Question

in a meter bridge balance point is found at a distance l1 with resistance R and S. when an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance l2. Find an expression for X in terms of l1, l2 and S.

Answer 1

FOR FIRST
R/S = L1/100-L1             [1]

WHEN X IS CONNECTED
R/SX/S+X = L2/100-L2           [2]

ON DIVIDING 1 AND 2
S+X/X  =  L1 ( 100-L2)/L2 (100-L1)

WE CAN CALCULATE X FROM ABOVE EQ

Answer 2

Answer:

Initially for R and S the balanced length is l1,

R/S = l1/(100-l1).......(1)

When X is connected in parallel,

S' = SX/S+X

Balance point is at l2,

R/S' = l2/100-l2

R(S+X)/SX=l2/100-l2.....(2)

Dividing equation 2 by 1 we get,

S+X/X = l2(100-l1)/l1(100-l2)

S/X + X/X = l2(100-l1)/l1(100-l2)

S/X + 1 = l2(100-l1)/l1(100-l2

S/X = [ l2(100-l1)/l1(100-l2) ] - 1

100l2 - 100l1/l1(100-l2) = S/X

X = l1(100-l2)S/100(l2-l1)