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
- A(100 n^2 n^2+4 )
- B(100 n^2 n^2+1 )
- C(400 1 n^2+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