Max Power Theorem

Discussion in 'Homework Help' started by sallamabirmail, Feb 10, 2009.

  1. sallamabirmail

    Thread Starter New Member

    Feb 10, 2009
    2
    0
    Can you help me I couldn't solve this problem.
    First If I try to find short circuit current then I find something like 1 amper.But how can I find open circuit voltage.By the way we have dependent source so we can't get rid of that source and easily find Rth right?.Can you help me to find this answer


    [​IMG]
    Sorry I forget that resistor value on Vx is 5 ohm and the other resistor is our RL and we are trying to find RL to find the max power of our circuit
     
  2. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Hi sallamabirmail,
    Let's mark the ends of the 5 ohm resistor as A and B.

    Now we open circuit the RL and now let the positive end of the independent source be at 0 v and negative be at -60v (i.e the bottom of Vx ,point B)

    The current (1 A)will flow only through the 5 ohm resistor (because of open circuit on the other side) producing a drop of 5 v.
    so the end A will be
    V at A = V at B + 5v = -55 Volts.
    Since V x = 5v ,the dependent source is 25 v.
    Now we know all voltages and applying KVL around that loop will give us the open circuit voltage, dividing which by the short circuit current (1 A)we will get the equivalent resistance as seen by the load resistor.
    Hope it helps
     
  3. sallamabirmail

    Thread Starter New Member

    Feb 10, 2009
    2
    0
    Sure It does ,thanks so the power on resistor 30 ohm is 7.5 watt right?
     
  4. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Yes thats what i get
     
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