Difficulty: Easy
Correct Answer: 1%
Explanation:
Introduction / Context:
Voltage regulation quantifies how much a power supply’s output voltage changes from no-load to full-load conditions. A smaller regulation percentage indicates a “stiffer” supply that holds its output more constant as current demand increases. This metric is widely used when comparing linear and switching regulators, or when validating a design against a specification sheet.
Given Data / Assumptions:
Concept / Approach:
When a load is connected, internal resistance and control-loop limitations cause a slight output drop. Regulation expresses that drop as a percentage referenced to the loaded voltage. Using the definition provided, compute the difference between no-load and full-load, then divide by the full-load voltage and multiply by 100 to convert to percent.
Step-by-Step Solution:
Compute the drop: ΔV = Vnl − Vfl = 20 − 19.8 = 0.2 V.Form the ratio to the full-load voltage: ΔV / Vfl = 0.2 / 19.8.Evaluate: 0.2 / 19.8 ≈ 0.010101…Convert to percent: 0.010101… * 100 ≈ 1.01%, which rounds to 1%.
Verification / Alternative check:
Some texts reference Vnl instead of Vfl in the denominator. Here the question explicitly specifies division by Vfl, removing ambiguity. If one mistakenly used Vnl, the result would be (0.2/20)*100 = 1.0%, which is essentially the same to two significant figures, reinforcing the answer of about 1%.
Why Other Options Are Wrong:
0%: would require Vnl = Vfl (no drop), which is not the case..1%: ten times too small; ignores the 0.2 V change relative to ~20 V.5%: five times too large given only a 0.2 V change.
Common Pitfalls:
Confusing which voltage to use in the denominator, mixing up units (mV vs V), or rounding too aggressively before the percent conversion. Always compute the difference first, then divide by the specified reference voltage, then multiply by 100.
Final Answer:
1%
Discussion & Comments