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,
Bob
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,
Bob
