The Power Budget for a Field Radio
A field unit must run a radio set continuously for 24 hours. The radio draws 2.5 A at 12 V. It is powered by a 12 V battery rated 100 Ah, but the battery is only 80% charged and should not be discharged below 20% of its rated capacity to avoid damage. A solar panel rated 60 W (at 12 V) is available; in the field it delivers only about 60% of its rated power for an average of 5 effective sunlight hours per day.
1. Calculate the radio's power draw and its 24-hour energy requirement in watt-hours.
Read the worked answer
Power P = V × I = 12 × 2.5 = 30 W. Energy over 24 h = 30 W × 24 h = 720 Wh. (Equivalently, charge needed = 2.5 A × 24 h = 60 Ah at 12 V, and 60 Ah × 12 V = 720 Wh - the two routes agree, which is a good consistency check.)
2. Determine the usable energy in the battery and whether it alone can power the radio for 24 hours.
Read the worked answer
Rated energy = 100 Ah × 12 V = 1,200 Wh. Currently 80% charged = 960 Wh stored. Usable window is from 80% down to 20% = 60% of rated = 0.6 × 1,200 = 720 Wh usable. Requirement is 720 Wh. So the battery is exactly at the limit - it can just cover 24 h, leaving zero margin and ending at the 20% floor. Any inefficiency, cold, or extra load would breach the limit. Verdict: technically possible but unsafe; the margin is nil.
3. Compute the solar panel's realistic daily contribution and state whether the combined system is safe.
Read the worked answer
Realistic panel power = 60 W × 0.6 = 36 W. Daily energy = 36 W × 5 h = 180 Wh. Net daily deficit = 720 − 180 = 540 Wh drawn from the battery. The battery's usable 720 Wh covers 720 / 540 ≈ 1.33 days - so it survives one day but will degrade and fail on the second. Not safe for sustained operation. Options: add a second panel (2 × 180 = 360 Wh, still short but extends range), reduce radio duty cycle, or bring a larger/more charged battery. The judgement: a single-day pass disguises a multi-day failure.