DC circuits (how to find a current through resistor using current divider?)

Discussion in 'Homework Help' started by solaris055, Aug 8, 2010.

  1. solaris055

    Thread Starter New Member

    Jul 20, 2010
    2
    0
    Can you help me with this problem ?
    i tried to solve it with superposition and it goes fine with calculating I7 powered only by E1 or E6, but when it comes to Is i am confused with finding the other currents that are used to calculate the voltage from A to B ( i tried using thevenin theorem) can anyone help me please????
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Well first of all the current divider rule is similar to the voltage divider, except that the numerator is the branch resistance that you're not interested in.

    That is, if you have R1 parallel with R2 being fed by Isrc, the current through R1 will be:
    Isrc * R2 / (R1 + R2)

    Converting Is into a voltage source doesn't work because R2 is in series with it. That gives you an infinite Thevenin resistance.

    Superposition does work for this, can you show your steps?
     
  3. solaris055

    Thread Starter New Member

    Jul 20, 2010
    2
    0
    powered by E1:

    R(4,5)=R4*R5/(R4+R5)

    I'7= V(AB)/(R(thevenin)+R7)

    R(thevenin)=((R3+R1)(R(4,5)+R6))/(R3+R1+R(4,5)+R6)

    V(AB)=E1-I1(R3+R4); I1=E1/(R3+R1+R(4,5)+R6)

    this is how i solved one part of the problem is it nearly accurate?
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Your notation is confusing.
    You say V(AB) when you mean the Thevenin voltage with R7 being open.
    That is not V(AB) as seen on the schematic, you need to indicate it somehow. Personally I would just call it Vthevenin, but something like V(AB)open would also work.

    Your equations for V(AB) and Rthevenin are correct.
    You could have saved yourself some time by applying the voltage divider rule to find V(AB).

    It's V(AB) = E1 * (R(4,5)+R6) / (R3+R1+R(4,5)+R6)
     
    solaris055 likes this.
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