circuit Kirchoff voltage law

Discussion in 'Homework Help' started by fran1942, May 12, 2011.

  1. fran1942

    Thread Starter Member

    Jul 26, 2010
    Hello, I have this circuit for which I have solved the current for the two loops indicated.
    I am now trying to understand how Kirchoff's voltage law applies to the right hand loop.

    eg. It starts off with 15V, then drops by 9V (9 ohm * 1A) (due to the parallel and series resistor combination represented by my diagram B). That brings it to 6V.
    But I am now struggling to understand how the remaining 6V is dropped.
    I have attempted to do the remaining drops on the lower right hand side of the diagram.
    I think it is like this:
    6v dropped by 1A * 2 ohm = 2V (6-2=4), and then (from loop 1) 2A * 2 Ohm = 4V (4-4=0 ?

    Can anyone please confirm if I am correct here.

    Thanks kindly for any help.
    • diag.jpg
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  2. jegues

    Well-Known Member

    Sep 13, 2010
    Do a KCL at the top node, the current through the branch with the 15V source will be 3amps.

    3A * 2Ω = 6V, is that the 6V you are looking for?
    fran1942 likes this.