Five equal resistaces each of resistance r are connected as shown in figure. A battery of V volt is connected between A and B. The current flowing in AFCEB will be
A
V/2r
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B
V/r
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C
3V/r
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D
2V/r
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Solution
The correct option is AV/2r A balanced wheatstone's bridge exist between A and B.
So resistance CD is no use because no current will flow.
Now as branch FD-DE and FC-CE are in series so these will become 2r and 2r.
As both of these(FDE-FCE) are in parallel, then the equivalent resistance will be 1Req=12r+12r Req=r
current throuth circuitV=IR I=V/Req=V/r
current through AFCEB=V/2r