Loading...

Saturday, December 13, 2008

Q) Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by 10%, by what percent does the balance-point shift (and in which direction) ?

A) Since a Metre-Bridge has wire of 100 cm, so if the 2 resistances are equal, the balance point is at 50 cm.

Now if the left resistance is increased by 10%, it becomes 1.1 times its original value.
So balance point L :
1.1 R/ R = L / (100 - L)
110 - 1.1 L = L
110 = 2.1 L
L = 52.4 cm approx.

So the balance point L has shifted right by 2.4 cm
So in terms of %, it has shifted right by (2.4 / 50 x 100) %
= 4.8 % to the right

________________________________________________________

1 comment:

chandu sekhar said...

How it becomes 1.1 times ?