Problem 5.5-1

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 R_t = 3 + 2 = 5 Ω

To find the Thévenin voltage v_t 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 v_t = R*I = 2*1 = 2V