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ABCD is a square where each side is a uniform wire of resistance 1Ω  . A point E lies on CD such that if a uniform wire of resistance 1Ω  is connected across AE and constant potential difference is applied across A and C then B and E are equipotential. Then, the ratio of lengths


  1. CEED=2
  2. CEED=12
  3. CEED=1
  4. CEED=2


Solution

The correct option is A CEED=2
We can rearrange the circuit as, with CE = 1–x and ED = x

Which is a balanced wheat stance network since B and E are at the same potential. Thus,
2+x1+x=11xx22x1=0
x=(21)Ω
CEDE=1xx=2221=2

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