The resistances in the two arms of the meter bridge are 5 and R respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l1. The resistance R is [figure]
- 10
- 15
- 20
- 25
Answer
(2) 15
Solution
Initial balance:
5R = l1100 - l1 ...(i)
When R is shunted with an equal resistance, the resistance becomes R2:
5R/2 = 1.6 l1100 - 1.6 l1 ...(ii)
Dividing (ii) by (i):
2 = 1.6 (100 - l1)100 - 1.6 l1
200 - 3.2 l1 = 160 - 1.6 l1
l1 = 25 cm
R = 5 (100 - 25)25 = 15
Answer: (2)