Choosing a Bulb for a Solar-Powered Chicken House
A poultry farmer wants to light a 4 m by 3 m chicken house at night using a 12 V battery charged by a 60 W solar panel. She has three bulbs: bulb A rated 12 V, 24 W; bulb B rated 12 V, 6 W; bulb C rated 240 V, 24 W. The battery has a capacity of 40 Ah and the panel produces useful output for about 5 hours per day. The house needs light for 6 hours each night. The farmer also wants to run a small 12 V fan rated 12 W for 2 hours per night. She asks which bulb to use and whether the system will cope.
1. Calculate the nightly energy demand for each option and determine whether the battery and panel can meet it. Use P = VI and energy in watt-hours.
Read the worked answer
Bulb C is unsuitable immediately. It is rated 240 V and the supply is 12 V. It would be very dim and would not give the intended light output. It is also a shock and fire risk if wired to a 12 V battery by someone who assumes it will work. Reject it.
Current drawn by each 12 V bulb: I = P/V.
- Bulb A: I = 24/12 = 2 A.
- Bulb B: I = 6/12 = 0.5 A.
Nightly energy:
- Bulb A: 24 W x 6 h = 144 Wh.
- Bulb B: 6 W x 6 h = 36 Wh.
- Fan: 12 W x 2 h = 24 Wh.
Battery capacity in watt-hours: 12 V x 40 Ah = 480 Wh. In practice a lead-acid battery should not be discharged below about 50%, so usable energy is about 240 Wh.
With bulb A: 144 + 24 = 168 Wh per night. That is within 240 Wh usable, but leaves only 72 Wh margin and deep cycling will shorten battery life. With bulb B: 36 + 24 = 60 Wh per night. That is comfortably within the usable capacity and leaves a large margin for cloudy days.
Panel check: 60 W x 5 h = 300 Wh per day in ideal conditions. Allowing for losses (say 70% efficiency), about 210 Wh per day. Bulb A plus fan (168 Wh) is close to this; bulb B plus fan (60 Wh) is easily covered.
Judgement: Bulb B is the better engineering choice: it meets the lighting need, keeps the battery above its safe depth of discharge, and leaves capacity for the fan and for cloudy days. Bulb A is feasible only if the fan is dropped or the panel and battery are enlarged.
2. The farmer notices the bulb gets dimmer over several nights and the fan slows. Suggest two likely causes and one measurement she could make to distinguish them.
Read the worked answer
Cause 1: The battery is not being fully recharged. If the panel is shaded, dirty, or the day is cloudy, the 60 W panel may deliver far less than 300 Wh. Over several nights the battery's state of charge falls, so its terminal voltage drops and both bulb and fan run dimmer and slower. This is a gradual, cumulative effect consistent with the description.
Cause 2: The battery is ageing or damaged. A lead-acid battery that has been deep-cycled repeatedly loses capacity. Its internal resistance rises, so under load the terminal voltage sags more than it should even when nominally charged. A poor connection or corroded terminal would give a similar effect and could also cause heating.
Measurement to distinguish them: Measure the battery terminal voltage with a voltmeter (a) with no load and (b) with the bulb and fan switched on.
- If the open-circuit voltage is already low (well below about 12.0 V for a 12 V lead-acid battery at rest) and rises little when charging, the battery is undercharged - check panel output and charging.
- If the open-circuit voltage looks reasonable but drops sharply when the load is applied, the battery has high internal resistance (ageing) or there is a bad connection.
She should also measure the current with an ammeter in series with the bulb; comparing it with the expected 0.5 A for bulb B shows whether the supply is delivering as it should. Cleaning terminals and checking panel shading are cheap first steps before replacing the battery.