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What exactly went wrong with the below method?

Here, in the same circuit I am getting a different value for the current in the mid wire only by rearranging the circuit.

Image 1

I was trying to solve this unbalanced Wheatstone bridge and found that the current (\$i\$) in the mid wire (in image 1) is 3 A. This result was achieved by using the node voltage method.

Then I rearranged the same circuit to the form shown in the second image.

Image 2

And at this point I flipped the right half of the circuit so as to obtain the equal resistors on the same side.

Image 3

Now I again rearranged the circuit as shown.

Image 4

Here, the final circuit is a balanced Wheatstone bridge and thus the current (\$i\$) must be 0 A.

So clearly something went wrong in there.

I posted a question before this one just to make sure that there is no mistake in rearranging the circuit in the way I have done.

The mid wire is a complete connected wire. The gap in the mid wire near the arrow was inevitable.


marked as duplicate by Voltage Spike, Finbarr, Dwayne Reid, Community Jan 3 at 14:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    \$\begingroup\$ this question was at first asked in physics stack exchange by me but it was unfortunately put on hold. link of first question -- physics.stackexchange.com/questions/450822/… \$\endgroup\$ – Feels awesome Dec 29 '18 at 7:14
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    \$\begingroup\$ Didn't we just go through all this, like a day ago? \$\endgroup\$ – jonk Dec 29 '18 at 8:54
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    \$\begingroup\$ that was a kind guy who posted my question again as the one that i posted was put on hold of some unknown reason . as the qs he posted has some errors I posted it again . \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:45
  • \$\begingroup\$ ya I have posted it again to get better ans \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:52
  • \$\begingroup\$ A Wheatstone bridge does not how the short in the middle it usually has a high impedance resistance from an amplifier \$\endgroup\$ – Voltage Spike Dec 30 '18 at 21:10

You start with a primitive circuit, and then apply a series of re-configurations, becoming progressively more sophisticated, until you arrive at the final level of sophistication - a single 4 ohm resistor and a source.

Along the way, the currents and voltages in/at the various conductors and nodes that you introduce/remove, change. It's not surprising. There are an infinite number of ways of transforming a single resistor into a bridge.

In 'simplifying' the original primitive bridge circuit you lose information that you cannot recover. You cannot rediscover the bridge if all you have to work on is a 36 volt source and a 4 ohm resistor.

  • \$\begingroup\$ so u that mean re-configuring the circuit is the mistake. \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:59
  • \$\begingroup\$ Not at all. It depends what you're doing it for. You lose the detail of the circuit when you simplify, and you can't go on a random path and expect to recreate it. \$\endgroup\$ – Chu Dec 29 '18 at 10:01
  • \$\begingroup\$ now it makes sense . 👍 \$\endgroup\$ – Feels awesome Dec 29 '18 at 10:04

Analyzing the circuit by using only Ohm's law and equivalent resistance we can calculate the current passing through each wire in the circuit:


It is clear that there's a 3A current going from node C to D.

Now if you switched R3 and R4 positions. then indeed no current will flow through the mid wire but the thing is, the current passing through the wire is not i. call it i2 as the circuit has indeed changed.

  • \$\begingroup\$ i agree with ur point . but we know that the first circuit(in my qs) is equivalent to the second circuit(in my qs) . so there is most probably no mistake the rearrangement .so where is the mistake ? \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:01
  • \$\begingroup\$ node C and node D dissappeared. so it's not the same circuit. \$\endgroup\$ – fhlb Dec 29 '18 at 9:12
  • \$\begingroup\$ no the node c and D has not disappeared it is present in the last circuit \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:41
  • \$\begingroup\$ 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D \$\endgroup\$ – fhlb Dec 29 '18 at 9:47
  • \$\begingroup\$ thats fine but what is the error in my rearrangement . ? \$\endgroup\$ – Feels awesome Dec 29 '18 at 9:49

THe upper cct has a bridge current of 12V/4Ω=3A but a total current of 36V /4.5Ω = 8A from (9//9=4.5Ω)

The lower circuit has a bridge current of 0A yet still a total current of 8A.


simulate this circuit – Schematic created using CircuitLab

These are not the same circuits.

  • \$\begingroup\$ but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible? \$\endgroup\$ – Feels awesome Dec 29 '18 at 7:48
  • \$\begingroup\$ THey total current is the same but not the same circuits with the R's swapped and bridged. Once side is 12V the other 24V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 29 '18 at 7:49
  • \$\begingroup\$ and the upper circuit has the current of 3 amp in the bridge . \$\endgroup\$ – Feels awesome Dec 29 '18 at 7:49
  • \$\begingroup\$ typo........... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 29 '18 at 7:50
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    \$\begingroup\$ why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes \$\endgroup\$ – Feels awesome Dec 29 '18 at 7:52

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