 and R2) is parallel to R1. Besides the source has 2 terminals.

What the heck did I just write. That's not a circuit. I'm multitasking here, so let me redo that. -_-

Okay, here it is.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-26162147.jpg?t=1285536201

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What you drew is that the branch (L1 + C), not R1, is parallel to (L2 + R2)!

hikaru1221 said:
What you drew is that the branch (L1 + C), not R1, is parallel to (L2 + R2)!

I thought the difference in current indicated a parallel circuit going into R1.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-26164217.jpg?t=1285537508

Wait. I see what you're saying now. Hold on.

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Now you got it. Then calculate the impedance from the circuit.

hikaru1221 said:
Now you got it. Then calculate the impedance from the circuit.

So, my terminals are correct?

If I'm looking at this the right way, there's one charge going "directly" through R1 and then the "other" charge going through L2 and R2. Those two charges constitute the I1 current through the R1, L1, and C series.

What's with "charges"? You mean "current"?

hikaru1221 said:
What's with "charges"? You mean "current"?

Yes. Current is simply the movement of the "charges" with respect to time.

Let me redraw the circuit in the standard way. See the picture attached.
Now can you calculate the impedance from the circuit?

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hikaru1221 said:
Let me redraw the circuit in the standard way. See the picture attached.
Now can you calculate the impedance from the circuit?

The answer the back of the book has is very long and nasty. I don't know how to get their answer.

Well if you don't remember how to calculate impedance, you'd better review your knowledge Hint: Use the complex forms of impedance:
$$Z_R = R$$
$$Z_L = jL\omega$$
$$Z_C = -\frac{j}{C\omega}$$

hikaru1221 said:
Well if you don't remember how to calculate impedance, you'd better review your knowledge Hint: Use the complex forms of impedance:
$$Z_R = R$$
$$Z_L = jL\omega$$
$$Z_C = -\frac{j}{C\omega}$$

Well, I put an equation for impedance in a previous post. The only difference is it's not in complex form. Why does it have to be in complex form?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25233241.jpg?t=1285475708

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That's not the right one. That formula is only true for the case that there are only 3 elements R, L, C in series with the source (RLC circuit). It doesn't apply to this case where the circuit is much more complicated.

hikaru1221 said:
That's not the right one. That formula is only true for the case that there are only 3 elements R, L, C in series with the source (RLC circuit). It doesn't apply to this case where the circuit is much more complicated.

Okay. Let me see what I can dig up in my Physics II book.

<h2>What is electrical analog?</h2><p>Electrical analog is a method of representing a physical system or process using electrical circuits. It involves creating an electrical circuit that mimics the behavior of the system being studied, allowing for easier analysis and understanding.</p><h2>What is impedance in electrical circuits?</h2><p>Impedance is the total opposition to the flow of an alternating current in an electrical circuit. It is a combination of resistance, inductance, and capacitance, and is measured in ohms.</p><h2>How is impedance calculated in a circuit?</h2><p>Impedance can be calculated using Ohm's Law, which states that impedance is equal to the voltage divided by the current. It can also be calculated using complex numbers, taking into account the phase difference between the voltage and current.</p><h2>What is the difference between resistance and impedance?</h2><p>Resistance is the opposition to the flow of direct current in a circuit, while impedance is the opposition to the flow of alternating current. Resistance is measured in ohms, while impedance is measured in complex numbers.</p><h2>Why is impedance important in electrical circuits?</h2><p>Impedance is important in electrical circuits because it affects the flow of current and the behavior of the circuit. It can also be used to analyze and design circuits, and can help determine the maximum power that can be transferred through a circuit.</p>

## What is electrical analog?

Electrical analog is a method of representing a physical system or process using electrical circuits. It involves creating an electrical circuit that mimics the behavior of the system being studied, allowing for easier analysis and understanding.

## What is impedance in electrical circuits?

Impedance is the total opposition to the flow of an alternating current in an electrical circuit. It is a combination of resistance, inductance, and capacitance, and is measured in ohms.

## How is impedance calculated in a circuit?

Impedance can be calculated using Ohm's Law, which states that impedance is equal to the voltage divided by the current. It can also be calculated using complex numbers, taking into account the phase difference between the voltage and current.

## What is the difference between resistance and impedance?

Resistance is the opposition to the flow of direct current in a circuit, while impedance is the opposition to the flow of alternating current. Resistance is measured in ohms, while impedance is measured in complex numbers.

## Why is impedance important in electrical circuits?

Impedance is important in electrical circuits because it affects the flow of current and the behavior of the circuit. It can also be used to analyze and design circuits, and can help determine the maximum power that can be transferred through a circuit.

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