What is the current that each circuit conductor to a 48 KW, 240-volt boiler will carry?

Study for the Independent Electrical Contractors (IEC) Year 2 Part 3 Test. Use flashcards and multiple choice questions with hints and explanations to prepare confidently. Get exam-ready now!

To determine the current that each circuit conductor to a 48 kW, 240-volt boiler will carry, you can use the power formula:

Power (P) = Voltage (V) × Current (I)

Rearranging the formula to solve for current gives us:

Current (I) = Power (P) / Voltage (V)

Substituting the given values into this equation:

• The power is 48 kW, which is 48,000 watts.

• The voltage is 240 volts.

So, you would calculate the current as follows:

I = 48,000 W / 240 V

I = 200 A

This calculation gives you the total current for the entire boiler. If the boiler is connected to a circuit with two conductors (assuming a single-phase connection), then each conductor would carry half of the total current. Therefore, dividing the total current by 2 provides the current per conductor:

200 A / 2 = 100 A (which aligns closely with one of the choices)

However, for a three-phase system, you would divide the total power by the square root of 3 and the line voltage to find the current per phase. Since the question does not clarify the system type