In this problem the first step is to find the equivalent resistor for
the Thévenin theorem. For this, we “turn off” the source, which means that the
voltage is zero.
We then have a 6Ω resistor in parallel with the 4Ω one. Which
gives R1 such that
1/R1 = 1/6 + 1/4 = 5/12 so R1 = 12/5 Ω
This R1 resistor is in series with the 2 Ω resistor, which gives
R2 such that
R2 = 12/5 + 2 = 22/5 Ω
And finally this R2 resistor is in parallel with the 20Ω one,
which gives Rt such that:
1/ Rt = 1/20 +5/22 = 122/440 =
61/220
Thus Rt = 220/61 Ω
To find the Thévenin’s voltage v t, we have to apply the
voltage divider (to the system with the 6Ω and the 4Ω resistors)
because the v t is equal to the open-circuit voltage.
We then have: v t = 4*v/(4+6) = 4*61/10 = 122/5 V