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Question

Five equal resistors each of resistance R are connected as shown in the Figure. A battery of voltage V is connected between A and B. The current flowing in AFCEB will be

A
VR
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B
V2R
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C
2VR
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D
3VR
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Solution

The correct option is B V2R

Rearranging the given circuit we get,


Since it is a balanced Wheatstone bridge, we can remove the middle resistance R. So network becomes

Here, Req for the branch AFCEB will be 2R.

iAFCEB=i1=V2R

Hence, option (b) is correct.
Why this question ?

Key concept:

Balancing condition of Wheatstone's
bridge is,

R1R2=R3R4




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