# Thevenins theorem?

Discussion in 'Homework Help' started by Intrud3r, Mar 10, 2009.

1. ### Intrud3r Thread Starter New Member

Mar 9, 2009
2
0
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

2. ### vvkannan Active Member

Aug 9, 2008
138
11
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.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
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.