Loaded divider sensitivity: A voltage divider uses three 1 kΩ resistors in series. Which load resistor value connected across the output (mid to ground) will cause the least change from the no-load output voltage?

Introduction / Context:
Any real voltage divider is affected by the load attached to its output node. The load forms a parallel combination with the lower leg of the divider, changing the division ratio and pulling the output away from the ideal value. To minimize loading error, the load resistance should be much larger than the divider leg it shunts.

Given Data / Assumptions:

• Three resistors of 1 kΩ in series form the divider (total 3 kΩ).
• A load is attached at the output node to ground.
• We compare different R_load values for least disturbance.

Concept / Approach:
Loading is least when R_load ≫ R_bottom. A very large load resistance draws negligible current and barely alters the equivalent of the lower branch, keeping the output close to the no-load value. Therefore, pick the largest R_load offered.

Step-by-Step Solution:

Verification / Alternative check:
Equivalent bottom leg with load: R_eq = (1 kΩ * R_load)/(1 kΩ + R_load). With 1 MΩ, R_eq ≈ 0.999 kΩ (negligible change). With 1 kΩ, R_eq = 0.5 kΩ (severe change). The math confirms the qualitative conclusion.

Why Other Options Are Wrong:

• 100 kΩ: Some loading occurs; output decreases slightly.
• 1 kΩ or 330 Ω: Strong loading; output sags significantly from no-load value.

Common Pitfalls:

• Assuming the divider output is independent of load unless buffered.
• Forgetting that parallel combinations reduce resistance, not increase it.

Final Answer:
1 MΩ