Question 3-18B3

Voltage regulation measures how well a power supply maintains a constant output voltage despite changes in the load current. A lower percentage indicates better regulation. This is critical in amateur radio to ensure stable operation of transceivers and sensitive digital circuits, which can be affected by voltage fluctuations. The formula for calculating percentage voltage regulation is: $$ \text{Percentage Regulation} = \frac{(V_{\text{no-load}} - V_{\text{full-load}})}{V_{\text{full-load}}} \times 100\% $$ Given the values: * $V_{\text{no-load}}$ (no load voltage) = 12 volts * $V_{\text{full-load}}$ (full load voltage) = 10 volts Substitute these values into the formula: $$ \text{Percentage Regulation} = \frac{(12\,V - 10\,V)}{10\,V} \times 100\% $$ $$ \text{Percentage Regulation} = \frac{2\,V}{10\,V} \times 100\% $$ $$ \text{Percentage Regulation} = 0.2 \times 100\% $$ $$ \text{Percentage Regulation} = 20\% $$ Thus, the percentage of voltage regulation is 20%. This means the output voltage drops by 20% relative to the full-load voltage when a load is applied. Options A, B, and D result from incorrect application of the formula, such as dividing by the no-load voltage or calculating the voltage drop incorrectly.