Efficiency of a power supply: If a supply delivers 0.6 W of output power while drawing 0.7 W of input power, what is its efficiency expressed as a percentage?

Introduction / Context:Efficiency quantifies how effectively a device converts input power to useful output power. In power electronics, higher efficiency means less waste heat, smaller heatsinks, and improved energy savings—critical in portable devices and large-scale systems alike.

Given Data / Assumptions:

• Output power P_out = 0.6 W.
• Input power P_in = 0.7 W.
• Efficiency η is defined as output/input in percentage.

Concept / Approach:The efficiency formula is η (%) = (P_out / P_in) * 100. Substitute the given numbers to compute the ratio and convert to percent. Ensure consistent units (both in watts).

Step-by-Step Solution:

Verification / Alternative check:If the input is 0.7 W and 85.7% is delivered as output, the rest is loss: 0.7 W * (1 − 0.857) ≈ 0.1 W dissipated as heat, which is consistent with 0.6 W output. The arithmetic closes properly.

Why Other Options Are Wrong:

• 8.57% and 4.28%: Off by a factor of 10 or more; would imply extremely poor conversion.
• 42.85%: Approximately half the correct value; likely from using (P_in − P_out)/P_in incorrectly.

Common Pitfalls:

• Mixing input and output in the denominator. Always divide by input power for efficiency.
• Rounding too early; keep sufficient precision until the final percentage is stated.

Final Answer:85.7%