Capacitance and Inductance questions

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Homework Statement


78lfvRG.jpg


Homework Equations




The Attempt at a Solution


Here are two questions. I'm stuck on question 1, and need someone to check question 2.

For the first question I see C2, C3, and C4 in parallel, so Ceq of those 3 are C2 + C3 + C4 = 3C

but now how do I continue? seems like the 3 remaining capacitors aren't in series or parallel..

Am I suppose to use Y-delta transformation?

______________________________________


For the 2nd question.

I see the top 3 inductors are in series and the botton 3 are in series.

Therefore L1 + L2 + L3 = 3L

at the bottom row, L4 + L5 + L6 = 3L

now we have 3L and 3L in parallel connection, 1/Leq = 1/3L + 1/3L

Leq = 3/2 L
 

Answers and Replies

  • #2
berkeman
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but now how do I continue? seems like the 3 remaining capacitors aren't in series or parallel..
What can you replace the voltage source with to keep making progress on combining capacitances here? Hint -- what is the AC impedance looking into a voltage source?
Leq = 3/2 L
Correct. :smile:
 
  • #3
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What can you replace the voltage source with to keep making progress on combining capacitances here? Hint -- what is the AC impedance looking into a voltage source?

Correct. :smile:
Replace voltage source? AC impedance?

I'm in a intro class, we only worked with DC so far :(
 
  • #4
berkeman
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I'm in a intro class, we only worked with DC so far :(
Well, capacitors and inductors are AC components, so it looks like you've moved on to the next chapter... :smile:
 
  • #5
berkeman
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Replace voltage source? AC impedance?
Replace the power supply with an AC short. Then what can you do to keep going...?
 
  • #6
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Replace the power supply with an AC short. Then what can you do to keep going...?
if the Battery became a short, C1 would be parraell with 3C , and then the Ceq of those 2 would be C + 3C = 4C.

Then 4C would be in series with C5, then 1/Ceq = 1/4C + 1/C

Ceq of entire circuit = 4/5 C
 
  • #7
berkeman
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if the Battery became a short, C1 would be parraell with 3C , and then the Ceq of those 2 would be C + 3C = 4C.

Then 4C would be in series with C5, then 1/Ceq = 1/4C + 1/C

Ceq of entire circuit = 4/5 C
Looks good to me. Can you check the answers online?
 
  • #8
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Replace the power supply with an AC short. Then what can you do to keep going...?
but in what scenario would a battery be replaced by a short? My textbook doesn't seem to cover this, it does say that a capacitor becomes open in a DC circuit, and an inductor becomes closed in a DC circuit.
 
  • #9
berkeman
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but in what scenario would a battery be replaced by a short? My textbook doesn't seem to cover this, it does say that a capacitor becomes open in a DC circuit, and an inductor becomes closed in a DC circuit.
When you are calculating equivalent impedances, it is common to replace a current source with an open, and a voltage source with a short. That's how the actual circuits for each behave anyway, so it's a reasonable simplification to make when finding equivalent impedances. You will probably be covering it in class soon, since you use it in this question.
 
  • #10
epenguin
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Seems to me it is needed to be able to solve this with elementary concepts.

That C5 might as well not be there – there is no way you can put a charge on it as one end is not connected to anything so there is nowhere a charge can come from to the isolated plate and therefore you cannot maintain a charge on the opposite one.

Then the potential difference is the same across C2, C3 and C4. Charges on each capacitor would therefore be proportional just to their capacitance.. The total charge is the sum of the three. So the sum the equivalent capacitance of the part 234 will be just the sum of the capacitances, which is in fact the general rule for parallel capacitors

Replacing then the three capacitors with one equivalent capacitance, we are left with two capacitors in series. Call the second capacitance C234. System there has a conductively isolated 'inside' part with two plates. The charge on one plate is equal but opposite to that on another, because all the electrons on one must have come from the other. So the two capacitors charges are equal to each other, call this charge Q.
The potential differences in series add up: Vtotal = V1 + V234 = Q1/C1 + Q234/C234. But the charges here, as we previously said, are equal so we just have
Vtotal = Q(1/C1 + 1/C234)
Comparing to the general formula V = Q/C you see that the equivalent capacitance C of this circuit is given by
1/C = (1/C1 + 1/C234)

So that's practically all the calculation, and all the general formula, from first principles.
 
  • #11
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Seems to me it is needed to be able to solve this with elementary concepts.

That C5 might as well not be there – there is no way you can put a charge on it as one end is not connected to anything so there is nowhere a charge can come from to the isolated plate and therefore you cannot maintain a charge on the opposite one.

Then the potential difference is the same across C2, C3 and C4. Charges on each capacitor would therefore be proportional just to their capacitance.. The total charge is the sum of the three. So the sum the equivalent capacitance of the part 234 will be just the sum of the capacitances, which is in fact the general rule for parallel capacitors

Replacing then the three capacitors with one equivalent capacitance, we are left with two capacitors in series. Call the second capacitance C234. System there has a conductively isolated 'inside' part with two plates. The charge on one plate is equal but opposite to that on another, because all the electrons on one must have come from the other. So the two capacitors charges are equal to each other, call this charge Q.
The potential differences in series add up: Vtotal = V1 + V234 = Q1/C1 + Q234/C234. But the charges here, as we previously said, are equal so we just have
Vtotal = Q(1/C1 + 1/C234)
Comparing to the general formula V = Q/C you see that the equivalent capacitance C of this circuit is given by
1/C = (1/C1 + 1/C234)

So that's practically all the calculation, and all the general formula, from first principles.

Maybe it is unclear if it is part of a bigger circuit, or if it is not connected to anything at all. But if it actually is part of a bigger circuit, then the Ceq would be 4/5C then, correct? and If it is actually isolated, then it Ceq would be 3/4 C.
 
  • #12
gneill
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That C5 might as well not be there – there is no way you can put a charge on it as one end is not connected to anything so there is nowhere a charge can come from to the isolated plate and therefore you cannot maintain a charge on the opposite one.
That's not the right way to look at it. The open terminals are where you're "looking from" to judge the equivalent capacitance of the network. You have to imagine what the behavior of the circuit would be if some tests were applied at those terminals. Such tests might include applying various voltage or current sources and measuring the results (current injected or voltage that appears across the terminals).

If open terminals nullified capacitance then an lone capacitor would not have a capacitance...
 
  • #13
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That's not the right way to look at it. The open terminals are where you're "looking from" to judge the equivalent capacitance of the network. You have to imagine what the behavior of the circuit would be if some tests were applied at those terminals. Such tests might include applying various voltage or current sources and measuring the results (current injected or voltage that appears across the terminals).

If open terminals nullified capacitance then an lone capacitor would not have a capacitance...
so regardless if the terminals are connected to anything or not, the Ceq of the circuit should be the same/unaffected?
 
  • #14
gneill
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so regardless if the terminals are connected to anything or not, the Ceq of the circuit should be the same/unaffected?
Yes.

An "equivalent value" implies that you are looking at the circuit from some particular port. In this case that's the open terminals of the circuit.
 
  • #15
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is there a way to solve number 1 without using source transformation?
 
  • #16
cnh1995
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is there a way to solve number 1 without using source transformation?
There is no need of source transformation there. Source transformation means conversion of a voltage source into equivalent current source and vice-versa.
 
  • #17
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There is no need of source transformation there. Source transformation means conversion of a voltage source into equivalent current source and vice-versa.
I see, is there a way to solve question1 without replacing the battery with a short circuit?
 
  • #18
cnh1995
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I see, is there a way to solve question1 without replacing the battery with a short circuit?
Equivalent capacitance between two points is the capacitance "seen" by a voltage source connected between those two points. If the equivalent capacitance is asked between the rightmost two points, you need to short the battery. If the equivalent capacitance is asked between the battery terminals, you'll have to open the battery terminals.
 
  • #19
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Equivalent capacitance between two points is the capacitance "seen" by a voltage source connected between those two points. If the equivalent capacitance is asked between the rightmost two points, you need to short the battery. If the equivalent capacitance is asked between the battery terminals, you'll have to open the battery terminals.
If I short the terminals on the right, I will have C2 + C3 + C4 + C5 = 4C, since they are in parallel. Then 4C and C1 are in series. Then Ceq would be 4/5 C.
 
  • #20
cnh1995
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Why short the terminals on the right? Between which two points is the equivalent capacitance asked?
 
  • #21
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Why short the terminals on the right? Between which two points is the equivalent capacitance asked?
I'm a bit confused by what you mean points, do you mean nodes?
I think it is asking for the Ceq between the top node and the ground?
 
  • #22
cnh1995
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I'm a bit confused by what you mean points, do you mean nodes?
Yes. Top node on the right and the ground..
 
  • #23
cnh1995
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The points between which the capacitance is asked are not shorted, but are kept open. It means a voltage source inserted between those two points will see the that equivalent capacitance.
 
  • #24
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Yes. Top node on the right and the ground..
If i open the battery then I would have this?
upload_2016-5-21_0-11-50.png
 
  • #25
cnh1995
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If i open the battery then I would have this?
View attachment 101028
Yes. But Ceq234 will be 3C and C5 is redundant (i.e. it is not a part of the circuit). So the net capacitance viewed from the battery terminals is C1 in series with C123.
 
  • #26
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Yes. But Ceq234 will be 3C and C5 is redundant (i.e. it is not a part of the circuit). So the net capacitance viewed from the battery terminals is C1 in series with C123.
oops, yes it should be 3C. But I'm still having trouble understanding why C5 is not part of the circuit. If I open the battery terminal, the C1 is identical to C5, but yet only C5 is redundant?
 
  • #27
cnh1995
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oops, yes it should be 3C. But I'm still having trouble understanding why C5 is not part of the circuit. If I open the battery terminal, the C1 is identical to C5, but yet only C5 is redundant?
If you are opening the battery, you are calculating the capacitance between the top-left node and ground i.e. capacitance viewed from the existing battery terminals. So, one terminal of C5 is connected to the middle node but the other terminal is hanging, not connected to the circuit. It won't form a closed path and hence it won't be charged by the existing battery. Hence, it is redundant.
 
  • #28
epenguin
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If open terminals nullified capacitance then an lone capacitor would not have a capacitance...

Surely it doesn't? It doesn't accept a charge.

Or else capacitance is undefined – since it is usually defined in terms of involving potential difference across its connected terminals?

Whatever - it to seems to me we are talking of definitions and debating what the question actually is. But if the question is the equivalent capacitance of that circuit, i.e. the charge you would measure e.g. by measuring the integrated current after you connected the potential source, then the loose capacitor C5 might just as well not be there.

Have you been given an answer? Does it correspond to the calculation I indicated?
 
  • #29
gneill
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Surely it doesn't? It doesn't accept a charge.

Or else capacitance is undefined – since it is usually defined in terms of involving potential difference across its connected terminals?
A solitary capacitor has two open terminals. As an ideal component it has a capacitance that does not depend upon anything but its geometry. You can connect those terminals to whatever equipment you wish in order to measure that capacitance, but in the end it boils down to the inherent geometry and laws of physics.
Whatever - it to seems to me we are talking of definitions and debating what the question actually is. But if the question is the equivalent capacitance of that circuit, i.e. the charge you would measure e.g. by measuring the integrated current after you connected the potential source, then the loose capacitor C5 might just as well not be there.
The source indicated in the circuit diagram plays no role in the equivalent capacitance as judged from the terminals. Place the entire circuit in a black box with only those two terminals showing. Then ask the question, "what is the equivalent capacitance of this component?"

When you set about to find the equivalent capacitance, one of the first things you do is suppress the sources. In this case the voltage supply is suppressed by replacing it with a short circuit. Then you find the equivalent capacitance looking into the open terminals.
Have you been given an answer? Does it correspond to the calculation I indicated?
Goonking had it right in post #6.
 
  • #30
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Surely it doesn't? It doesn't accept a charge.

Or else capacitance is undefined – since it is usually defined in terms of involving potential difference across its connected terminals?

Whatever - it to seems to me we are talking of definitions and debating what the question actually is. But if the question is the equivalent capacitance of that circuit, i.e. the charge you would measure e.g. by measuring the integrated current after you connected the potential source, then the loose capacitor C5 might just as well not be there.

Have you been given an answer? Does it correspond to the calculation I indicated?
I have no answer key to the questions, but after asking for clarification, the terminals on the right aren't connected to anything, if that changes anything.
 

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