Can You Combine Current Sources Question

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Discussion Overview

The discussion revolves around the combination of current sources in a circuit analysis context, specifically addressing whether current sources can be combined and how to approach solving a circuit with resistors and dependent sources. Participants explore the implications of circuit configuration on current flow and the application of Kirchhoff's Current Law (KCL).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express uncertainty about combining current sources, suggesting that it depends on the circuit configuration, particularly when dependent sources are involved.
  • There is a proposal to first combine resistors before addressing the current sources, though this approach is questioned based on the circuit's drawing.
  • Participants discuss the need for clarity in the circuit diagram, with one suggesting that the current path appears to short circuit the resistors.
  • Some participants recommend using KCL to derive relationships between currents and voltages in the circuit, indicating a methodical approach to solving the problem.
  • There are conflicting interpretations regarding the current through specific resistors, with participants providing different equations based on their understanding of the circuit.
  • One participant expresses skepticism about the simplicity of a proposed solution, questioning if it is too straightforward.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether current sources can be combined in this specific case, and there are multiple interpretations of the circuit behavior and current flow through the resistors.

Contextual Notes

Limitations include the unclear circuit diagram and the potential for misinterpretation of current paths, which may affect the application of KCL and the resulting equations.

eatsleep
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http://imgur.com/09Rj0ND
2. V=IR
3. My real question is if I can combine the current sources. I'm pretty sure you cannot. I have combined the resistors in parallel, but I am not sure what to do next
 
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eatsleep said:
http://imgur.com/09Rj0ND



2. V=IR



3. My real question is if I can combine the current sources. I'm pretty sure you cannot. I have combined the resistors in parallel, but I am not sure what to do next

It depends on the situation, but in this case you cannot, IMO. There are current branches off of the top node, and the right side source is a dependent source.

If you have two independent current sources in parallel, you would usually be able to combine them.
 
eatsleep said:
http://imgur.com/09Rj0ND

2. V=IR

3. My real question is if I can combine the current sources. I'm pretty sure you cannot. I have combined the resistors in parallel, but I am not sure what to do next
Display of your image.
attachment.php?attachmentid=58919&stc=1&d=1369174508.png
 

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berkeman said:
It depends on the situation, but in this case you cannot, IMO. There are current branches off of the top node, and the right side source is a dependent source.

If you have two independent current sources in parallel, you would usually be able to combine them.


In order to solve this problem, am I correct in first combining the resistors? Then does Ib turn into 1Amp?
 
eatsleep said:
In order to solve this problem, am I correct in first combining the resistors? Then does Ib turn into 1Amp?

Not the way the circuit is drawn. Can you post the original circuit, so we can check that this drawing is correct?

The way it is drawn is a bit confusing. You show a wire carrying a current Ib, but that wire shorts the top of the two resistors, so there is no wire segment that can carry a current...
 
berkeman said:
Not the way the circuit is drawn. Can you post the original circuit, so we can check that this drawing is correct?

The way it is drawn is a bit confusing. You show a wire carrying a current Ib, but that wire shorts the top of the two resistors, so there is no wire segment that can carry a current...

This is the original drawing, from my teacher
 
eatsleep said:
This is the original drawing, from my teacher

That's unfortunate.

Well, you can still try to solve it, and hopefully it will yield the correct answer. Write KCL equations for the left node (to the left of the arrow "Ib") and for the right node, first assuming that they are at different voltages. Solve the right side equation to get a relationship between the voltage and the current Ib. Then assume that the two voltages are the same, and solve for the current Ib.

Seems kind of clunky. I honestly do not know why you were given a circuit drawn like that...
 
berkeman said:
That's unfortunate.

Well, you can still try to solve it, and hopefully it will yield the correct answer. Write KCL equations for the left node (to the left of the arrow "Ib") and for the right node, first assuming that they are at different voltages. Solve the right side equation to get a relationship between the voltage and the current Ib. Then assume that the two voltages are the same, and solve for the current Ib.

Seems kind of clunky. I honestly do not know why you were given a circuit drawn like that...

Thank you for the help!
 
Using Kirchhoff's Current Law:
What is the current through the 2kΩ resistor? (in terms of Ib) ?

What is the current through the 5kΩ resistor? (in terms of Ib) ?​
 
  • #10
SammyS said:
Using Kirchhoff's Current Law:
What is the current through the 2kΩ resistor? (in terms of Ib) ?

What is the current through the 5kΩ resistor? (in terms of Ib) ?​

2kΩ resistor 10Ib?

5kΩ resistor 8Ib-1?
 
  • #11
eatsleep said:
2kΩ resistor 10Ib?
Yes.


5kΩ resistor 8Ib-1?
No.

1 Ampere into the junction. Current of Ib out.

Those together with current through 5kΩ , whichever direction you choose.
 
  • #12
SammyS said:
Yes.


No.

1 Ampere into the junction. Current of Ib out.

Those together with current through 5kΩ , whichever direction you choose.

I'm sorry, so the current through the 5kΩ is Ib-1?
 
  • #13
eatsleep said:
I'm sorry, so the current through the 5kΩ is Ib-1?
Yes.

Upward.
 
  • #14
Will solving for Vr1 by doing 5(Ib-1)=2(10Ib), solving for Ib, then plugging that backing to 5(Ib-1) work? seems too easy
 
  • #15
eatsleep said:
Will solving for Vr1 by doing 5(Ib-1)=2(10Ib), solving for Ib, then plugging that backing to 5(Ib-1) work? seems too easy
Yes.

It should be easy to check the results.
 

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