Question

(a) In a meter bridge [Fig. 3.27], the balance point is found to be at39.5 cm from the end A, when the resistor Y is of 12.5 W.Determine the resistance of X. Why are the connections betweenresistors in a Wheatstone or meter bridge made of thick copperstrips? (b) Determine the balance point of the bridge above if X and Y areinterchanged.(c) What happens if the galvanometer and cell are interchanged atthe balance point of the bridge? Would the galvanometer showany current?

Answer 1

Given: The distance of balance point from end A is 39.5 cm and the resistance of the resistor Y is 12.5 Ω.

a)

Consider the given figure below.

The balance condition is given as,

R 1 R 2 = 100− l 1 l 1

Where, the distance of balance point from end A is l 1 and the resistance of resistor Y is R 2 and the resistance of resistor X is R 1 .

By substituting the given values in the above expression, we get

R 1 12.5 = 100−39.5 39.5 R 1 =1.532×12.5 R 1 =8.2 Ω

Thus, the resistance of resistor X is 8.2 Ω and to minimize the resistance the connection between resistors in a Wheatstone bridge or meter bridge is made of thick copper strips and that is not taken into consideration in the bridge formula.

b)

If X and Y are interchanged, then l 1 and 100− l 1 will get interchange.

The distance of balance point from point A is given as,

d=100− l 1 .

By substituting the given values in the above equations, we get

d=100−39.5 =60.5 cm

Thus, the distance of balance point from end A is 60.5 cm.

(c)

At the balance point of the bridge when the galvanometer and cell are interchanged then the galvanometer will show no deflection.

Thus, there will be no current flow through the galvanometer.