Effect of loading on a voltage divider: The no-load output of a divider is 12 V. After connecting a load across the output, what happens to the output voltage?

Introduction / Context:
Voltage dividers only hold their ideal output when the output node is unloaded or connected to an infinite resistance (open circuit). Any finite load forms a parallel path with one divider resistor, reducing effective resistance and thereby reducing the output voltage. Recognizing this behavior is fundamental when designing sensor interfaces and bias networks.

Given Data / Assumptions:

• No-load (open-circuit) divider output: 12 V.
• A load is subsequently connected across the output.
• Passive resistive divider; no buffering amplifier.

Concept / Approach:
With a load present, the bottom leg of the divider becomes R_bottom || R_load, which is less than R_bottom alone. The voltage division ratio V_out/V_in = R_eq_bottom / (R_top + R_eq_bottom) decreases because R_eq_bottom decreases. Therefore, V_out must fall below the no-load value.

Step-by-Step Solution:

Verification / Alternative check:
A numeric example (e.g., R_top = R_bottom and R_load = 1 MΩ) shows a modest drop from 12 V to a slightly lower value; a smaller load resistance causes a more pronounced drop, confirming the general rule.

Why Other Options Are Wrong:

• increases: Only an active circuit (e.g., amplifier) could raise the output beyond the divider's set ratio.
• remains the same: True only for infinite load (no load), which is not the case here.
• becomes zero: Would require a short circuit, not merely a finite load.

Common Pitfalls:

• Forgetting that adding a load changes the divider ratio unless the source has zero output resistance and the load is infinite.

Final Answer:
decreases