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Difficulty: MediumExpansion Cards and Power Supplies

An IT specialist is assembling a dedicated virtualization host system with the following estimated power requirements:

- Processor (CPU): 140 W140\text{ W}
- Auxiliary graphics card (GPU): 160 W160\text{ W}
- Quad-port PCIe network expansion card: 45 W45\text{ W}
- Four enterprise SATA SSDs: 10 W10\text{ W} each (40 W40\text{ W} total)
- Motherboard, system RAM, and chassis fans: 95 W95\text{ W}

To ensure operational efficiency and maintain power headroom, the organization's policy requires that the total system power draw under full load must not exceed 60%60\% of the power supply unit's (PSU) maximum rated capacity. Which of the following is the minimum PSU wattage rating required to meet these specifications?

  1. A
    480 W480\text{ W}
  2. B
    576 W576\text{ W}
  3. 800 W800\text{ W}Answer
  4. D
    320 W320\text{ W}

Answer

The minimum power supply unit rating required is 800 W800\text{ W}.
The total power draw of the system under peak load is 480 W480\text{ W} (140 W+160 W+45 W+40 W+95 W140\text{ W} + 160\text{ W} + 45\text{ W} + 40\text{ W} + 95\text{ W}). To ensure that 480 W480\text{ W} does not exceed 60%60\% of the power supply's total capacity, the required minimum rating is calculated as 480 W0.60=800 W\frac{480\text{ W}}{0.60} = 800\text{ W}.

Step-by-Step Solution

1
Calculate the total power consumption of all system components under peak load.
Total peak load = 140 W (CPU)+160 W (GPU)+45 W (NIC)+40 W (4x SSDs)+95 W (Motherboard/RAM/Fans)=480 W140\text{ W} \text{ (CPU)} + 160\text{ W} \text{ (GPU)} + 45\text{ W} \text{ (NIC)} + 40\text{ W} \text{ (4x SSDs)} + 95\text{ W} \text{ (Motherboard/RAM/Fans)} = 480\text{ W}.
Accurately summing all component power draws establishes the baseline load requirement.
2
Apply the 60%60\% maximum continuous workload constraint to determine the required total PSU rating capacity.
Minimum PSU Wattage = Total Peak LoadMax Load Percentage=480 W0.60=800 W\frac{\text{Total Peak Load}}{\text{Max Load Percentage}} = \frac{480\text{ W}}{0.60} = 800\text{ W}.
Dividing the peak load by 0.600.60 ensures the 480 W480\text{ W} demand consumes exactly 60%60\% of the PSU rating.

Key Concept

Power supply unit (PSU) wattage sizing with continuous load headroom calculation.
Estimated Time:1m 30s
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Topics

Expansion Cards and Power Supplies

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