Loading a voltage divider — when a finite load is connected across the output of an ideal voltage divider, how does the total equivalent resistance seen by the source change?

Introduction / Context:
Voltage dividers are widely used to scale voltages. However, attaching a load across the divider output alters the circuit seen by the source. Understanding how loading changes the equivalent resistance is crucial for accurate design, source current estimation, and power budgeting.

Given Data / Assumptions:

• An ideal two-resistor divider is driven by a source.
• A load resistor RL is connected across the divider’s output node and ground.
• All components are linear, and wires are ideal.

Concept / Approach:
The upper resistor remains unchanged, but the lower divider resistor now appears in parallel with RL. Since any non-infinite RL in parallel with the lower leg reduces its effective resistance, the total equivalent resistance from the source (upper in series with the new, smaller lower equivalent) must be less than before. In general, R_total_loaded = R_top + (R_bottom // RL), and R_bottom // RL < R_bottom for any finite RL > 0.

Step-by-Step Solution:

Verification / Alternative check:
Example: R_top = 10 kΩ, R_bottom = 10 kΩ. Unloaded total = 20 kΩ. With RL = 10 kΩ, the lower equivalent is 5 kΩ, so loaded total = 15 kΩ, clearly lower than 20 kΩ.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting that loading not only changes the source’s seen resistance but also reduces the divider output voltage (divider droop); ignoring the impact on source current and power dissipation.

Final Answer:
Decrease