Thevenins Theoram With 5 Resistors Im Stuck!!

Discussion in 'Homework Help' started by donniemateno, Dec 14, 2009.

  1. donniemateno

    Thread Starter New Member

    Dec 14, 2009
    1
    0
    wondering if any one could give me a hand. i been set this circuit for homework and i have to find the current across Rl. im getting 0.00277 amps as a answer and it doesnt seem right to me. im getting abit thrown off by the 2 resistors in series (R2 and R3) after ive removed rl. i dont get how im meant to work out the voltage that i needed to work out the final current. any help or step by step kinda help for me would b much appriciated. this is for my motor vehicle degree homework
     
  2. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Why are you thrown off by R2 and R3? They are simply series resistors which you can add up. Thus, you can replace both those resistors with a single resistor which has a value of 69Ω (22Ω + 47Ω).

    Austin
     
  3. Nanophotonics

    Active Member

    Apr 2, 2009
    365
    3
    Remove RL, calculate the open circuit voltage across AB, then short the supply voltage and calculate the equivalent resistance looking into the circuit at AB. Get the Thevenin's equivalent circuit and re-insert RL as in a series circuit, find the current using ohm's law.
     
  4. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    I'm not sure this is right, I am only learning this stuff at the moment, but here is my working.

    The voltage at the node above R2 and R3 = Thevenin voltage because no current flows through R4. By KCL at that node: (10-V1)/100 = V1/69

    This solves to V1 and therefore Vth = 4.08

    To find Rth, I always tend to join the output terminals so R4 is in parallel with the 69ohm combination. Then add the other resistor in parallel. I'm not sure if this is correct, i.e the 69ohm should be in parallel with R1, and then add R4 in parallel. Fortunately they are both the same in this example!

    (100//69)//100

    I evaluate this to around 29ohm.

    Therefore I make hte current through the load 0.003965A.

    Different to yours, but I'm not 100% sure I am right.
     
  5. Nanophotonics

    Active Member

    Apr 2, 2009
    365
    3
    You got the Thevenin's voltage right, but not the resistance. You're probably confusing with Norton's theorem maybe.

    In Thevenin's theorem, you short circuit the voltage source. That gives you R1 || (R2+R3), and the whole in series with R4. This is the Thevenin's equivalent resistance. Note that you're looking into the circuit at the terminal AB.
     
  6. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    Great, thanks for clearing that up.
     
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