The circuit shown here is used to compare the EMFs of two cells $E_{1}$ and $E_{2} (E_{1} > E_{2})$ . The null point is at $C$ when the galvanometer is connected to $E_{1}$ . When the galvanometer is connected to $E_{2}$ , the null point will be

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E 1 = ϕ × l 1 … i E 2 = ϕ × l 2 … i i Given, E 2 < E 1 . Therefore, l 2 < l 1 by comparing ( i ) and ( ii ) . Hence balance point will shift towards left of C .

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