Question

In a meter bridge (given figure), the balance point is found to be at$39.5cm$ from the end $A,$ when the resistor $S$ is of $12.5\Omega .$ Determine the resistance of $R.$ Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

$2)$ In a meter bridge (given figure), the balance point is found to be at$39.5cm$ from the end $A,$ when the resistor $S$ is of $12.5\Omega .$ Determine the balance point of the bridge below if $R$ and $S$ are interchanged.


$3)$

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

Answer 1

$1)$ Given,balance point from end $A,l_1=39.5cm$
resistance of the resistor $S=12.5\Omega$

From the condition for balance of metre bridge

$\frac{R}{S}=\frac{l_1}{100-l_1}$

$R=S\times \frac{l_1}{100-l_1}$

$R=S\times \frac{l_1}{100-l_1}$

$R=12.5\times \frac{39.5}{100-39.5}=8.2\Omega$

Resistance of the resistor $R=8.2\Omega$

The connection between the resistors in a Wheatstone or meter bridge is made of thick copper strips to minimize the resistance of connection so that we can neglect that resistance in the bridge formula.

$2)$ If $R$ and $S$ are interchanged, then the balance point of the bridge from end $A$ and $C$ will also be interchanged.

The balance point of the bridge will be at

$100-l_1$ from end $A.$

The balance point from end $A=100-l_1$

$100-39.5=60.5cm$


$3)$ If we interchange galvanometer and cell, the balance point of meter bridge will not be changed. Therefore, no current would flow through the galvanometer. Hence, the galvanometer will show zero deflection.