What capacity in amperes is needed from a storage battery to operate a 50-watt transmitter for 6 hours, with a continuous load of 70% of 40 A and an emergency light load of 1.5 A?

To determine the required capacity in ampere-hours for the storage battery to operate a 50-watt transmitter for 6 hours, we need to calculate the total current draw during this time.

First, calculate the current needed for the 50-watt transmitter using the formula:

[ I = \frac{P}{V} ]

Assuming a typical operating voltage (which may vary, but let’s assume 12 volts for common battery-operated systems), the current drawn by the transmitter would be:

[ I = \frac{50 \text{ watts}}{12 \text{ volts}} = 4.17 \text{ amps} ]

Next, consider the continuous load of 70% of 40 A:

[ Continuous Load = 0.7 \times 40 \text{ A} = 28 \text{ A} ]

When combined, the total continuous load during operation becomes:

[ Total Current = 4.17 \text{ amps} + 28 \text{ amps} = 32.17 \text{ amps} ]

Next, add the emergency light load of 1.5 A:

[ Total Current with Light = 32.17 \text{ amps} + 1.