Q.Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.

[!TLDR] Compare the open-circuit balancing length (proportional to full emf) against the loaded balancing length (proportional to terminal voltage across a known external resistance R …

This is potentiometer use (B) of Section 9.4.2. A potentiometer wire AB is driven by a cell of emf $\varepsilon$ through a key $K_1$ and rheostat, A at higher potential than B. The test cell (emf $\varepsilon_1$, internal resistance $r_1$ to be found) connects to the wire through a galvanometer G and jockey J, and a resistance box R is connected across the SAME test cell through a second key $K_2$. With $K_1$ closed and $K_2$ open, the circuit is simply the driving cell, the test cell and the wire; the null point at length $l_1$ corresponds to the test cell's FULL emf (it is effectively open-circuit, drawing no current, since the potentiometer branch draws none at balance):

Now $K_2$ is also closed, so the resistance box R is connected across the test cell, which now delivers a current I through R; the null point shifts to a SHORTER length $l_2$, corresponding to the terminal potential difference $V=IR$ across the now-loaded cell:

$V = k\,l_2$ …

Find the open-circuit balancing length (full emf) and the loaded balancing length (terminal voltage across known R), then combi …

Forgetting that the SECOND balance length corresponds to the loaded TERMINAL voltage V=IR, not the full emf -- both keys must be cons …