Probable Effect Concentration Quotients (PECQS) Practice Test

For an outlet supplying loads of 250 W, 15 W, 100 W, and 200 W, what is the minimum size of conductor required?

1. 1.5 sq mm

2. 2.0 sq mm

3. 2.5 sq mm

4. 3.0 sq mm

The correct answer is: 2.0 sq mm

To determine the minimum size of conductor required for the given loads, we first need to establish the total load that will be supplied by the outlet. In this case, the total power is the sum of the individual loads: - 250 W - 15 W - 100 W - 200 W Adding these together gives a total power of 250 + 15 + 100 + 200 = 565 W. Next, we can use this total power to determine the current that will flow through the conductor. Assuming the outlet operates at a typical voltage of 230 V (common in many regions), we can calculate the current using the formula: Current (A) = Power (W) / Voltage (V) Substituting the values gives: Current = 565 W / 230 V = 2.46 A Now we must consider the appropriate size of the conductor that can safely carry this current. In electrical installations, standards dictate that the minimum size of conductors must be determined not only by the current they will carry but also by factors such as permissible voltage drop, heat dissipation, and safety margins. The most commonly used standards specify that for general wiring, a conductor size of 1.5 sq