**(a)** In a metre bridge [shown in the figure], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

**(b)** Determine the balance point of the bridge above if X and Y are interchanged.

**(c)** What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

#### Solution

A metre bridge with resistors X and Y is represented in the given figure.

**(a)** Balance point from end A, l_{1} = 39.5 cm

Resistance of the resistor Y = 12.5 Ω

Condition for the balance is given as,

`"X"/"Y" = (100 - "l"_1)/("l"_1)`

X = `(100 - 39.5)/39.5 xx 12.5` = 8.2 Ω

Therefore, the resistance of resistor X is 8.2 Ω.

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

**(b) **If X and Y are interchanged, then l_{1} and 100 − l_{1 }get interchanged.

The balance point of the bridge will be 100 − l_{1 }from A.

100 − l_{1 }= 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

**(c) **When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.