Potential divider with equal resistances Q and wire Steps: - Battery emf 6 V connects in series with Q (resistance R) and uniform wire (length L, resistance R), so total resistance 2R and current I = 6 V / 2 R = 3 V / R. - Fixed point Y is at junction between Q and wire; potential at Y is I R = 3 V above negative terminal. - Point P is at distance l from Y along wire toward negative terminal; resistance Y to P is (l / L) R. - Potential drop Y to P is I (l / L) R = 3 (l / L) V, so PD between fixed point (at Y) and P is 3 (l / L) V, linear in l from 0 to 3 V as l from 0 to L. Why B is correct: - B shows linear increase from 0 V at l = 0 to 3 V at l = L, matching PD = (R / total R) × emf × (l / L) from Ohm's law in series …