Loading a two-resistor divider: A voltage divider uses two 12 kΩ resistors in series. Which load resistor connected across the lower 12 kΩ will disturb (change) the divider output the most?

Introduction / Context:
Voltage dividers provide a fraction of a source voltage, but adding a load across the output forms a parallel combination with the lower leg, changing the effective divider ratio. The smaller the load, the stronger the loading effect and the bigger the output error. This is critical when designing reference taps for sensors or ADCs.

Given Data / Assumptions:

• Divider resistors: 12 kΩ (upper) and 12 kΩ (lower).
• Load resistor connected across the lower 12 kΩ.
• We compare candidate load values for how much they alter the output.

Concept / Approach:

The equivalent lower leg becomes R_eq = (R_lower * R_load) / (R_lower + R_load). A smaller R_load yields a smaller R_eq, pulling the output down more and causing greater error. Therefore, among options, the smallest load value creates the largest disturbance.

Step-by-Step Solution:

Verification / Alternative check:

Compute quickly: 12 kΩ ‖ 12 kΩ = 6 kΩ (output drops to 1/3 of source instead of 1/2). For 24 kΩ, R_eq = 8 kΩ (output = 8/(12+8) = 0.4 of source), smaller disturbance than 1/3 vs 1/2 shift.

Why Other Options Are Wrong:

Higher load resistances draw less current, hence less loading. 1 MΩ is nearly open, minimally affecting the divider.

Common Pitfalls:

Assuming equal percent change for any load; ignoring that the output node resistance is the parallel result of the lower leg and load.

Final Answer:

12 kΩ