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image ***1***

I was trying to solve this unbalanced Wheatstone bridge and found that the current $i$ in the mid wire is 9A .This result was achieved by using node voltage method.

image ***2***

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

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.

So clearly something went wrong in there.

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

It was answered by the user PhillipWood as being correct and was unfortunately put on hold.

So can someone please point out what exactly went wrong with the above method.

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    \$\begingroup\$ Wheatstone bridges are OPEN between the two arms, not shorted. \$\endgroup\$ – Scott Seidman Dec 28 '18 at 18:53
  • \$\begingroup\$ @ScottSeidman Yes, but since the current in the wire will be zero, it should not matter whether it is open or not. \$\endgroup\$ – harshit54 Dec 28 '18 at 18:56
  • \$\begingroup\$ But you change the resistors position in the bridge. So this is why the circuit has changed and you got a different result. \$\endgroup\$ – G36 Dec 28 '18 at 18:59
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    \$\begingroup\$ When you move the resistors around, you change what the whole circuit is. So even if it's "the same branch", it's a branch in a different circuit. So you can't expect the current to stay the same. (Note: in circuit theory it is the same circuit but the wire isn't actually a branch. Give the wire some nominal resistance \$r\$, maybe 1 mOhm, and now when you move the resistors around it isn't the same circuit any more) \$\endgroup\$ – The Photon Dec 28 '18 at 19:14
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    \$\begingroup\$ This question was also asked on Physics. \$\endgroup\$ – rob Dec 28 '18 at 19:48
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Below; just another way to see what's been written in comments.

schematic

simulate this circuit – Schematic created using CircuitLab

Your first claim in your question is wrong. The current \$i\$ in the top schematic you provide isn't \$9\:\text{A}\$. Instead, it is \$3\:\text{A}\$ (obviously.) You can easily see this by comparing the currents entering and leaving.

(Swapping \$R_1\$ with \$R_2\$ in the above, rightmost case would arrange things so that the currents flowing into and out of the left side would be balanced. Same with the currents flowing into and out of the right side. So in that case there would be no need for a current between the left and right sides because ... well, because you've changed the question being asked now.)


Now, let's adjust the above, rightmost schematic:

schematic

simulate this circuit

Again, without having to introduce any new resistors, you can see the obvious. The \$9\:\text{A}\$ flowing downward from above must obviously also flow along the wire connecting top to bottom. There's no other way for it.

(Swapping \$R_1\$ with \$R_2\$ in this last case makes no difference. The same current sum still must flow through them to ground. So in this case you can swap \$R_1\$ and \$R_2\$, or swap \$R_3\$ and \$R_4\$, without changing the current arriving and leaving through the narrow vertical wire connecting the pairs.)

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