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2i-6i1+4i2=0 3i+4i1-9i2=0 5i-3i1-5i2=0 check i1

  1. Sep 25, 2010 #1
    prob.jpg In this diagram(made at last) if i assume the curren direction as shown, I get 3 equations:
    2i-6i1+4i2=0
    3i+4i1-9i2=0
    5i-3i1-5i2=0
    check i1 is i subscript 1.(i dunno how to write subscript).
    if i put in matrix form( i want to solve via matrix method only).
    i am getting i=0 i1=0 i2=0, used cramer,s rule.
    also tried row echeleon form. but didn't get any solution.
    am i right in giving curret direction? is there any wrong convection? am i forgetting any rule.
    in book the author has taken x y z(x and y in leftmost two branch AND Z IN MIDDLE branch).
    he gets a good matrix with 0,0,2 in right hand side after "=" sign.(i m not writing those equations)
    tell me what is the wrong with me?plz neede within 1 day!!!!!!!!!!
     
  2. jcsd
  3. Sep 25, 2010 #2
    Re: Kvl

    I'm sorry but that diagram is hard to read. I can tell you though that the system of equations represent voltages in a loop that sum to zero. Your diagram shows a 2v battery and your system of equations do not show this voltage.

    That maybe one of the problems...
     
  4. Sep 25, 2010 #3
    Re: Kvl

    That means i should write
    equation=2 volts?
    But kvl says "the algebric sum of the products of currents and resistances in each of the conductor in any closed path(mesh) in a network plus the algebric sum of the emfs in that path is zero"
    taking this def. Along the rhombhus path. I don't get any emf. There's only one emf i.e outermost.
     
  5. Sep 25, 2010 #4

    vk6kro

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    Re: Kvl

    The voltages across either two of the outer series resistors equals the supply voltage.

    So, you would get I1 * 1 + (I1 - I2) * 2 = 2 ...........Where the final 2 is the supply voltage
     
  6. Sep 26, 2010 #5
    Re: Kvl

    Yes u have done right thing.
    I1*1-(I1-I2)=2. I UNDERSTAND.
    BUT WHEN WE USE KVL, THEN WE CHOOSE A LOOP.THEN WE TRAVEL AROUND THAT LOOP ADD THE VOLTAGE DROPS. WE TAKE PROPER SIGN CONVENTION. THAT IS IF WE TRAVEL IN DIRECTION OF ASSUMED CURRENT THEN WE TAKE VOLTAGE DROP AS NEGATIVE AND VICE VERSA. TAKING THESE THINGS AND THREE LOOPS I HAVE WRITTEN 3 EQUATIONS. EACH EQUALS TO 0. SO I DON'T GET ANY RESULT. WHAT WRONG HAVE I DONE?
     
  7. Sep 27, 2010 #6

    vk6kro

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    Science Advisor

    Re: Kvl

    You haven't done anything wrong, but the currents in the circuit must depend on the supplied voltage, so you have to include equations that have this voltage in them.

    There are two resistors at the top and two at the bottom that are across 2 volts. So you can make equations using that information.

    The total current from the 2 V supply in your circuit is 1.068 amps.
     
  8. Sep 28, 2010 #7
    Re: Kvl

    Fine. Thanks all.
     
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