What is the maximum length of a No. 16 cable to be installed from a bus to the equipment in a 28-volt system with a 25-ampere intermittent load and a 1-volt drop?

To determine the maximum length of a No. 16 cable in a 28-volt system with a 25-ampere intermittent load and a 1-volt drop, it's essential to consider the voltage drop and the resistance of the wire over the length of the cable.

No. 16 wire has a resistance of about 4.016 ohms per 1,000 feet. The formula to calculate the voltage drop in a single direction is:

[ \text{Voltage Drop} = \text{Current (Amperes)} \times \text{Resistance (Ohms)} \times \text{Length (Feet)} ]

Since the voltage drop is typically calculated for a round trip (to the load and back), the effective length considered in this case would be twice the length of the cable.

Setting up the equation for the maximum allowable voltage drop (1 volt):

[ 25 , \text{A} \times \left(\frac{4.016 , \text{Ohms}}{1000 , \text{feet}}\right) \times 2L = 1 , \text{volt} ]

This leads to:

[ 25 \times 0.004016