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Parallel Circuit with current source and dependent current source. HELP!

Discussion in 'Electronics Homework Help' started by Juan Valedon, Aug 22, 2016.

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  1. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    Hey! I've been trying a lot to solve this circuit with no correct results. How can I make it?

    Here is a photo of what I already tried several times. I need to find voltage on the 8 Ohm's Resistance and the current i0.

    My equivalent circuit are correct? My equation is correct?

    Any hint to find the voltage? All I know is that I need to find Voltage in the node on top the 8 Ohm's Resistance.

    PD: I am learning circuits by my own. I am not enrolled in any course at the moment.
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I've deleted your other thread. It doesn't matter how you're learning, I'm sure what you want is assistance, but not complete answers.

    I'm assuming that the final image is the problem itself.

    The first question for you is: Is there any ambiguity in the direction of current flow in any of the components?

    The second question is: What is the sum of the currents flowing through the two resistors and the dependent current source?

    Can you write an equation for the current through the 8 ohm resistor in terms of that through the 2 ohm resistor?

    Having done this, can you see how to get the answer?
     
  3. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    For the second question: i1 - i2 - i3 - i4 = 0 Should I need to find voltage for the resistance of 2Ω?

    I always want to find V0 in node of top of the resistance of 8Ω. I need all the component to find it or just two or three?

    This is how I divide the circuit...Is that correct?
    Check the attachment.
     

    Attached Files:

  4. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    The book says that the answers are 8V and i0 = 4A
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Please answer my first question first.
     
  6. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    I think so. Because the sum of all the currents are equal to 0.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Ok, you're wrong. Lets see why.

    Firstly, how many nodes are there?

    Is there any ambiguity in the direction of current flow in the independent current source?

    What can you say about the direction of current flow through both of the resistors? (I.e. Can they ever be in different directions?)

    Next, ignoring the dependant current source, what is direction of current flow through the 2ohm resistor?

    Knowing the direction of the current flow through the 2 ohm resistor, is this the same or opposite to that through R2?

    Given all of that, what can you say about the direction of current flow in the components other than the independent current source?
     
  8. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    There is two nodes.

    There is no ambiguity, the current flow goes up.

    No! The current never go in different directions.

    They both go down (+ to -)

    Opposite?

    Opposite?

    Let me write the equations and show you.
     
  9. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    I got this equation:

    6A - i0 - i0/4 - (8Ω)i0 = 0; therefore 6A = (1 + 1/4 + 8Ω)i0 = 0
     
  10. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    This can be a possible answer?
     

    Attached Files:

  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You're getting closer, but ohms should not appear.

    You've got the relationship between the current through the 2ohm resistor and the dependant current source, but you need to look more carefully at the current through the 8ohm resistor.

    What can you state about the voltage across both the resistors?
     
  12. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    Common Voltage
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Correct

    Correct

    Correct

    Correct

    Wrong. You are saying the current flow through the 8ohm resistor is opposite to that through the 2 ohm resistor.

    Haven't you already said the current flows downward in R2? are you saying it flows against the arrow in the dependant source?

    Now, it could do that if the direction of current flow changed in R2 (because a -ve current multiplied by a constant is also negative) but there is no suggestion that this will happen.

    Ok, your equation (whilst wrong) is kinda right, but if you don't have the signs right you will get the wrong answer.
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Ok, so what is the current in the 8 ohm resistor in terms of the current in the 2 ohm resistor?
     
  15. Ratch

    Ratch

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    It is a simple problem in node analysis. Do you know how to do node analysis? It looks like you are making hard work of it.

    Ratch
     
  16. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    I already practice a example problem with only one node. One dependent current source, one resistant and finally a current source. I dont know why is very hard to sole this excercise.
    So this equation should be: 6 = i0 + (i0/4) + i0?
     
  17. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    I am tired of practicing this "easy" exercise. I am stuck here for several days. Can you show me the correct equations? It would be really appreciated.
     
  18. Juan Valedon

    Juan Valedon

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    Aug 21, 2016
    Maybe my answers are right, but the book doesn't write the decimal points.
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    We don't give the answer.

    However, once you can determine the current through the 8 ohm resistor in terms of that through the 2 ohm resistor you are able to write the equation which will give you the answer.

    This should be trivial, but if you can't do it in your head, try calculating the current through a 2ohm and an 8ohm resistor with (say) 8V, 16V, and 24V across them. The pattern should be obvious.

    The rest of the stuff we've been doing is to ensure you get the signs right in your equation. This is also vital or you'll make silly mistakes leading to the wrong answer.

    If you need us to tell you what to do then you're not prepared to tackle this problem and you need to go back and work on simpler stuff first.
     
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Highly unlikely.
     
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