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 ??

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  • \$\begingroup\$ What is the voltage across the 5 ohm resistor when RL is removed? \$\endgroup\$
    – Andy aka
    Commented Oct 2, 2021 at 17:43
  • 1
    \$\begingroup\$ Question has been re-opened, now that a solution attempt has been added. \$\endgroup\$
    – SamGibson
    Commented Oct 2, 2021 at 18:22
  • \$\begingroup\$ 5.44R in series with 6R resistor. \$\endgroup\$
    – G36
    Commented Oct 2, 2021 at 18:29
  • \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Commented Oct 2, 2021 at 18:39

1 Answer 1


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:

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


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