What capacity in amperes of storage battery is required to operate a 50 W emergency transmitter for 6 hr.? Assume a continuous transmitter load of 70% of the key-locked demand of 40 A, and an emergency light load of 1.5 A.
• Ampere-hours (Ah) as a measure of battery capacity (current × time) • Relationship between power (watts), voltage, and current: P = V × I • How to combine multiple loads (transmitter + emergency light) into a total current demand
• First, what is the average current draw of the transmitter if it has a key-locked demand of 40 A and a continuous load of 70% of that? • Once you know the transmitter’s average current and the emergency light current, how do you find the total current the battery must supply? • After finding the required ampere-hours, how do the answer choices’ wording about greater than vs. at least affect which one(s) could be correct?
• Carefully compute the transmitter load as 0.70 × 40 A, not 70 A or 40 A alone • Multiply the total current (in amperes) by the time in hours (6 hr) to get ampere-hours • Compare your numeric result against 177 Ah and decide whether the requirement is to be equal to, greater than, or strictly greater than this value
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