Question

In potentiometer experiment, if l₁ is the balancing length for e.m.f. of cell of internal resistance r and l₂ is the balancing length for its terminal potential difference when shunted with resistance R then :

विकल्प

• `l_1=l_2((R+r)/R)`

• `l_1=l_2((R)/(R+r))`

• `l_1=l_2((R)/(R-r))`

• `l_1=l_2((R-r)/R)`

Answer 1

`l_1=l_2((R+r)/R)`

The internal resistance of a cell is given as

`r= ((l_1-l_2)/l_2)R`

→`r/R=(l_1-l_2)/l_2`

∴ `rl_2=Rl_1-Rl_2`

∴`l_2(R+r)=Rl_1`

∴`l_1=l_2(R+r)/R`