If a 5.5 sq.mm. TW wire inside a conduit pipe is connected with eighty (80) percent loading, what is the maximum allowable loading that it can sustain?

If a 5.5 sq.mm. TW wire inside a conduit pipe is connected with eighty (80) percent loading, what is the maximum allowable loading that it can sustain?

• A. 15 amperes
• B. 20 amperes
• C. 25 amperes
• D. 30 amperes

Answer: (D)

To determine the maximum allowable loading for a 5.5 sq.mm. TW wire when it is under 80% loading, it's essential to first understand the current-carrying capacity of the wire based on standard guidelines, which often refer to the American Wire Gauge (AWG) or local electrical codes. Typically, a 5.5 sq.mm. TW wire has a maximum current-carrying capacity of approximately 30 amperes. When the wire is loaded to 80% of its capacity, we need to calculate 80% of the full capacity to find the maximum load it can sustain without overheating or causing safety hazards. Calculating 80% of 30 amperes yields: \[ 30 \text{ A} \times 0.80 = 24 \text{ A} \] This indicates that the maximum loading under 80% condition should not exceed 24 amperes. Given the options provided, 25 amperes would be slightly above this calculated limit and therefore not safe. Selecting 30 amperes assumes a full loading, which exceeds the intended 80% condition. Thus, D, as the maximum safe allowable loading under these particular guidelines, is indeed aligned with the principles of electrical safety and wire

To determine the maximum allowable loading for a 5.5 sq.mm. TW wire when it is under 80% loading, it's essential to first understand the current-carrying capacity of the wire based on standard guidelines, which often refer to the American Wire Gauge (AWG) or local electrical codes.

Typically, a 5.5 sq.mm. TW wire has a maximum current-carrying capacity of approximately 30 amperes. When the wire is loaded to 80% of its capacity, we need to calculate 80% of the full capacity to find the maximum load it can sustain without overheating or causing safety hazards.

Calculating 80% of 30 amperes yields:

[ 30 \text{ A} \times 0.80 = 24 \text{ A} ]

This indicates that the maximum loading under 80% condition should not exceed 24 amperes. Given the options provided, 25 amperes would be slightly above this calculated limit and therefore not safe.

Selecting 30 amperes assumes a full loading, which exceeds the intended 80% condition. Thus, D, as the maximum safe allowable loading under these particular guidelines, is indeed aligned with the principles of electrical safety and wire