Question

Consider the meter bridge circuit without neglecting end corrections. If we used $100\Omega$ and $200\Omega$ resistances in place of $R$ and $S$ , respectively, we get null deflection at $l_{1}=33.0\,cm$ . If we interchange the resistances, the null deflection is found to be at $l_{2}=67.0\,cm$ . If $\alpha $ and $\beta $ are the end correction, then the value of $\alpha +\beta $ should be,

Answer 1

The first end correction in terms of resistance,
$\alpha =\frac{R_{2} l_{1} - R_{1} l_{2}}{R_{1} - R_{2}}=\frac{\left(200\right) \left(33\right) - \left(100\right) \left(67\right)}{100 - 200}=1\,cm$
The second end correction in terms of resistance,
$\beta =\frac{R_{1} l_{1} - R_{2} l_{2}}{R_{1} - R_{2}}-100$
$=\frac{\left(33\right) \left(100\right) - \left(200\right) \left(67\right)}{100 - 200}-100=1\,cm$ .
Now, the sum of the end corrections,
$\alpha +\beta =1+1=2$ .