When an emergency transmitter uses 325 watts and a receiver uses 50 watts, how many hours can a 12.6 volt, 55 ampere-hour battery supply full power to both units?

Study for the Federal Communications Commission (FCC) Element 3 Test. Use flashcards and multiple choice questions with hints and explanations. Get prepared for your exam today!

To determine how long a 12.6 volt, 55 ampere-hour battery can supply power to both an emergency transmitter and a receiver, first calculate the total power consumed by both devices.

The transmitter uses 325 watts, while the receiver uses 50 watts. By adding these values together, the total power consumption is:

325 watts + 50 watts = 375 watts.

Next, to find out how long the battery can supply this power, we need to understand the capacity of the battery in watt-hours. The battery’s capacity in ampere-hours (Ah) is given as 55 Ah at 12.6 volts. To convert this capacity into watt-hours (Wh), you multiply the amp-hour capacity by the voltage of the battery:

Capacity in watt-hours = Voltage × Ampere-hour capacity

= 12.6 volts × 55 Ah

= 693 watt-hours.

Now, divide the total watt-hours of the battery by the total power consumption:

Battery life (in hours) = Total capacity in watt-hours / Power consumption in watts

= 693 Wh / 375 W

≈ 1.848 hours.

This value rounded is approximately 1.8 hours. Therefore, the correct answer, which represents the