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I am trying to find the thevenin resistance value and the Vth for maximum power transfer. I have tried to find Rth by calculating R3 and R2 as parallel resistance and same for R4 and R5 and then adding them with R1 but I am not sure if I have done it the right way. For Vth I have tried using KVL but I think I have calculated wrong current values using current divider rule. I calculated the total current by diving V1 and R1. I hope someone can help. I have attached the picture of the circuit.

Circuit diagram

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  • \$\begingroup\$ Since it's homework I'd suggest you to try a bit harder still. What values did you get? Why are they wrong? Do you have values for R1/2/3/4/5 or is it a theory exercise? Can you please draw your thevenin equivalents for resistor calculation? \$\endgroup\$ – Andrés Jun 14 '17 at 14:33
  • \$\begingroup\$ Let's check if you're on the right path at least. If you didn't have R1, only R2/3/4/5, what would your Rth be? \$\endgroup\$ – Andrés Jun 14 '17 at 14:42
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    \$\begingroup\$ Possible duplicate of Maximum power transfer for unknown resistance \$\endgroup\$ – Olin Lathrop Jun 22 '17 at 11:23
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I post this as an answer so I can plot a circuit equivalent.

schematic

simulate this circuit – Schematic created using CircuitLab

So. There we can see the same circuit (without the supply, we want to calculate R equivalent as viewed from your two nodes.

If R1 weren't there, your equivalent would be (R2+R4)//(R3+R5).

What happens if we add R1? Well, if the bridge is balanced, meaning both dividers provide the same voltage, R1 would be doing nothing. That's not the case here.

Let's make it the lazy way first. Wanna know the resistance? We can put a voltage source and check the current, or put a current source and check the voltage. Let's click simulate circuit and see.

Put 1A, we get 4.713V. That would mean 4.713ohms.

Put 1V we get 212.2mA, as expected 4.713 ohms.

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  • \$\begingroup\$ No, it's definitely not the sum of all R's. You need to solve the circuit. This is homework so I will not hand feed you the calculations. But you already have a simulation which gives you the result. You can only sum them when are in series, and there's nothing in between them. That would mean that with your pencil you can start at NODE1, and without lifting go through all the resistors only once and end at NODE2. Then they would all be in series and you can add them up \$\endgroup\$ – Andrés Jun 14 '17 at 15:00
  • \$\begingroup\$ Dude, this is not guessing. If the result is 4.713ohms.. does any of your guesses give you that value? Then it's not correct. I'm afraid you'll need to do the calculations, there's no easy way out. \$\endgroup\$ – Andrés Jun 14 '17 at 15:05
  • \$\begingroup\$ Then that means that's not the answer. \$\endgroup\$ – Andrés Jun 14 '17 at 15:09
  • \$\begingroup\$ Theoretically correct? No. You need to provide valid reasons to neglect R1, if you do. If R1 was a smaller value, then your result would be way off. \$\endgroup\$ – Andrés Jun 14 '17 at 15:22
  • \$\begingroup\$ The smaller R1, the more it affects the circuit. If R1 tend to infinity it's as if it was not there. The problem is that this is not about trying, but solving. If the simulator can solve it, you can. Put a 1V voltage source in between Node1 and Node2 and solve the circuit in your favorite manner. Then you can get the current going through it and calculate the equivalent resistance. \$\endgroup\$ – Andrés Jun 14 '17 at 15:27

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