Find the Thevenin equivalent

NOTE: The easiest method to solve this is almost certainly source transforms and resistor combinations. I will work it that way first, but also show the more general solution using v_oc and i_sc.

First, transform the 60 V source and the 10 W resistor.

Next, combine the 10 W and 40 W resistors in parallel.

Now transform the 6 A source and 8 W resistor.

Finally, combining the two resistors in series yields the Thevenin equivalent circuit.

First find v_oc

First note that since there is no connection to the right side of the 8 W resistor, there is no current through it, thus no voltage across it. An immediate implication of this is that v_oc equals the voltage across the 40 W resistor. Since the 10 W and 40 W resistors are effectively in series (no current through the 8 W thus the current is the same through the 10 W and 40 W resistors) we can use voltage division to find the voltage across the 40 W resistor (+ reference at top), thus v_oc.

Thus we have verified that V_Th = 48 V, since v_oc = V_Th.

Next, find i_sc. I will use mesh analysis.

Mesh a:

10 i_a + 40 (i_a - i_b) - 60 = 0

50 i_a - 40 i_b = 60 (equation 1)

Mesh b:

8 i_b + 40 (i_b - i_a) = 0

- 40 i_a + 48 i_b = 0 (equation 2)

Solving the two equations yields

i_a = 3.6 A

i_b = 3 A

But

i_sc = i_b = 3 A

so now we can calculate R_Th

R_Th = v_oc / i_sc = 48 / 3 = 16 W

Thus verifying R_Th.