# Potentiometer analysis

As I was studying about potentiometer I came across how to use it to compare emf of two cells.

It was written my text book that if current I flows through potentiometer wire then

$$\mathcal{E}= v\cdot L$$

where $$v$$ is potential drop per unit length. Now

$$\mathcal{E}_1= v\cdot L_1$$

and

$$\mathcal{E}_2= v\cdot L_2$$

But on what basis did we take $$v$$ as a constant $$v=\mathcal{E}/L$$? If current through battery is zero then

$$\mathcal{E}=\mathrm{voltage}=IR/L$$

$$=I\rho/A,\quad \mathrm{(\rho=resistivity)}$$

Thus $$v$$ depend on $$I$$ and if we change $$\mathcal{E}$$ then $$I$$ must change and thus $$v$$ will also change. But here $$v$$ is taken as constant.

My book