In this problem the first step is to find the equivalent resistor for
the Thévenin theorem. For this we “turn off” the sources, which means that the
current is equal zero and the voltage is zero in the two sources.
We then have, on the left, a 3 Ω resistor in parallel with a 6
Ω one, which gives Re = 1/3 + 1/6 = 3/6 = 1/2 so Re = 2 Ω. This Re1
is in series with the other 3 Ω resistor so we finally get Rt
= 3 + 2 = 5 Ω
To find the Thévenin voltage vt we can transform the voltage
source on the left into a current one. The new current source will be of
Is = V/R = 12/3 = 4A, and the resistor Rs = 3 Ω unchanged.
We have
The two current sources give only one, of 1A (the currents are in
opposite directions) and the two resistors in parallel give Re = 2Ω as
stated in the first calculation.
We have thus:
This current source is equivalent to a voltage source such that the new
voltage is Thévenin’s voltage vt = R*I = 2*1 = 2V