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This question is an exact duplicate of:

schematic

simulate this circuit – Schematic created using CircuitLab

Hi,

I am trying to find out what should be the value of resistor connected across node 'ab' for maximum power transfer. So far I have tried working out by hand and I don't think I have got the right values. I have shorted out voltage source to find thevenin resistance and considered R1 parallel with (R2+R3) and (R4+R5). Can anyone please help me to find out equivalent thevenin circuit.

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marked as duplicate by PlasmaHH, Olin Lathrop, JRE, Voltage Spike, Chris Stratton Jun 22 '17 at 17:02

This question was marked as an exact duplicate of an existing question.

  • \$\begingroup\$ @Andyaka Yes it is, but I never got to any final conclusion so I posted the question again! \$\endgroup\$ – Hamza Hussain Jun 22 '17 at 10:21
  • \$\begingroup\$ @HamzaHussain Do not cross post, ever. It angers the internets. If there is a problem with the first question then edit it. Its your job to write a good question that conforms the standards of this site. \$\endgroup\$ – Voltage Spike Jun 22 '17 at 15:54
  • \$\begingroup\$ @laptop2d I did post it in a good format. But if people like you don't want to answer then maybe they shouldn't bother posting a comment either. \$\endgroup\$ – Hamza Hussain Jun 22 '17 at 15:58
  • \$\begingroup\$ @HamzaHussain I'm commenting because you cross posted, it is better to edit the first question. I should bother because if I don't then I'll vote to close both questions. \$\endgroup\$ – Voltage Spike Jun 22 '17 at 16:04
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    \$\begingroup\$ @HamzaHussain I don't take the time to help people that don't read the rules and guidelines of the site. You are the one coming here for help, the least you could do is have some respect for the volunteers on this site and the SE.EE community \$\endgroup\$ – Voltage Spike Jun 22 '17 at 16:11
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In order to find out Thevenize circuit, you want to do following things.

  1. Temporary remove the load resistance(which is temoved here) whose current is required.
  2. Find out the open circuit voltage Voc which appears across the 2 terminals from where resistance has been removed. It is called Thevenin voltage Ath.
  3. Compute the Thevenin resistance, Rth of whole network as looked into from these 2 terminals(a and b) after all voltage sources have been removed leaving behind their internal resistances(if any) and current sources have been replaced by open circuit ie. infinite resistance.
  4. Replace the entire network by a single Thevenin source, whose voltage is Vth or Voc and whose internal resistance is Rth or Ri.
  5. Connect load resistance back to its terminals from where it was removed.
  6. Finally calculate current flowing through load resistance Rl using equation, I= Vth/(Rth+Rl) or I= Voc/(Ri+Rl).

Maximum power transfer

According to maximum power transfer theorem, the maximum power Pmax=((V^2)/4Rl)=((V^2)/4Ri)

Or

We can simply say that when internal resistance of source,Ri= load resistance Rl, then the maximum power transfer happens...

According to me you want to put Thevenin's resistance, Rth across ab. Since the question is not providing information about internal resistance of V1, we can consider internal resistance zero while Thevenizing circuit (Some articles also saying we can make this consideration theoretically). But the system do not exists practically since there is no voltage source with zero internal resistance.

Sorry for typos. This answer may be false. Please critically think.

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  • \$\begingroup\$ Thank you @Arun, I know the steps but the problem is that the circuit looks very complicated and it is very hard to identify which resistor is parallel and which one is in series to each other after shorting out the voltage source. Another issue is find Vth the thevenin voltage across ab as I am finding it hard to understand how and where to apply KVL in this circuit. \$\endgroup\$ – Hamza Hussain Jun 22 '17 at 14:41
  • \$\begingroup\$ @Hazama Hussain After short circuit V1, R1, (R2+R3) and (R4+R5) becomes parallel as you mentioned in question. Since in this question V1 has no internal resistance(since it is not given in question, consider it as zero). We know that, maximum power transfer occurred at Rl(resistance between a and b) becomes internal resistance of V1. That is resistance between a and b also zero or there is only a short circuit with a thick wire between a and b(but practically there is no voltage source can have zero internal resistance. But theoretically it is possible). I don't know it is correct. \$\endgroup\$ – Arun Jun 22 '17 at 15:07
  • \$\begingroup\$ Arun, you should BOLD-TYPE the conclusion \$R_i = R_l\$. This is so fundamental it is worth to memorize it in a brain circuit of its own. \$\endgroup\$ – Janka Jun 22 '17 at 15:34
  • \$\begingroup\$ @Janka I have no idea how to conclude. But I concluded as above. I don't think my conclusion is completely correct. \$\endgroup\$ – Arun Jun 22 '17 at 17:02
  • \$\begingroup\$ This answer started out OK, but then turned wrong at the end. No, the maximum power transfer is NOT with a short. Think about it. By definition, the short has no voltage across it, so dissipates no power regardless of how much current. Conversely, a open dissipates no power since the current is always 0. The maximum power into a load is when that load has the same impedance as the source it is connected to. V1 may have 0 impedance, but the impedance driving A-B is certainly not 0, due to the resistors in the circuit. \$\endgroup\$ – Olin Lathrop Jun 22 '17 at 17:07
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Try a WYE-DELTA transformation for \$R_1\$,\$R_2\$ and \$R_3\$ once you short your source. This should allow you to reconfigure and collapse the circuit to give you a value for \$R_{TH}\$. https://en.wikipedia.org/wiki/Y-Δ_transform

For \$V_{TH}\$ (Open circuit voltage), find the voltages at point \$a\$ and \$b\$ using your voltage divider rules. And use the one you used as your reference point in the \$R_{TH}\$ calculation as your point of low potential, subtracting it from the other.

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One way to solve this is to reduce the output from A-B to a Thevenin source. Once you've done that, the load resistance for maximum output power is just the Thevenin resistance.

There are various ways to reduce this network to a Thevenin source. You could crank away at it, incorporating one more resistor each step. Or you can analyze the circuit you have to find the open circuit voltage and the short circuit current. That gives the you the Thevenin parameters almost directly. Only a single divide needed.

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  • \$\begingroup\$ Whoever downvoted this, please explain what you think is wrong. \$\endgroup\$ – Olin Lathrop Jun 22 '17 at 17:04

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