Effect of removing a load: In a simple two-resistor voltage divider, if the load resistance connected to the output is removed (opened), how does the current drawn from the source change?

Introduction / Context:
Voltage dividers are sensitive to loading. Adding a load draws extra current from the source; removing that load reduces the current. Understanding this helps in power budgeting and prevents overloading sources or causing unexpected voltage sag at the divider output.

Given Data / Assumptions:

• A two-resistor divider connected to a source.
• A load resistor was attached to the output node and is then removed (open circuit).
• Resistors are ideal; DC conditions.

Concept / Approach:

Total current from the source equals the sum of the divider's standing current plus any load current. With the load connected, I_source = I_divider + I_load. Removing the load sets I_load to zero, so the source current must decrease to I_divider only. The divider's standing current remains, unless the entire divider is opened (which is not the case here).

Step-by-Step Solution:

Verification / Alternative check:

Consider limits: R_load → ∞ (removed) means no output current; only divider current flows. Conversely, a small R_load dramatically increases I_source. These edge cases confirm the trend.

Why Other Options Are Wrong:

Increases would require adding a load, not removing it. Remains the same is false unless the load current was negligible (not stated). Is cut off implies zero current, which would require opening the divider itself, not just removing the load.

Common Pitfalls:

Confusing divider current with load current; assuming removing the load opens the entire circuit; overlooking that the divider path still exists.

Final Answer:

decreases