Circuit Theorem Analysis: Thevenin and Source Tranformation

Discussion in 'Homework Help' started by EE_Bob, Aug 2, 2009.

  1. EE_Bob

    Thread Starter Member

    Jul 21, 2009
    Hello, this is yet another source transformation problem from Alexander and Sadiku's "Fundamentals of Electric Circuits"

    It is simple to obtain the circuit in Fig 4.21A however combining the 2 ohm resistors in parallel is something which I am uncomfortable with. The voltage Vx is across the 2nd 2 ohm resistor as shown in Fig 4.21A. I realize that because both resistors are in parallel that they will share the same voltage. However I do not understand how the terminals for the voltage Vx may still be shown after the resistors are combined and the 3 amp current source is transferred to a voltage source. Please explain why this is allowed.

    The problem is shown below

    Also a much shorther question. The circuit below is a problem that involves Thevenin circuit analysis. Getting the Thevenin resistance is simple enough. I would like to know if the Thevenin voltage Vth as shown below is as easily calculated by assuming Vth=12(i1-i2), where Vth is across the the 12 ohm resistor. I assume that if a short were placed over the output terminals that the 1 ohm resistor would be in parallel with the 12 ohm allowing the voltage drop over the resistors to be equal. Nodal analysis might easily lend a hand however I just want to be sure that my reasoning as stated above is correct.

    Thanks again,