Question

When a resistor of $5\Omega$ is connected across the cell, its terminal potential difference is balanced by $150cm$ of potentiometer wire and when a resistance of $10\Omega$ is connected across the cell, the terminal potential difference is balanced by $175cm$ same potentiometer wire. Find the balancing length when the cell is in open circuit and the resistance of the cell.

Answer 1

$r=\left(\frac{l-l_2}{l_2}\right)R_1$

$r=\left(\frac{l-l_2}{l_1}\right)R_2$

$r=\left(\frac{l-150}{150}\right)5$

$r=\left(\frac{l-175}{175}\right)10$

$\frac{5l-5\times 150}{150}=\frac{10l-175\times 10}{175}$

$7(5l-5\times 150)=6(10l-175\times 10)$
$7l-1050=12l-2100$

$1050=5l$

$l=210cm$

$r=\frac{210-150}{150}\times 5$

$=\frac{60}{150}\times 5=2\Omega$