What is the value of the current (i) and what is the right way to do it?












4














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.






current circuit-analysis wheatstone-bridge nodal-analysis







share|improve this question









New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 3




    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/…
    – Feels awesome
    2 days ago








  • 2




    Didn't we just go through all this, like a day ago?
    – jonk
    2 days ago






  • 1




    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 .
    – Feels awesome
    2 days ago






  • 3




    Possible duplicate of What's wrong with this Wheatstone bridge?
    – fhlb
    2 days ago










  • ya I have posted it again to get better ans
    – Feels awesome
    2 days ago
















4














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.






current circuit-analysis wheatstone-bridge nodal-analysis







share|improve this question









New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    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/…
    – Feels awesome
    2 days ago








  • 2




    Didn't we just go through all this, like a day ago?
    – jonk
    2 days ago






  • 1




    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 .
    – Feels awesome
    2 days ago






  • 3




    Possible duplicate of What's wrong with this Wheatstone bridge?
    – fhlb
    2 days ago










  • ya I have posted it again to get better ans
    – Feels awesome
    2 days ago














4












4








4


1





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.






current circuit-analysis wheatstone-bridge nodal-analysis







share|improve this question









New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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.






current circuit-analysis wheatstone-bridge nodal-analysis




current circuit-analysis wheatstone-bridge nodal-analysis






share|improve this question









New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 2 days ago









Elliot Alderson

4,7641918




4,7641918






New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Feels awesome

294




294




New contributor




Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    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/…
    – Feels awesome
    2 days ago








  • 2




    Didn't we just go through all this, like a day ago?
    – jonk
    2 days ago






  • 1




    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 .
    – Feels awesome
    2 days ago






  • 3




    Possible duplicate of What's wrong with this Wheatstone bridge?
    – fhlb
    2 days ago










  • ya I have posted it again to get better ans
    – Feels awesome
    2 days ago














  • 3




    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/…
    – Feels awesome
    2 days ago








  • 2




    Didn't we just go through all this, like a day ago?
    – jonk
    2 days ago






  • 1




    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 .
    – Feels awesome
    2 days ago






  • 3




    Possible duplicate of What's wrong with this Wheatstone bridge?
    – fhlb
    2 days ago










  • ya I have posted it again to get better ans
    – Feels awesome
    2 days ago








3




3




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/…
– Feels awesome
2 days ago






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/…
– Feels awesome
2 days ago






2




2




Didn't we just go through all this, like a day ago?
– jonk
2 days ago




Didn't we just go through all this, like a day ago?
– jonk
2 days ago




1




1




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 .
– Feels awesome
2 days ago




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 .
– Feels awesome
2 days ago




3




3




Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
2 days ago




Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
2 days ago












ya I have posted it again to get better ans
– Feels awesome
2 days ago




ya I have posted it again to get better ans
– Feels awesome
2 days ago










3 Answers
3






active

oldest

votes


















4














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.






share|improve this answer























  • so u that mean re-configuring the circuit is the mistake.
    – Feels awesome
    2 days ago










  • 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.
    – Chu
    2 days ago












  • now it makes sense . 👍
    – Feels awesome
    2 days ago



















1














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



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.






share|improve this answer





















  • 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 ?
    – Feels awesome
    2 days ago












  • node C and node D dissappeared. so it's not the same circuit.
    – fhlb
    2 days ago










  • no the node c and D has not disappeared it is present in the last circuit
    – Feels awesome
    2 days ago










  • 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
    – fhlb
    2 days ago










  • thats fine but what is the error in my rearrangement . ?
    – Feels awesome
    2 days ago



















0














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.





schematic





simulate this circuit – Schematic created using CircuitLab



These are not the same circuits.






share|improve this answer























  • but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
    – Feels awesome
    2 days ago










  • 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
    – Tony EE rocketscientist
    2 days ago










  • and the upper circuit has the current of 3 amp in the bridge .
    – Feels awesome
    2 days ago










  • typo...........
    – Tony EE rocketscientist
    2 days ago






  • 1




    why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
    – Feels awesome
    2 days ago











Your Answer





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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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.






share|improve this answer























  • so u that mean re-configuring the circuit is the mistake.
    – Feels awesome
    2 days ago










  • 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.
    – Chu
    2 days ago












  • now it makes sense . 👍
    – Feels awesome
    2 days ago
















4














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.






share|improve this answer























  • so u that mean re-configuring the circuit is the mistake.
    – Feels awesome
    2 days ago










  • 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.
    – Chu
    2 days ago












  • now it makes sense . 👍
    – Feels awesome
    2 days ago














4












4








4






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.






share|improve this answer














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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Chu

5,1352611




5,1352611












  • so u that mean re-configuring the circuit is the mistake.
    – Feels awesome
    2 days ago










  • 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.
    – Chu
    2 days ago












  • now it makes sense . 👍
    – Feels awesome
    2 days ago


















  • so u that mean re-configuring the circuit is the mistake.
    – Feels awesome
    2 days ago










  • 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.
    – Chu
    2 days ago












  • now it makes sense . 👍
    – Feels awesome
    2 days ago
















so u that mean re-configuring the circuit is the mistake.
– Feels awesome
2 days ago




so u that mean re-configuring the circuit is the mistake.
– Feels awesome
2 days ago












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.
– Chu
2 days ago






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.
– Chu
2 days ago














now it makes sense . 👍
– Feels awesome
2 days ago




now it makes sense . 👍
– Feels awesome
2 days ago













1














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



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.






share|improve this answer





















  • 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 ?
    – Feels awesome
    2 days ago












  • node C and node D dissappeared. so it's not the same circuit.
    – fhlb
    2 days ago










  • no the node c and D has not disappeared it is present in the last circuit
    – Feels awesome
    2 days ago










  • 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
    – fhlb
    2 days ago










  • thats fine but what is the error in my rearrangement . ?
    – Feels awesome
    2 days ago
















1














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



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.






share|improve this answer





















  • 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 ?
    – Feels awesome
    2 days ago












  • node C and node D dissappeared. so it's not the same circuit.
    – fhlb
    2 days ago










  • no the node c and D has not disappeared it is present in the last circuit
    – Feels awesome
    2 days ago










  • 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
    – fhlb
    2 days ago










  • thats fine but what is the error in my rearrangement . ?
    – Feels awesome
    2 days ago














1












1








1






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



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.






share|improve this answer












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



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.







share|improve this answer












share|improve this answer



share|improve this answer










answered 2 days ago









fhlb

799721




799721












  • 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 ?
    – Feels awesome
    2 days ago












  • node C and node D dissappeared. so it's not the same circuit.
    – fhlb
    2 days ago










  • no the node c and D has not disappeared it is present in the last circuit
    – Feels awesome
    2 days ago










  • 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
    – fhlb
    2 days ago










  • thats fine but what is the error in my rearrangement . ?
    – Feels awesome
    2 days ago


















  • 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 ?
    – Feels awesome
    2 days ago












  • node C and node D dissappeared. so it's not the same circuit.
    – fhlb
    2 days ago










  • no the node c and D has not disappeared it is present in the last circuit
    – Feels awesome
    2 days ago










  • 2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
    – fhlb
    2 days ago










  • thats fine but what is the error in my rearrangement . ?
    – Feels awesome
    2 days ago
















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 ?
– Feels awesome
2 days ago






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 ?
– Feels awesome
2 days ago














node C and node D dissappeared. so it's not the same circuit.
– fhlb
2 days ago




node C and node D dissappeared. so it's not the same circuit.
– fhlb
2 days ago












no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
2 days ago




no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
2 days ago












2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
– fhlb
2 days ago




2 new nodes appeared but C and D are clearly note there. C by definition is the node joining R2,R4 and node D
– fhlb
2 days ago












thats fine but what is the error in my rearrangement . ?
– Feels awesome
2 days ago




thats fine but what is the error in my rearrangement . ?
– Feels awesome
2 days ago











0














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.





schematic





simulate this circuit – Schematic created using CircuitLab



These are not the same circuits.






share|improve this answer























  • but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
    – Feels awesome
    2 days ago










  • 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
    – Tony EE rocketscientist
    2 days ago










  • and the upper circuit has the current of 3 amp in the bridge .
    – Feels awesome
    2 days ago










  • typo...........
    – Tony EE rocketscientist
    2 days ago






  • 1




    why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
    – Feels awesome
    2 days ago
















0














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.





schematic





simulate this circuit – Schematic created using CircuitLab



These are not the same circuits.






share|improve this answer























  • but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
    – Feels awesome
    2 days ago










  • 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
    – Tony EE rocketscientist
    2 days ago










  • and the upper circuit has the current of 3 amp in the bridge .
    – Feels awesome
    2 days ago










  • typo...........
    – Tony EE rocketscientist
    2 days ago






  • 1




    why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
    – Feels awesome
    2 days ago














0












0








0






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.





schematic





simulate this circuit – Schematic created using CircuitLab



These are not the same circuits.






share|improve this answer














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.





schematic





simulate this circuit – Schematic created using CircuitLab



These are not the same circuits.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Tony EE rocketscientist

62k22193




62k22193












  • but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
    – Feels awesome
    2 days ago










  • 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
    – Tony EE rocketscientist
    2 days ago










  • and the upper circuit has the current of 3 amp in the bridge .
    – Feels awesome
    2 days ago










  • typo...........
    – Tony EE rocketscientist
    2 days ago






  • 1




    why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
    – Feels awesome
    2 days ago


















  • but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
    – Feels awesome
    2 days ago










  • 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
    – Tony EE rocketscientist
    2 days ago










  • and the upper circuit has the current of 3 amp in the bridge .
    – Feels awesome
    2 days ago










  • typo...........
    – Tony EE rocketscientist
    2 days ago






  • 1




    why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
    – Feels awesome
    2 days ago
















but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 days ago




but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 days ago












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
– Tony EE rocketscientist
2 days ago




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
– Tony EE rocketscientist
2 days ago












and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 days ago




and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 days ago












typo...........
– Tony EE rocketscientist
2 days ago




typo...........
– Tony EE rocketscientist
2 days ago




1




1




why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 days ago




why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 days ago










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Feels awesome is a new contributor. Be nice, and check out our Code of Conduct.












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