The answer is Vth=0.839 V and Rth=3.77 ohm but I can't get the right answer. Please give me the steps to find it.

what i did to find Rth: removed RL and 6V, 6 & 8 ohm are in parallel, which gives eq. eq is in series with 12 ohm, which gives eq2. eq2 is in parallel with 5, which gives Rth = 3.77 ohm

for Vth, I'm having difficulty with identifying what is in series/parallel. we will remove RL first, then find for Vth. 5 and 12 are in series (eq=17). 17 and 8 are in parallel which gives 5.44 ohm. Now after this, how is 5.44 ohm and 6 in parallel? i dont understand that. can someone explain. and afterwards, how do I find Vth value from all I've calculated? I can't understand that part. why are we going to do (5.44 | 6)*5/17 ??

• What is the voltage across the 5 ohm resistor when RL is removed? Oct 2, 2021 at 17:43
• Question has been re-opened, now that a solution attempt has been added. Oct 2, 2021 at 18:22
• 5.44R in series with 6R resistor.
– G36
Oct 2, 2021 at 18:29
• But you do not have to combine 6R resistor together with 5.44. You can use a voltage divider equation to find the voltage across 5.44R resistor (the voltage across 8R resistor and across 5R+12R ) V = 6V * 5.44/(6 + 5.44) = 2.85314685V. Now you can use the voltage divider equation again to find the voltage across 5 ohm resistor. Because now you know the voltage across 8 ohm resistor and this voltage must be the same across 17 ohm resistor.
– G36
Oct 2, 2021 at 18:39

For working out the Thevenin resistance, your approach was wrong.

Look at the following schematics which I believe accurately describe your approach to find the Thevenin resistance. With $$\R_{_\text{L}}\$$ disconnected and $$\V_1\$$ shorted, you transition from the left-side to the right-side, below:

simulate this circuit – Schematic created using CircuitLab

Here, you take $$\R_1\$$ in parallel with $$\R_2\$$, first. That combination will then be in series with $$\R_3\$$. And that combination will be in parallel with $$\R_4\$$.

So the Thevenin resistance, seen by $$\R_{_\text{L}}\$$, will be:

$$R_{_\text{TH}}=\left[\left(R_1\mid\mid R_2\right)+R_3\right]\mid\mid R_4$$

(You should be able to see that $$\R_4\$$ is directly in parallel with $$\R_{_\text{L}}\$$, anyway. So you should know, right away, that whatever answer you come up with -- that this answer must end with putting $$\R_4\$$ in parallel with the immediately prior result.)

For working out the Thevenin voltage, your approach that got $$\5.44\:\Omega\$$ is workable.

simulate this circuit

First, sum up $$\R_3\$$ and $$\R_4\$$, as you indicated. Then put that result in parallel with $$\R_2\$$. That will, in fact, result in $$\R_{_\text{A}}=\left(R_3+R_4\right)\,\mid\mid\, R_2=5.44\:\Omega\$$.

$$\R_{_\text{A}}\$$ now makes a voltage divider with $$\R_1\$$, with the resulting voltage being $$\V_{_\text{A}}=V_1\cdot\frac{R_{_\text{A}}}{R_1+R_{_\text{A}}}\$$.

You now just want the voltage across $$\R_4\$$, which will be less than $$\V_{_\text{A}}\$$. But $$\R_3\$$ and $$\R_4\$$ are just one more voltage divider, though you want the upper-half voltage (across $$\R_4\$$), so this will be $$\V_{_\text{TH}}=V_{_\text{A}}\cdot\frac{R_4}{R_3+R_4}\$$.

So the Thevenin voltage, seen by $$\R_{_\text{L}}\$$, will be:

\begin{align*} V_{_\text{TH}}&=V_1\cdot\frac{R_{_\text{A}}}{R_1+R_{_\text{A}}}\cdot\frac{R_4}{R_3+R_4} \\\\ &= V_1\cdot\frac{\left(R_3+R_4\right)\,\mid\mid\, R_2}{R_1+\left[\left(R_3+R_4\right)\,\mid\mid\, R_2\right]}\cdot\frac{R_4}{R_3+R_4} \end{align*}