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AP EAMCET201922 Apr 2019Morning ShiftPhysicsCurrent ElectricityActual

(n ) identical resistance are taken in which ( n 2 ) resistors are joined in series in the left gap and the remaining ( n 2 ) resistances are joined in parallel in the right gap of a metre bridge. Balancing length in ( cm ) is

Options

  1. A(100 n^2 n^2+4 )
  2. B(100 n^2 n^2+1 )
  3. C(400 1 n^2+4 )
  4. D(400 1 n^2+1 )

Correct answer

A. (100 n^2 n^2+4 )

Step-by-step solution

Meter bridge is shown in the figure below, When ( n 2 ) resistances are joined in series in left gap each of resistance (R₁ ), then equivalent resistance in left gap. (R= R₁ 2 + R₁ 2 + R₁ 2 + n 2 times = R₁ n 2 ) When ( n 2 ) resistors are joined in parallel in right gap, then the equivalent resistance in right gap. ( aligned 1 S & = 1 R₁ + 1 R₁ + 1 R₁ + . n 2 times = n 2 R₁ S & = 2 R₁ n aligned ) If (l ) be the balancing length in the meter bridge wire, then ( aligned & R S = l 100-l R₁ n 2 2 R₁ n = l 100-l & n^2

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