Loaded voltage divider analysis: A divider uses two 100 kΩ resistors across a 12 V source. If a 1 MΩ load is connected at the midpoint to ground (i.e., across the lower 100 kΩ), what is the resulting output voltage?

Introduction / Context:
Real-world voltage dividers are rarely unloaded. When a load is attached to the output node, it forms a parallel path with one of the divider resistors, altering the effective resistance and reducing the output voltage. This is a classic example of source-loading and Thevenin-equivalent thinking.

Given Data / Assumptions:

• Top resistor R_top = 100 kΩ from +12 V to the output node.
• Bottom resistor R_bottom = 100 kΩ from output node to ground.
• Load R_load = 1 MΩ from output node to ground (in parallel with R_bottom).
• All components are ideal; DC operation.

Concept / Approach:
First, find the equivalent of the bottom leg: R_eq_bottom = (R_bottom || R_load). The output is then V_out = V_in * R_eq_bottom / (R_top + R_eq_bottom). Because R_load is large relative to 100 kΩ, loading is modest but not negligible, pulling the output below the open-circuit 6 V level.

Step-by-Step Solution:

Verification / Alternative check:
Open-circuit output would be 6 V. A finite load in parallel with 100 kΩ must reduce the output; 5.7 V is close to, but slightly below, 6 V, which fits expectations. The Thevenin perspective around the midpoint yields the same result.

Why Other Options Are Wrong:

• 6 V: That is the no-load value; it ignores loading.
• 12 V: Impossible for a passive divider without amplification.
• 0.57 V: Off by a factor of 10 from the correct loaded output.

Common Pitfalls:

• Treating the load as separate rather than in parallel with the lower resistor.
• Arithmetic mistakes when handling ratios with kilo-ohm and mega-ohm values.

Final Answer:
5.7 V