Impedance Matrix of a Potential Divider.

In summary, the equations for this circuit are: Vi = I1Ra + (I1-I2)Rb, Vo = (I1-I2)Rb. The impedance matrix is: \left[ \begin{array}{c} Vi \\Vo \end{array} \right] = \left[ \begin{array}{cc} (Ra+Rb) & -Rb \\ Rb & -Rb \end{array} \right]\left[ \begin{array}{c} I1\\I2\end{array} \right]. The matrix equation that relates current to voltage is: I=A*V.
  • #1
CraigH
222
1

Homework Statement


This question should be quite easy, I'm really struggling though.

http://img32.imageshack.us/img32/5241/potdiv.png

1) Write down two simultaneous equations for this circuit that describe the voltage as a function of current

2)Write down the impedance matrix that relates voltage to current

3)Determine the matrix equation that relates current to voltage

Homework Equations



V=IR

The Attempt at a Solution



1)
Vi = I1Ra + (I1-I2)Rb
Vo = (I1-I2)Rb

2)

[itex] \left[ \begin{array}{c} Vi \\Vo \end{array} \right] = \left[ \begin{array}{cc} (Ra+Rb) & -Rb \\ Rb & -Rb \end{array} \right]\left[ \begin{array}{c} I1\\I2\end{array} \right][/itex]

3)

I don't understand what is being asked here, and it's worth a lot of marks. But so is the last section, so it can't just be the same thing again. What do you think it is asking?
 
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  • #2
An impedence matrix seems like a very sophisticated way of talking about a potential divider...

Put "simply" the voltage experienced at the load is:

[itex]

\frac{V_{\text{in}}R_B}{R_A+R_B}

=

V_{\text{out}}

[/itex]
 
  • #3
Another good point to make is that unless the terminals on the right are connected no current I2 will flow.. as there is no path to ground. And if they are shorted then the load is in parallel with no load... and so all of the current will flow through RA only. And if the load on the right is non-zero -> then the question is rather different isn't it. Its rather late here and it's been a long day so I may have fundamentally misunderstood the question or misapplied circuit analysis. Will have another look tomorrow.
 
  • #4
Its okay I've got it now. I misread the question, part 3 is "relates current to voltage" and part 2 is "relates voltage to current".

So for part 3 I had to invert the impedance matrix to get the admittance matrix A . And then

I= A * V

A matrix does seem like an overly complicated way to find the currents in a potential divider, I guess it could be used to create simulations or something though. I think this was just an arbitrary example to test our matrix knowledge.

And yes I've also been thinking that no current will flow through I2 if the terminals are not connected. This confused me a bit but I think my answer is still correct.

I think that Vo will be connected by a load, e.g a 10 ohm resistor, but my matrix equation will still be true for any load as the question seems to imply that you already know all the voltages and currents and you just want to work out the resistances.

Like I said though, It's just an arbitrary matrix question.
 
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  • #5
Im glad that you've solved it. This is a strangely mathematical way of talking about circuit analysis. It is very probably true and yet slightly weird at the same time!
 
  • #6
In your schematic you have the I2 current going the wrong way according to the convention usually used. See this page:

http://fourier.eng.hmc.edu/e84/lectures/ch2/node4.html

Notice how the currents are all going into the network (two-port), not out. The way you have done it the 2,2 element of your matrix is negative. That element should be the resistance looking into the network (from the output) with the input open circuited, and it shouldn't be a negative resistance.

Did you draw that schematic yourself, or did you paste from the actual homework assignment? I would check with the instructor about this.
 
  • #7
Thanks for the link, its good stuff :)

The diagram was taken straight from a past exam paper.

I think however the impedance matrix is still correct, or at least it is correct in the eyes of my tutor. Like I mentioned earlier this was from my maths exam (as part of an electronic engineering degree), so the question is just an arbitrary matrix question.

I think that arrow showing [itex]V_{0}[/itex] is the wrong way around, as I think that [itex]V_{0}[/itex] is not a voltage source, but a voltage drop across a load between the two nodes of this port. In this case I2 is going in the correct direction, and I will just end up with a negative value for [itex]V_{0}[/itex] if I am to calculate it.
 
  • #8
Your tutor makes the rules for homework problems, so you need to do what gets you the good grade.

But, once you're out of that class, be aware that the elements on the main diagonal of a Z matrix should all be (by convention) positive when the elements of the circuit are just plain old resistors.
 
  • #9
Thank you I will keep that in mind.
 

1. What is the Impedance Matrix of a Potential Divider?

The Impedance Matrix of a Potential Divider is a mathematical representation of the electrical impedances of each component in a potential divider circuit.

2. Why is the Impedance Matrix important?

The Impedance Matrix allows us to analyze the behavior of a potential divider circuit and calculate the voltage across each component. It is crucial for designing and troubleshooting circuits.

3. How is the Impedance Matrix calculated?

The Impedance Matrix is calculated by finding the reciprocal of each component's impedance and arranging them in a matrix format. The matrix can then be used to solve for the voltages in the circuit.

4. What factors affect the Impedance Matrix?

The Impedance Matrix is affected by the values of the components in the potential divider circuit, as well as the frequency of the input signal. Changes in these factors can alter the impedance values and therefore the matrix.

5. How is the Impedance Matrix used in practical applications?

The Impedance Matrix is used in various practical applications, such as designing audio circuits, analyzing signal transmission in communication systems, and calculating the voltage distribution in power supply circuits.

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