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Finding voltage in a branch

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    Using KCL at the top node of the branch whose voltage I'm supposed to find, I get this equation
    5 - 1- i2 + 6 = 0

    i2 is the current in third branch from left, whose direction has been chosen arbitrarily by me. Solving this gives me i2 = 10. And hence voltage through the third branch from left is 10*10. i.e 100. Since the branch in which voltage is asked is parallel to the branch of whose voltage I just found, they both have the same voltage, and thus, the answer to the question is 100. But that's not the answer they gave. Where am I going wrong?
     

    Attached Files:

  2. jcsd
  3. Jun 24, 2013 #2
    The two 10 ohm resistors are in parallel. Replace with an equivalent resistor.
     
  4. Jun 24, 2013 #3
    Okay, how do I do it without an equivalent resistor? I know what you mean. But I'm reading a new book, and they haven't reached equivalent resistors yet. They expect us to do it without the knowledge of it.

    Also, equivalent resistor won't help here since I have to find voltage through a particular branch
     
  5. Jun 24, 2013 #4
    Also, can someone explain the passive sign convention? My problems are always going wrong because of signs.

    Please take a look at the attachment with his post too. How did he apply can I use my own direction for current? Because on doing so, my answer doesn't match.

    What I did was took i6 in the opposite direction of what the author has taken. Hence all the current would be flowing into the node. That would give

    i6 + 2ix + 0.024 + ix = 0

    Also, why is one of his voltages negative? If current is passing through the 2k ohm resistor the way he has shown, according to the sign convention, shouldn't the voltage be positive?
     

    Attached Files:

    Last edited: Jun 24, 2013
  6. Jun 24, 2013 #5
    In Node analysis always take the current arriving at Node as positive. For unkown currents take them arriving. For known currents take them positive if arriving and negative if leaving. It should be like

    i1+i2+....+in = 0
     
  7. Jun 24, 2013 #6
    I know node analysis, but like I said the book expects us to do using only KVL and KCL.
    Also, please check the book above.

    This is from the book you recommended for circuit analysis.
    Great book.
     
  8. Jun 24, 2013 #7
    Also, I have followed the convention of every current arriving at a node to be taken as positive and every current leaving as negative. Answers still don't match.
     
  9. Jun 24, 2013 #8
    I haven't seen your work though. Let me have a look. But if you are doing this way, the answer should match. Meantime you can try mesh analysis in this problem!
     
  10. Jun 24, 2013 #9
    The two resistora are in parallel. I studied series-parallel combinations in highschool only.

    They are asking to find the Voltage between two points. You can there are two resistors between those two points. You can replace them by one!
     
  11. Jun 24, 2013 #10
    I know. Even I have studied parallel and series circuits.

    Assume the question to be Find the voltage using KCL and KVL without applying equivalent resistance rules.
     
  12. Jun 24, 2013 #11
    In the 2k ohm resistor the Voltake taken negative is correct. Remember ohms Law, it says V= IR, where I is by convention taken as current flowing from positive to negative terminal. Here in th 2k ohm resistor the current is flowing from 'negative to positive'.

    And you KCL equation is right. Use your equation.
     
  13. Jun 24, 2013 #12
    For such problem i will give you a hint.

    Hint: Ohms Law ;)
     
  14. Jun 24, 2013 #13
    Okay, thanks! Got it.
     
  15. Jun 24, 2013 #14
    You are welcome.
     
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