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    Hello. I am strugling with iths problem. After applying KCL and solving for nodes, i get that the difference is 10V. I tried it countless times and different ways but still can not get the difference of 2V(as b part wants). Help?!:rolleyes:
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    (Original post by Patrick.Mccullen)


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    Hello. I am strugling with iths problem. After applying KCL and solving for nodes, i get that the difference is 10V. I tried it countless times and different ways but still can not get the difference of 2V(as b part wants). Help?!:rolleyes:
    The best you can do here is to show the calculation where you get 10V.
    There are a number of ways of setting up Kirchhoff's currents for this circuit, each will give a slightly different set of equations depending on how you set up the currents and their directions.
    They will all, of course, give the same answer. 2V.
    If you can show your working, it will be possible to see where you have either made a mistake in the setup, or in the solving the actual equations.
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    (Original post by Stonebridge)
    The best you can do here is to show the calculation where you get 10V.
    There are a number of ways of setting up Kirchhoff's currents for this circuit, each will give a slightly different set of equations depending on how you set up the currents and their directions.
    They will all, of course, give the same answer. 2V.
    If you can show your working, it will be possible to see where you have either made a mistake in the setup, or in the solving the actual equations.
    Ahhh ive got it!!! but now i am unsure about part c. this is my calculations:

    1. R(total)=(20+30)II(10+40)=25 ohms
    2. I(total)=50volts/25ohms=2amps
    3.(this is the hard part) Well because both of the branches have the same resistance of 50ohms i assume that current will split in 2. So 1 amp will go through 20ohm resistor and reach nod B and 1 amp will go through 10 ohm resistor reaching nod A. I get V(b)=20Volts V(a)=10Volts V(thevenin)=V(a)-V(b)=10Volts.

    4. R(thevenin) is the same as R(total) unless im not seeing something correctly.

    Are these calculations allright?
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    (Original post by Patrick.Mccullen)
    Ahhh ive got it!!! but now i am unsure about part c. this is my calculations:

    1. R(total)=(20+30)II(10+40)=25 ohms
    2. I(total)=50volts/25ohms=2amps
    3.(this is the hard part) Well because both of the branches have the same resistance of 50ohms i assume that current will split in 2. So 1 amp will go through 20ohm resistor and reach nod B and 1 amp will go through 10 ohm resistor reaching nod A. I get V(b)=20Volts V(a)=10Volts V(thevenin)=V(a)-V(b)=10Volts.

    4. R(thevenin) is the same as R(total) unless im not seeing something correctly.

    Are these calculations allright?
    The Thevenin voltage is correct. It's the voltage across AB when there is an open circuit there.
    The Thevenin resistance is the resistance you would see between A and B if the voltage source was short circuited. You usually need to redraw the circuit to make it easier to see what's going on.
    You can also find it from Rth = Vth/I where I is the current that flows in the circuit with A and B short circuited. That would not be 2A, as that is the current with AB open.
    Theory here
    http://hyperphysics.phy-astr.gsu.edu.../thevenin.html
 
 
 
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