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Find the current and the voltage; the Circuit Problem

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the circuit diagram, see attachment. a. Caluclate the current through the 22 Ohm resistor. b. What is the voltage across ab?


    2. Relevant equations
    Kirchoff's rule, Ohm's Law


    3. The attempt at a solution

    I am not sure, but it seems to me that the two upper meshes are in parrallel, and the 3rd one is in series with the two upper. Please, tell me if I am wrong. I know how to solve the problem, I am just confused that the left side of the diagram is like in parrallel, and the right side seems like the lower mesh connects with another to in series. Is it correct?

    a.
    1st mesh:-
    =>22Ωi2 -14Ωi1 = 8V-12V
    =>14Ωi1 - 22Ωi2 = 4V -------------(i)
    2nd mesh:-
    14Ωi1 + 6Ω(i1+i2) + 6Ω(i1+i2) = 12V+18V
    =>26i1 + 12i2 = 30 V-----------(ii)
    By (i) x 12V + (ii) x 22Ω :- we have
    =>i1 = 0.96 amp
    Thus i2 = 0.43 amp
    b. V = i x R
    =>V = (i1+i2) x (R3 +R4)
    =>V = (0.96 A+ 0.43A) x (6Ω+6Ω)
    =>V = 16.68 Volt
     

    Attached Files:

  2. jcsd
  3. Jul 18, 2012 #2

    SammyS

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    I wouldn't describe the situation that way, in terms of series & parallel.

    --- but your equations are essentially correct.

    Hello Kate02. Welcome to PF !

    Multiply equation (i) by 12, not by 12 V. --- no units.

    Multiply equation (i) by 22, not by 22 Ω .

    Your answers for the currents are correct to two decimal places.



    When you found the voltage across AB, you forgot to include the 18 V battery.
     
  4. Jul 18, 2012 #3

    TSny

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    Hello. Note that you could redraw the circuit as shown below and it would be entirely equivalent to your circuit. If you think of it as made of three branches, the three branches are in parallel (but, of course, none of the individual resistors are in parallel).
     

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  5. Jul 19, 2012 #4
    Thank you, guys! I will solve it as you suggested, and we'll see if it works!

    So, I just added the 18 V battery to the equation for the part b) and got 34.68 V, right? I hope so. Correct me if I am wrong!
     
    Last edited: Jul 19, 2012
  6. Jul 20, 2012 #5

    SammyS

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    Not correct.

    The current through the two 6Ω resistors produces a voltage drop, right to left, which is the direction of i1 and i2 through them.
     
  7. Jul 20, 2012 #6
    Ah, now, I see! So, 16.68V-18V= -1.32V, right?
     
  8. Jul 20, 2012 #7

    SammyS

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    Yes, that's the voltage at point B relative to the voltage at point A .
     
  9. Jul 22, 2012 #8
    Great! Thanks a lot!
     
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