Consider this Circuit which is to be converted to *Equivalent Thevenin's Circuit*.

[![Full Circuit][1]][1]



Now I tried Solving this using two methods and got contradictory result.
We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$: -

**Method -$1$**: - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$.
Hence $V_{Th}=80+30=110V$

>**Thevenin's  Circuit**

[![Thevenin's Circuit][2]][2]




**Method -$2$**: - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

>**Solution**

![Solution][3]

[![Solution][4]][4]

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

  [1]: https://i.sstatic.net/A2GABhN8.png
  [2]: https://i.sstatic.net/pEGkJmfg.jpg
  [3]: https://i.sstatic.net/ykA695j0.jpg
  [4]: https://i.sstatic.net/fz8dWUe6.jpg