Current Relationship

Discussion in 'Homework Help' started by tquiva, Dec 16, 2010.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    Problem:
    [​IMG]

    I first began by denoting the very top middle node as A.
    At node A, I applied KCL:

    Σi = I - I_R - I_L
    => I = I_R + I_L

    So does this mean that I_R = I_L ? And the Answer is i?
     
  2. tyblu

    Member

    Nov 29, 2010
    199
    16
    How did you get I_R = I_L from I = I_R + I_L? You need to define I_R and I_L in terms of R using Ohm's law, V = I \times R. Just pretend the bottom terminal is ground (0V).
     
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  3. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    With the top node as A and bottom node a B, I = I_R + I_L simplifies to:
    I = (V_A - V_B) / (2R) + (V_A - V_B) / (R)
    = (V_A- V_B + 2V_A - 2V_B) / (2R)
    = 3V_A - 3V_B / (2R)

    Where would I go from here?
     
  4. tquiva

    Thread Starter Member

    Oct 19, 2010
    176
    1
    I also tried current division:

    I_R = [(2/3)R]/[2R] * I_total = V_AB / (2R)

    I_L = [(2/3)R]/[R] * I_total= V_AB / R

    So I_R > I_L

    Is this correct?
     
  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,157
    I_L > I_R

    It's a parallel circuit with three equal R's. Two of which are in one leg. I_L is .667 of I leaving I_R at .333 of I.
     
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  6. tyblu

    Member

    Nov 29, 2010
    199
    16
    I_R = \frac{V_A-V_B}{2R}\\<br />
I_L = \frac{V_A-V_B}{R}\\<br />
\frac{I_L}{I_R} = \frac{\left(V_A-V_B\right)/R}{\left(V_A-V_B\right)/2R}<br />
 = \frac{2R}{R} = 2<br />
    I_L is 2x larger than I_R.

    Another way to look at this is that current is inversely proportional to resistance, or I \propto \frac{1}{R}. This gives:
    I_L \propto 1/R \\<br />
I_R \propto 1/2R \\<br />
\frac{I_L}{I_R} \propto 2R/R = 2<br />
     
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