A meter bridge with two resistances R1 and R2 as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 resistance is connected in parallel to R2. The values of resistances R1 and R2 are . [figure]
- R2 = 8 , R1 = 163
- R2 = 12 , R1 = 123
- R2 = 4 , R1 = 43
- R2 = 16 , R1 = 163
Answer
(1) R2 = 8 , R1 = 163
Solution
Point P is the end of the bridge wire on the side of R1. At balance: R1R2 = l100 - l
First balance (l = 40 cm):
R1R2 = 4060 R1 = 23R2 ...(i)
With 16 in parallel to R2: R2' = 16R216 + R2
Second balance (l = 50 cm): R1 = R2'
23R2 = 16R216 + R2
2(16 + R2) = 48
R2 = 8
From (i): R1 = 23 8 = 163
Answer: (1)