I've been posed with with following question: I thought that using a thevenin replacement circuit would be best, because then i wouldn't have to do nodal analysis every single time for the circuit. However i've run into a slight hickup. I've calculated the Thevenin resistance quite easily to be 8ohms. However i am stuck on trying to figure out the Thevenin equivalent voltage. My working out is as followed. I calculated the current in each loop. These were obtained from the following 3 equations: -24 + 8(i1) + 8(i1-i2) = 0 i2 = -5 i3 = 8(i3-i2) + 8(i3) + 8 = 0 subbing i2 into them i get: i1 = -1A i2 = -5A i3= -3A This just means i've assumed the wrong current flow and in reality it is flowing the other way correct? -V1 = 4A(in direction of i2)(8ohm) = 24 , v1 = -24 -V2 = 2(in direction of i2)(8ohm) = 16, v2 = -16 -V1 + V(therevin) +V2 = 0 V1 - V2 = V(therevin) -24 +16 = -8V Which would imply that the terminals are the other way correct? i.e V2 = positive & v1 = negative ?? Could you please tell me where i went wrong. how i should be approaching it, and what the answer is, so that when i try it again i'll have something to compare it to. I'm absolutely stumped with this one Best Wishes, A very confused and troubled student
Hello Intrud3r, Seeing the circuit first i thought using superposition would prove lot more easier. why have you calculated v1 as -v1 = 4 x 8 instead of +v1(v1 is at a higher potential than the ground) and you have made a multiplication mistake there.
You got: i1 = -1A i2 = -5A i3= -3A But it looks like you got a multiplication (and sign) wrong. It should be: -V1 = -4A(in direction of i2)(8ohm) = -32, v1 = 32 So Vth is 48 volts.