using nodal analysis and place a test voltage, V on the node which is on the 2Amp source
current into node = current out of node
2A = (V-12)/6 + V/10 and solve for V
once you had determined value of V , use voltage divider to obtain voltage drop on the 4ohm resistor (which is Vab)...
Eth is measured from voltage drop of the 6.8kohm resistor and the 6v source so discard the 1.2kohm resistor and RL
now your circuit has only two meshes instead of three so by kvl the equations below were derived. I1 mesh current on the left and I2 mesh current on the right
1. 7.8kI1 -...
To find Rth :
short out 9v voltage supply and open RL resistor
the independent 3Vx source remains.
Introduce voltage source Vth from RL terminal
and Isc flowing into the 4ohm resistor from Vth
using KVL with mesh currents both are clockwise,
The below equation is derived :
<<...
I is directly proportional to Vs
therefore I=BVs where B is a constant
the initial values were I=10mA when Vs=100V
10m=B(100)
B=(10m)/100
B=0.1m
thus I=(0.1m)Vs
Vs=25 I=(0.1m)*25 = 2.5mA
Vs=-12 I=(0.1m)*(-12)...