In order for a firm voltage divider to operate properly, the load resistance value should be at least __________ times greater than resistance value of the voltage divider bleeder resistor.

• A. 2
• B. 5
• C. 10
• D. 20

Answer: (C)

Loading effects are the key idea here. When you connect a load to the output of a voltage divider, it draws current and effectively sits in parallel with the bottom resistor. This changes the bottom resistance to Rbottom_eff = (R2 || RL), which shifts the divider ratio and alters the output voltage away from the no-load value.

To keep the output close to the intended fraction of Vin, the load must be much larger than the bleeder (bottom) resistor. The larger the load resistance relative to R2, the less it disturbs the divider. A practical and widely used guideline is to have the load at least ten times larger than the bleeder resistor. This typically keeps the loading error within a few percent. If the load is only 2x or 5x, the output voltage deviates more noticeably; 20x would reduce the error even further, but tenfold is a convenient and common compromise that works well in many designs.

For example, with a bottom resistor of 1 kΩ, a top resistor chosen so the no-load output is about the desired fraction, and a load of 10 kΩ, the bottom path becomes about 0.909 kΩ, causing a small drop in output toward the load. The tenfold rule ensures the change remains modest, making ten times the bleeder resistance the best answer among the options.