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 $l_1=33.0cm$ . If we interchange the resistances, the null deflection is found to be at $l_2=67.0cm$ . If $\alpha$ and $\beta$ are the end correction, then the value of $\alpha +\beta$ should be,

Answer 1

Answer: 2

The first end correction in terms of resistance,
$\alpha =\frac{R_2{l}_1-R_1{l}_2}{R_1-R_2}=\frac{\left(200\right)\left(33\right)-\left(100\right)\left(67\right)}{100-200}=1cm$
The second end correction in terms of resistance,
$\beta =\frac{R_1{l}_1-R_2{l}_2}{R_1-R_2}-100$
$=\frac{\left(33\right)\left(100\right)-\left(200\right)\left(67\right)}{100-200}-100=1cm$ .
Now, the sum of the end corrections,
$\alpha +\beta =1+1=2$ .