what is the value of the current (i )and what is the right way to do it?
can someone please point out what exactly went wrong with the below method.
here,in a same circuit I am getting to different value for the current in the mid wire only by rearranging the circuit.
I was trying to solve this unbalanced Wheatstone bridge and found that the current (i) in the mid wire (in image 1) is 3 amp .This result was achieved by using node voltage method.
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.
now I again rearranged the circuit as shown.
Here, the final circuit is a balanced Wheatstone bridge and thus the current (i) must be 0 amp
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
asked 3 hours ago
Feels awesome
183
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Feels awesome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
can someone please point out what exactly went wrong with the below method.
here,in a same circuit I am getting to different value for the current in the mid wire only by rearranging the circuit.
I was trying to solve this unbalanced Wheatstone bridge and found that the current (i) in the mid wire (in image 1) is 3 amp .This result was achieved by using node voltage method.
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.
now I again rearranged the circuit as shown.
Here, the final circuit is a balanced Wheatstone bridge and thus the current (i) must be 0 amp
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
asked 3 hours ago
Feels awesome
183
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.
1
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
3 hours ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour 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
55 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago
add a comment |
can someone please point out what exactly went wrong with the below method.
here,in a same circuit I am getting to different value for the current in the mid wire only by rearranging the circuit.
I was trying to solve this unbalanced Wheatstone bridge and found that the current (i) in the mid wire (in image 1) is 3 amp .This result was achieved by using node voltage method.
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.
now I again rearranged the circuit as shown.
Here, the final circuit is a balanced Wheatstone bridge and thus the current (i) must be 0 amp
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
asked 3 hours ago
Feels awesome
183
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.
can someone please point out what exactly went wrong with the below method.
here,in a same circuit I am getting to different value for the current in the mid wire only by rearranging the circuit.
I was trying to solve this unbalanced Wheatstone bridge and found that the current (i) in the mid wire (in image 1) is 3 amp .This result was achieved by using node voltage method.
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.
now I again rearranged the circuit as shown.
Here, the final circuit is a balanced Wheatstone bridge and thus the current (i) must be 0 amp
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
asked 3 hours ago
Feels awesome
183
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 3 hours ago
Feels awesome
183
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 3 hours ago
Feels awesome
183
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 3 hours ago
Feels awesome
183
asked 3 hours ago
Feels awesome
183
183
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.
1
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
3 hours ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour 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
55 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago
add a comment |
1
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
3 hours ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour 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
55 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago
1
1
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
3 hours 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
3 hours ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour 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
55 mins 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
55 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago
add a comment |
3 Answers
3
active
oldest
votes
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, 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.
answered 44 mins ago
Chu
5,1052611
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
add a comment |
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.
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
|
show 9 more comments
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.
answered 1 hour ago
fhlb
789721
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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
|
show 2 more comments
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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, 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.
answered 44 mins ago
Chu
5,1052611
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
add a comment |
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, 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.
answered 44 mins ago
Chu
5,1052611
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
add a comment |
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, 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.
answered 44 mins ago
Chu
5,1052611
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, 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.
answered 44 mins ago
Chu
5,1052611
edited 41 mins ago
answered 44 mins ago
Chu
5,1052611
answered 44 mins ago
Chu
5,1052611
answered 44 mins ago
Chu
5,1052611
5,1052611
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
add a comment |
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins ago
so u that mean re-configuring the circuit is the mistake.
– Feels awesome
41 mins 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
39 mins 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
39 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
now it makes sense . 👍
– Feels awesome
36 mins ago
add a comment |
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.
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
|
show 9 more comments
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.
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
|
show 9 more comments
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.
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
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.
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
edited 2 hours ago
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
answered 2 hours ago
Tony EE rocketscientist
61.7k22193
61.7k22193
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
|
show 9 more comments
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours ago
but hey both are same circuits ,right ?.and we still get two different values of current hiw is it possible?
– Feels awesome
2 hours 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 hours 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 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
and the upper circuit has the current of 3 amp in the bridge .
– Feels awesome
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
typo...........
– Tony EE rocketscientist
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
why arent they equivalent circuits ? I think , I have only rearranged them without any mistakes
– Feels awesome
2 hours ago
|
show 9 more comments
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.
answered 1 hour ago
fhlb
789721
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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
|
show 2 more comments
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.
answered 1 hour ago
fhlb
789721
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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
|
show 2 more comments
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.
answered 1 hour ago
fhlb
789721
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.
answered 1 hour ago
fhlb
789721
answered 1 hour ago
fhlb
789721
answered 1 hour ago
fhlb
789721
answered 1 hour ago
fhlb
789721
789721
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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
|
show 2 more comments
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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins 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
1 hour 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
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
node C and node D dissappeared. so it's not the same circuit.
– fhlb
1 hour ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins ago
no the node c and D has not disappeared it is present in the last circuit
– Feels awesome
59 mins 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
53 mins 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
53 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
thats fine but what is the error in my rearrangement . ?
– Feels awesome
51 mins ago
|
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Feels awesome is a new contributor. Be nice, and check out our Code of Conduct.
Feels awesome is a new contributor. Be nice, and check out our Code of Conduct.
Feels awesome is a new contributor. Be nice, and check out our Code of Conduct.
Feels awesome is a new contributor. Be nice, and check out our Code of Conduct.
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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
3 hours ago
Didn't we just go through all this, like a day ago?
– jonk
1 hour 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
55 mins ago
Possible duplicate of What's wrong with this Wheatstone bridge?
– fhlb
51 mins ago
ya I have posted it again to get better ans
– Feels awesome
48 mins ago