Can You Combine Current Sources Question

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SUMMARY

This discussion focuses on the combination of current sources in electrical circuits, specifically addressing the challenges of combining independent and dependent current sources. Participants confirm that while independent current sources can typically be combined in parallel, the presence of dependent sources complicates this process. The conversation emphasizes the importance of applying Kirchhoff's Current Law (KCL) to analyze the circuit correctly and derive relationships between currents and voltages in the circuit.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with circuit analysis techniques
  • Knowledge of independent and dependent current sources
  • Basic concepts of resistors in parallel
NEXT STEPS
  • Study the application of Kirchhoff's Current Law in complex circuits
  • Learn about the characteristics of independent vs. dependent current sources
  • Explore techniques for analyzing circuits with multiple resistors in parallel
  • Investigate common circuit analysis software tools for simulation
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Electrical engineering students, circuit designers, and professionals involved in circuit analysis and design will benefit from this discussion, particularly those working with current sources and resistor networks.

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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.
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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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