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enter image description here Where does the Vth=2.571 come from? I get the resistance is (4+6)4/(10+4)+5. This is how Hayt's solved example pattern right before would look like if applied here:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Since the \$6\:\Omega\$ and \$4\:\Omega\$ resistors are in series, you can add them to get \$10\:\Omega\$ (eliminating one of them by shorting it.) Then you just have a resistor divider network with your voltage source which computes out a \$V_\text{TH}\$ and \$R_\text{TH}\$ result. You are then left with a very simple series resistance loop with \$7\:\Omega+R_\text{TH}\$ and your equivalent voltage source, \$V_\text{TH}\$. The current just "falls out" from there. \$\endgroup\$ – jonk May 27 '20 at 7:01
  • \$\begingroup\$ @jonk and what is that calculation which is different from the formula I wrote above. Just write the numbers and Ithink I can figure it out from there. \$\endgroup\$ – user5389726598465 May 27 '20 at 7:05
  • \$\begingroup\$ \$V_\text{TH}=9\:\text{V}\cdot\frac{4\:\Omega}{4\:\Omega+10\:\Omega}\$ and \$R_\text{TH}=\frac{4\:\Omega\cdot 10\:\Omega}{4\:\Omega+10\:\Omega}\$. So, \$I_\text{TOTAL}=\frac{V_\text{TH}}{R_\text{TH}+5\:\Omega+2\:\Omega}\$. It is that easy. \$\endgroup\$ – jonk May 27 '20 at 7:07
  • \$\begingroup\$ @jonk. so 9/14 represents the current in the loop and that same current can apply to the resistor load. \$\endgroup\$ – user5389726598465 May 27 '20 at 7:08
  • \$\begingroup\$ Once you Thevenize your resistor divider (after adding the 4 Ohm and 6 Ohm resistor, together), yeah. It's that easy. You have a Thevenin source voltage followed by a Thevenin series resistance followed by two more resistors, also in series as part of a circuit or loop. All done, then. \$\endgroup\$ – jonk May 27 '20 at 7:10
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I can't really reproduce what you were calculating, maybe you can explain. The steps you need to take to get the current through the 2\$\Omega\$ resistor

  1. Calculate the voltate drop across the 4\$\Omega\$ by merging \$(2+5)||4 = 2.55\Omega\$ which then form a voltage divider with the other two resistors and gives a voltage across the 4\$\Omega\$ of \$9V/12.55*2.55=1.82V\$
  2. Knowing the voltage across the 5\$\Omega\$ + 2\$\Omega\$ you can calculate the current \$I_{2\Omega}\$ as \$1.82V/7\Omega=261.24mA\$ (without the rounding you get the 260.8mA from your solution

The bad thing is that this has nothing to do with Thevenin. To solve it using Thevenin you would remove your load resistor from the circuit

schematic

simulate this circuit – Schematic created using CircuitLab

You can now calculate the output voltage or \$V_{th}\$ as \$9V/14\Omega*4\Omega=2.571V\$.

Then short circuit your voltage supply to calculate \$R_{th}\$ as $$ ((4\Omega+6\Omega) || 4\Omega) + 5\Omega = 7.8571\Omega $$

and finally your load current is \$V_{th}/(R_{th}+R_{load}=260.8mA\$

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  • \$\begingroup\$ yes, I think you're just simplifying the circuit. The book states to use Thevenin open circuit the 2ohm and short the voltage. Do you know how to calculate it from there? \$\endgroup\$ – user5389726598465 May 27 '20 at 5:59
  • \$\begingroup\$ I don't understand the 9/14 ohms*4. Can you explain that? \$\endgroup\$ – user5389726598465 May 27 '20 at 6:39
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    \$\begingroup\$ As R4 is open the 9V just get distributed among R1, R2 and R3 proportionally and the voltage on the output end of R4 equals the voltage across R2 \$\endgroup\$ – po.pe May 27 '20 at 6:45
  • \$\begingroup\$ But why use the current from the loop 9/14, that is a different branch. \$\endgroup\$ – user5389726598465 May 27 '20 at 6:56
  • \$\begingroup\$ There is no other current if R4 is open. Maybe you can read allaboutcircuits.com/textbook/direct-current/chpt-10/… \$\endgroup\$ – po.pe May 27 '20 at 7:51

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