Question

A potential drop per unit length along a wire is 5 × 10⁻³ V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.

Answer 1

Data: K = 5 × 10^-3 `"V"/"m"`, L = 216 cm = 216 × 10^-2 m

E = KL

∴ E = 5 × 10^-3 × 216 ×10^-2

= 1080 × 10^-5

= 0.01080 V

The emf of the cell is 0.01080 volL