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The question asks to find the thevnin equivalent of the given network as viewed from a set of specific terminals. Now I have a slight doubt. All the questions I have done on thevnin are of two types. In the first one, a network is given and two terminals are kept open across Which any suitable load can be connected. Finding the thevnin equivalent of such a circuit is pretty straightforward. And second type, in which a network is given and voltage or power or current across a specific resistor is of interest. In such types also, what we have to do is to disconnect the resistor of interest and carry on the same steps as in first case. Here, I stumbled upon this question which seems to be a misfit of the two types I am acquainted with. Also removing the 50 ohms resistor from terminal xx' yields wrong answer. In the answer key, the 50 ohm resistor has been included and infact the voltage across it has taken as the thevnin equivalent voltage. I'm confused about what's wrong. enter image description here

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  • \$\begingroup\$ To find \$V_{th}\$ of this network, apply superposition: determine the voltage across x an x' (if this is the considered output) when the 88-V source is replaced by a short circuit while the 1-A source is alive. Then repeat the calculation when the 1-A source is open-circuited and the 88-V source is back. The sum of the answers is \$V_{th}\$. For \$R_{th}\$, open-circuit the 1-A source, replace the 88-V source by a short circuit and determine the resistance seen from x an x' (if this is the considered output). Keep series and || associations in the symbolic answer. Good luck! \$\endgroup\$ – Verbal Kint Jul 20 '17 at 12:13
  • \$\begingroup\$ And while doing so, do I have to remove the 50 ohms resistor from the xx' terminal(if this is the considered output terminal) and then using superposition find the thevnin equivalent voltage at open circuited xx' terminal or do I have to let the 50 ohm resistor remain there \$\endgroup\$ – YOGENDRA SINGH Jul 20 '17 at 12:20
  • \$\begingroup\$ If you remove the \$50\;\Omega\$ resistance, then put it back in || with the final result. This resistor is part of the circuit and must remain in place. \$\endgroup\$ – Verbal Kint Jul 20 '17 at 12:24
  • \$\begingroup\$ Exactly ! There is where from the doubt ensues. If I remove the 50 ohm resistor, the thevnin voltage at the open circuit xx' terminals will certainly differ from the value of the thevnin voltage when 50 ohms is there. \$\endgroup\$ – YOGENDRA SINGH Jul 20 '17 at 12:34
  • \$\begingroup\$ Look, this is quite simple actually: in your head, remove the 1-A generator and short the 80-V source. Now, imagine placing an ohm-meter across the \$50\;\Omega\$ resistance. What would you measure? In other words, if you inspect the circuit across the x and x' connections, you "see" the following resistive combination: 50 || 10 || (20 + 40) and this is your Thévenin resistance. \$\endgroup\$ – Verbal Kint Jul 20 '17 at 15:37
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The Thévenin voltage is obtained by calculating the voltage across terminals x and x'. There are many ways to get there. The easiest and fastest way is to apply superposition. You have two sources that you will alternately turn off during an intermediate calculation. First, turn the 88-V off. Turning a voltage source off or reducing its voltage to 0 V is identical to replacing it with a short circuit: replace the 88-V source by a wire and calculate the voltage between x and x' while the 1-A source biases the network. To get there, calculate the voltage at y and y' then involve a resistive divider to determine the voltage between x and x'. No line of algebra, just inspect the circuit. Once this is done, turn the 1-A current source off which is similar to reducing it to 0 A or open-circuiting it: remove it from the circuit and the 88-V source remains alone. Calculate the voltage across x and x', again using a resistive divider. Now, sum up the two voltages obtained during these intermediate steps, you have the voltage between x and x'. This is \$V_{th}\$. Without disclosing the calculation steps - this is your job : ) a quick sim will let you know if you have it right, \$V_{th}=69.3\;V\$:

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For the output resistance, you must turn all sources off to calculate it. This is one way of determining \$R_{th}\$ but there are other ways. This one is simple. You see that there is already a resistance across x and x'. This is the \$50-\Omega\$ resistor. Therefore, remove this resistance in your head and determine the resistance you "see" between the connecting terminals. This is \$R?\$ in the below sketch. Once \$R?\$ is determined (it is easy, it is a series-parallel association), the Thévenin resistance is simply \$R_{th}=50 || R?\$. If you bias x and x' with a 1-A test source \$I_T\$ as shown in the sketch, the voltage developed across the current source is the resistance you want (\$7.32\;\Omega\$).

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  • \$\begingroup\$ That's an answer I'd call "Go the extra mile". \$\endgroup\$ – Harry Svensson Jul 20 '17 at 16:48
  • \$\begingroup\$ But don't we remove the load resistor everytime we apply thevnin onto a network ? And then find the voltage across this open circuit and the equivalent resistance seen from the terminls ? \$\endgroup\$ – YOGENDRA SINGH Jul 20 '17 at 16:49
  • \$\begingroup\$ Ok, assume you have a black box feeding a load made of a resistance. If you want to characterize the equivalent Thévenin generator feeding this resistance, then yes, you have to calculate the unloaded voltage \$V_{th}\$ and \$R_{th}\$ without the load resistance. Then, the Thévenin equivalent network feeds the load. But in your case, there is no load as the \$50-\Omega\$ resistance is part of the circuit for which you want the Thévenin equivalent circuit. \$\endgroup\$ – Verbal Kint Jul 20 '17 at 17:34
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    \$\begingroup\$ @Harry Svensson, yes, this extra mile should be left for the student legs : ) \$\endgroup\$ – Verbal Kint Jul 20 '17 at 17:39
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According to the thevenin theorem, we have to calculate the open terminal volatage and the equivalant thevenin resistance. That means whatever element is there will not be included in the calculation .

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