Voltage drop across supply wires requires higher power supply p.d.

Steps:

• Each wire is 800 m long with resistance 800 × 0.05 Ω/m = 40 Ω.
• Total wire resistance (series for go and return): 40 Ω + 40 Ω = 80 Ω.
• Voltage drop in wires: 0.05 A × 80 Ω = 4 V (using motor current).
• Minimum supply p.d.: 16 V (for motor) + 4 V (drop) = 20 V.

Why D is correct:

• The supply p.d. equals motor p.d. plus I × total wire resistance, per Ohm's law, ensuring 16 V at the motor.

Why the others are wrong:

• A: Subtracts drop from 16 V, ignoring need to compensate.
• B: Ignores voltage drop entirely.
• C: Uses incorrect total resistance (e.g., one wire only).

Final answer: D