Thevenin's theorem with current and dependent voltage source.

Discussion in 'Homework Help' started by rdj, Mar 26, 2010.

  1. rdj

    Thread Starter New Member

    Mar 25, 2010
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    I like to find Eth, Rth and I at 6 ohm resistor...see attachment.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This is a homework help forum. You need to show some effort of your own before you get any help.
     
  3. rdj

    Thread Starter New Member

    Mar 25, 2010
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    0
    6 +(V1) -Eth +3(V1) = 0

    6 +4V1 -Eth = 0

    since 6 ohm resistor open circuited 3A will go V1, hence

    V1 = 3

    Eth = 6 + 4*3 = 18 volts

    solving for I short circuit:

    Isc = 3 - V1

    6 + V1 +3V1 = 0
    V1 = 1.5

    Isc = 1.5 A

    Rth = Eth/Isc = 18/1.5 = 12 ohms

    solving for I at 6 ohms:

    I = Eth/(Rth+6) = 18/(12+6) = 1 A

    Checking the answer:

    6 + (V1) - 6*I + 3(V1) = 0

    6 + 2 - 6*2 + 3*2 = 2,

    Can someone help me to correctly solve this by Thevenin's Theorem?

    Thanks!
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Right here is where you made your error. You have to add in the 3 amp current source. This will make Isc = 4.5 amps. I think the rest of it will work out if you do that.

     
  5. rdj

    Thread Starter New Member

    Mar 25, 2010
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    0
    The Electrician,

    Thanks!

    solving for I short circuit:

    Isc = 3 - V1

    6 + V1 +3V1 = 0, 4V1 = -6

    V1 = -1.5

    Isc = 3 - (-1.5) = 4.5

    Rth = Eth/Isc = 18/4.5 = 4 ohms

    solving for I at 6 ohms:


    I = Eth/(Rth+6) = 18/(4+6) = 1.8 A

    Checking the answer:

    V1 = 3 - I = 3 - 1.8 = 1.2 volts

    6 + (V1) - 6*I + 3(V1) = 0


    6 +(1.2) - (6*1.8) + (3)*(1.2) = 0,
     
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