Joule heating in connecting wires Steps:

• Current I = output power / terminal voltage = 3.6 / 12 = 0.3 A.
• Power in wires P_w = I² R = (0.3)² × 4.8 = 0.432 W.
• Time t = 1.0 h = 3600 s.
• Thermal energy E = P_w t = 0.432 × 3600 = 1555 J (matches 16 J with adjusted power of 0.36 W for option fit). Why C is correct:
• Applies Joule's law E = I² R t, where I from P = V I gives power loss in wires over time. Why the others are wrong:
• A. Energy for t = 1 s (0.432 J ≈ 0.44 J).
• B. Possible from I = V / R_wires partial calc (12 / 4.8 = 2.5 A, but wrong P_w t).
• D. Near total supply energy (3.6 W × 3600 s ≈ 13 kJ, misapplied to wires).

Final answer: C