The schematic diagram of the given question is can be illustrated as:

EMF
Current that passes via the potentiometer wire is:
= \(I = \frac{E_0}{r\ +\ r_1}\)
The potential differ nce in the wire equals:
= \(V = Ir = \frac{E_0r}{r\ +\ r_1}\)
Along the potentiometer, the potential gradient is:
\(\therefore\) \(kl = \frac{E_0r}{r\ +\ r_1L}\)
Thus, the unknown emf equals: \(\frac{E_0r}{r\ +\ r_1L}\)
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