Find the Equivalent Impedance of the Network

[rtareen]

TL;DR Summary

Problem is from Faissler's Intro to Modern Electronics Chapter 8. We are asked to find the equivalent impedance of the network. Nothing is in series or parallel so I don't know what to do.

We are asked to find the equivalent impedance of the network. Nothing is in series or parallel so I don't know what to do. We cannot use mesh equations or Thevenin's theorem because is no current or voltage source. So I have no clue.

 

[f95toli]

I believe the issue here is not this particular circuit, but that you've haven't quite understood what "in parallel" and "in series" means.

What is the definition of two elements being connected in parallel?
In series?

 

[rtareen]

I believe the issue here is not this particular circuit, but that you've haven't quite understood what "in parallel" and "in series" means.

What is the definition of two elements being connected in parallel?
In series?

I know in series means if a current flows through the first component, that entire current must flow through the second. Parallel is a little vague. I think they have the same voltage across them they are in parallel. But since there is no voltage source here that definition would lead to all three being in parallel. I don't know any general definition that always works. Can you teach me the definiton?

 

[tech99]

It does depend on frequency of course. Is a frequency specified?

 

[rtareen]

It does depend on frequency of course. Is a frequency specified?

Yes 440 Hz

 

[tech99]

There are formulas for converting series combinations to parallel and vice versa. You need to apply this to R and L to obtain the series equivalent, then the combination is in series with C. The two series reactances can then be combined into one by addition.

 

[rtareen]

There are formulas for converting series combinations to parallel and vice versa. You need to apply this to R and L to obtain the series equivalent, then the combination is in series with C. The two series reactances can then be combined into one by addition.

I know the formulas, but I don't know how to tell that R and L are in series?

 

[tech99]

L and R are in parallel. First of all, find the reactance of L at 440Hz. Then do the conversion to of L and R in parallel to series elements. Then ask again about the next step!

 

[rtareen]

L and R are in parallel. First of all, find the reactance of L at 440Hz. Then do the conversion to of L and R in parallel to series elements. Then ask again about the next step!

How do you know that they are in parallel?

 

[tech99]

R and L have both their ends connected together, they share the same voltage.

 

[rtareen]

R and L have both their ends connected together, they share the same voltage.

Yes but doesn't having a capacitor or even an input/output wire in between disrupt that? I deleted my work it was completely off

 

[DaveE]

Once you are comfortable understanding series and parallel resistance equivalents. Look at this video about how to use the substitution for series and/or parallel resistors to simplify more complex combinations of impedances. It's the same process with inductors and capacitors. Initially just call everything "Z" (Z1, Z2, Z3, etc.), some complex impedance value, and treat it like it's a resistor. After you have solved the circuit (or part of it, if you like), then you can substitute the complex value for each "Z"; like $Z=\frac{1}{jωC}$ for a capacitor, etc.
https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/v/circuits-part-4

 

[tech99]

Yes but doesn't having a capacitor or even an input/output wire in between disrupt that? Anyways here is my work so far.

https://www.physicsforums.com/attachments/281897

This looks in the right ballpark, which I roughly estimated as 5000 + j5000. But resistance should not be negative. The next move is to connect this in series with the capacitor. So next find the reactance of the capacitor at 440Hz.

 

[tech99]

Once you are comfortable understanding series and parallel resistance equivalents. Look at this video about how to use the substitution for series and/or parallel resistors to simplify more complex combinations of impedances. It's the same process with inductors and capacitors. Initially just call everything "Z" (Z1, Z2, Z3, etc.), some complex impedance value, and treat it like it's a resistor. After you have solved the circuit (or part of it, if you like), then you can substitute the complex value for each "Z"; like $Z=\frac{1}{jωC}$ for a capacitor, etc.
https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/v/circuits-part-4

Unfortunately, the parallel combination involves conversion to admittances, adding admittances and then converting back to impedances, which is a complication for the OP. That is why I suggested just using the series-parallel conversion formulas.

 

[rtareen]

Unfortunately, the parallel combination involves conversion to admittances, adding admittances and then converting back to impedances, which is a complication for the OP. That is why I suggested just using the series-parallel conversion formulas.

Thanks. I'll use what I got but flip the sign of the real part. So $Z_{RL} = 4540 + 4990j$. Then we compute the capacitors impedance $Z_C = -j(1/(wC) = -j (\frac{1}{2\pi (440)(10^{-7}}) = -3620j$. Then we use

$Z_{eq} = Z_{RL} + Z_{C} = 4540 + (4990 -3620)j = 4540 + 1370j~\Omega$

However the book, which I don't trust, said the answer is 1157 -419j ohms. This book is full of mistakes though. I've already caught three. What do you think?

 

[DaveE]

Thanks. I'll use what I got but flip the sign of the real part. So $Z_{RL} = 4540 + 4990j$. Then we compute the capacitors impedance $Z_C = -j(1/(wC) = -j (\frac{1}{2\pi (440)(10^{-7}}) = -3620j$. Then we use

$Z_{eq} = Z_{RL} + Z_{C} = 4540 + (4990 -3620)j = 4540 + 1370j~\Omega$

However the book, which I don't trust, said the answer is 1157 -419j ohms. This book is full of mistakes though. I've already caught three. What do you think?

I like your answer better. I found 4542.5 - 1361.9j

Good work!

Edit: Oops typo: it's 4542.5 + 1361.9j

 

[rtareen]

I like your answer better. I found 4542.5 - 1361.9j

Good work!

Woohoo !

 

[DaveE]

Maybe it's time for a different book?

 

[rtareen]

Maybe it's time for a different book?

I'm only using this book because its supposedly used in this electronics class at my university. If you have any recommendations I wouldn't mind checking out some others.