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Some problems on Kirchhoff's law- pls help

  1. Apr 27, 2006 #1

    app

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    Some problems on Kirchhoff's law- pls help!!!

    hello.i am stuck with some kirchhoff's law problems. i have attached the problem along with this post. now, all the cells may or may not give current. but atleast the 15V cell has to discharge. (the longer line represents the positive end of the cell.) i'm having problems identifying, in which way current will flow in the circuit? which cell will be discharging and which will not? where will the current flow after starting from the 15V cell. if i could identify the currents, then applying the kirchhoff's loop rule is allright. pls help, i'm really confused. i'm a new member of PF, a good forum indeed. the text above each of the cells denotes the emf and the internal voltage. the two black boxes are resistance boxes with their resistance values given at the side. pls help...
     

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    Last edited: Apr 27, 2006
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  3. Apr 27, 2006 #2

    daniel_i_l

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    normally you set the direction of the current as going from the + to the -. But really you don't have to do it this way, you could say that the direction in both the top and the bottom are clockwise and then adjust the minus signs afterwards if that's easier.
     
  4. Apr 30, 2006 #3

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    how silly, wht an answer by daniel! everyone knows tht current flows frm positive to negative, i'm not that dumb! cant anyone show the path of current in the circuit? then only is it possible to apply kirchoffs law...
     
  5. Apr 30, 2006 #4

    nrqed

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    Daniel's answer is NOT silly!! The answer is that you put *arbitrary* dorections for all the currents, solve the equations (conservation of current at the nodes, sum of voltages equal to zero in loops) and at the very end if any of your value of current is negative, it tells you that the direction you had chosen (randomly) at the beginning was incorrect. If the sign of a current is positive, then it tells you that your guess was the right one.
     
  6. Apr 30, 2006 #5

    Curious3141

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    Way to get people to want to help you. Smooth, dude.
     
  7. Apr 30, 2006 #6

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    I'm SORRY !

    hey daniel, i'm sorry for calling tht answer silly. actually i didnt get the meaning at first. but now i got it. THANKS A LOT FOR HELPING. PLEASE DONT MIND, I'M REALLY SORRY.
     
  8. Apr 30, 2006 #7

    Curious3141

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    That's better. :approve:

    In all the following, the directions are indicated by the order of the letters.

    OK, divide the circuit into 2 clockwise loops DBACD and DCEFD. You can also analyse the big loop FBAEF but this you can solve the problem without it, it would be helpful in confirming your answer though.

    Now label the current going thru DB as [tex]I_1[/tex], and that going through EF as [tex]I_2[/tex].

    Then the current going through DC is ? (use Kirchoff's 2nd law here).

    Then use Kirchoff's first law in summing up voltages around each of the loops I mentioned. When going from the positive side to the negative side of a cell of voltage V, you add MINUS V (or subtract V).

    When going across a resistor R following the labelled direction of current I, you add MINUS IR (or subtract IR). When going across the same resistance opposite to the labelled direction of current, add IR instead.

    In this way, set up a system of 2 simultaneous equations in [tex]I_1[/tex] and [tex]I_2[/tex] and solve for it. You can use the big loop to confirm the answer is correct. If you do this the same way I did (as in the above), you will find that one of the current values comes out negative, meaning the "actual" conventional current in that part of the circuit flows the opposite way from the way we had assumed.

    For the other bit, the question is a little ambiguous, but I'll assume they want us to find the potential difference between points A and F [tex]V_{AF}[/tex] Here you don't need a complete loop, all you need to do is to treat the point A as if it contains a voltage source and F as if it's a ground point.

    Then, taking the path ABDF, [tex]V_{AF} - 15V - (-I_1)(1\ohm) = 0[/tex]

    and you can use the previously determined value of [tex]I_1[/tex] to get [tex]V_{AF}[/tex]. To check the answer, use the path ACEF.
     
    Last edited: Apr 30, 2006
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