ITPEC FE Morning October 2023 Question 22
A power supply unit has a rated output power of 500 watts and an efficiency of 80%. In order to obtain an output power of 500 watts from this power supply unit, how many watts is the minimum input power?
Explanation
625 W — efficiency is output over input, so the input must be the output divided by the efficiency.
- •Efficiency = output ÷ input, so input = output ÷ efficiency
- •Input: 500 ÷ 0.8 = 625 W
- •The 125 W difference is lost as heat inside the unit
Check: 80% of 625 W is exactly 500 W, so no smaller input can deliver the rated output.
Answer (c)
Why not others:
- •(a) 100 — takes 20% of 500 W, which is roughly the loss, not the input
- •(b) 400 — multiplies by the efficiency, 500 × 0.8; that is what comes out of a 500 W input, less than required
- •(d) 900 — would deliver 900 × 0.8 = 720 W, more than needed, so it is not the minimum
Key rule: With an efficiency below 100%, the input is always larger than the output: divide by the efficiency, never multiply.
Hardware · AI-written, may contain errors
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.