I have found the Rth=10 ohms, However i can't find the thevenin's voltage(between a &b)
Well, we are trying to analyze the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\text{I}_2+\text{I}_7\\
\\
\text{I}_2=\text{I}_4+\text{I}_6+\text{I}_8\\
\\
\text{I}_5=\text{I}_4+\text{I}_6+\text{I}_7\\
\\
\text{I}_1=\text{I}_5+\text{I}_9\\
\\
\text{I}_3=\text{I}_\text{k}+\text{I}_9\\
\\
\text{I}_3=\text{I}_\text{k}+\text{I}_8
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\
\\
\text{I}_5=\frac{\text{V}_3}{\text{R}_5}\\
\\
\text{I}_6=\frac{\text{V}_2-\text{V}_4}{\text{R}_6}\\
\\
\text{I}_7=\frac{\text{V}_1-\text{V}_3}{\text{R}_7}
\end{cases}\tag2
$$
And we also know that \$\text{V}_3-\text{V}_4=\text{V}_\text{p}\$.
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$
\begin{cases}
\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_7}\\
\\
\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\frac{\text{V}_2-\text{V}_4}{\text{R}_6}+\text{I}_8\\
\\
\frac{\text{V}_3}{\text{R}_5}=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\frac{\text{V}_2-\text{V}_4}{\text{R}_6}+\frac{\text{V}_1-\text{V}_3}{\text{R}_7}\\
\\
\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_3}{\text{R}_5}+\text{I}_9\\
\\
\frac{\text{V}_2}{\text{R}_3}=\text{I}_\text{k}+\text{I}_9\\
\\
\frac{\text{V}_2}{\text{R}_3}=\text{I}_\text{k}+\text{I}_8
\end{cases}\tag3
$$
Now, we can solve for using your values:

$$\text{V}_\text{th}=\lim_{\text{R}_7\to\infty}\left(\text{V}_1-\text{V}_3\right)=10\space\text{V}\tag4$$
$$\text{I}_\text{th}=\lim_{\text{R}_7\to0}\text{I}_7=1\space\text{A}\tag5$$
$$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=10\space\Omega\tag6$$

Where I used the following Mathematica-codes:
In[1]:=Vi = 30;
Vp = 20;
Ik = 5;
R1 = 10;
R2 = 10;
R3 = 10;
R4 = 20;
R5 = 10;
R6 = 20;
FullSimplify[
 Solve[{I1 == I2 + I7, I2 == I4 + I6 + I8, I5 == I4 + I6 + I7, 
   I1 == I5 + I9, I3 == Ik + I9, I3 == Ik + I8, I1 == (Vi - V1)/R1, 
   I2 == (V1 - V2)/R2, I3 == V2/R3, I4 == (V2 - V3)/R4, I5 == V3/R5, 
   I6 == (V2 - V4)/R6, I7 == (V1 - V3)/R7, V3 - V4 == Vp}, {I1, I2, 
   I3, I4, I5, I6, I7, I8, I9, V1, V2, V3, V4}]]

Out[1]={{I1 -&gt; 5/(10 + R7), I2 -&gt; -(5/(10 + R7)), I3 -&gt; 3, 
  I4 -&gt; (5 + R7)/(20 + 2 R7), I5 -&gt; 2 + 5/(10 + R7), 
  I6 -&gt; (25 + 3 R7)/(20 + 2 R7), I7 -&gt; 10/(10 + R7), I8 -&gt; -2, 
  I9 -&gt; -2, V1 -&gt; 30 - 50/(10 + R7), V2 -&gt; 30, 
  V3 -&gt; 20 + 50/(10 + R7), V4 -&gt; 50/(10 + R7)}}

In[2]:=FullSimplify[
 Limit[(30 - 50/(10 + R7)) - (20 + 50/(10 + R7)), R7 -&gt; Infinity]]

Out[2]=10

In[3]:=FullSimplify[Limit[10/(10 + R7), R7 -&gt; 0]]

Out[3]=1

In[4]:=FullSimplify[%2/%3]

Out[4]=10