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Circuit analysis help please?

  1. May 4, 2013 #1
    I have two problems and as you have seen ive been posting some exercises like this lately because I have 150 exercises to solve and I cant solve only a FEW of them
    1. The problem statement, all variables and given/known data

    For the first one :We have the circuit in the figure.We have to find i.If its not clear,the current on the right top says 3i,and the one in the left just i.
    http://i.imgur.com/kQdKOnQ.jpg
    The second one : This exercise is an alien! Its more difficult than the first one ,can you please give me a hint?http://oi41.tinypic.com/4vsi2p.jpg

    2. Relevant equations
    KCL,KVL.


    3. The attempt at a solution

    For the first one :I dont really know what to do about this one.The current across 10 Ohm is i1,now I write the KCL and I have i=3i+i1 here I have that i1=-2i.I find the voltage across 10 Ohm and it is -20 V.I am thinking about finding the resistance across 3i,so I can find the whole resistance of the whole circuit and then 25/ (Equivalent resistance) but the resistance is negative :/
     
  2. jcsd
  3. May 4, 2013 #2

    NascentOxygen

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    Try again, what is the correct expression for the voltage across the 10Ω?
     
  4. May 4, 2013 #3
    V across 10 Ohm= -2i * 10 Ohm=-20 V...
     
  5. May 4, 2013 #4

    NascentOxygen

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    Do you know the value of variable i?
     
  6. May 4, 2013 #5
    No,I dont...sorry it is -20i Volt
     
  7. May 4, 2013 #6

    NascentOxygen

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    So now you know the current through each resistor (in terms of variable i). Add them up and what do you get?
     
  8. May 4, 2013 #7
    No,I know the voltage across 10 Ohm in terms of variable i,not the current...
     
  9. May 4, 2013 #8

    SammyS

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    First of all, terminology:
    Voltage is dropped across a resistor or other circuit element.

    Current passes through a resistor or other circuit element.​

    attachment.php?attachmentid=58476&stc=1&d=1367692950.jpg

    You have a current of " i " flowing through the branch with the 25V voltage source and 6Ω and 4 Ω resistors .

    You have a current of 3i through the branch which has only the current source. That's the diamond shaped object with the arrow in it.

    Use KCL to find the current through the 10 Ω resistor. (This will be in terms of " i " at this point.)

    Then use KVL for the appropriate loop to give you an equation with only the variable, i .
     

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  10. May 4, 2013 #9
    Omg thanks :D How about the second one?
     
  11. May 4, 2013 #10

    SammyS

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    attachment.php?attachmentid=58479&stc=1&d=1367694910.jpg

    Similar idea, but a little more complicated.

    One fixed current source of 5 Amperes.

    One variable current source. For this, the current is v/4, where v is the voltage drop across the 6Ω resistor.


    I would let the variable, I, represent the current through the 6Ω resistor.
     

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  12. May 5, 2013 #11
    Yes,but I dont have enough data to apply KCL at the node above 6 ohm..
     
  13. May 5, 2013 #12

    NascentOxygen

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    Yes you do, providing you write each current in terms of variable v.

    For starters, how much current flows to ground through the 6Ω resistor (in terms of v)?
     
  14. May 5, 2013 #13
    I understand it completely now,thanks.For the exercise before this,did you get i=4.5 A?
     
  15. May 5, 2013 #14

    NascentOxygen

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    Aren't you to find v?
     
  16. May 5, 2013 #15
    I have posted two exercises here,I mean the first one,where I have to find i...
     
  17. May 5, 2013 #16

    NascentOxygen

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    For the solution to the first question, I did not get i=4.5A
     
  18. May 5, 2013 #17
    What was the i that you got? I will figure out it again..
     
  19. May 5, 2013 #18

    NascentOxygen

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    You should be able to check your answer. Work your way around the circuit substituting the value you have, and all voltages and currents should tie in with Kirchoff's laws. If they all check out, then your answer is correct!
     
  20. May 5, 2013 #19

    SammyS

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    I thought you said you had this figured out.

    Anyway, please show how you got this answer.

    What was your loop equation?
     
  21. May 6, 2013 #20
    The final loop equation : -25 V - 4i -20i -6i=0 ..
     
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