Loaded vs. unloaded divider — if a certain loaded voltage divider produces 12 V at its output, what happens when the load is removed (open-circuit)? Does the output decrease or increase relative to 12 V?

Introduction / Context:
Voltage dividers sag under load because current through the load creates an additional drop across the source (Thevenin) resistance. Removing the load (open-circuiting the output) eliminates that sag. This question addresses whether the output decreases or increases when the load is taken away.

Given Data / Assumptions:

• A divider or source modeled by V_th in series with R_th.
• Loaded output measured as 12 V with some R_L attached.
• Then the load is removed (R_L → ∞), measuring open-circuit output.

Concept / Approach:
The loaded output is V_load = V_th * (R_L / (R_th + R_L)). When R_L is finite, V_load < V_th (if R_th > 0). As R_L → ∞ (load removed), the divider current into the load goes to zero and the output approaches V_th, which is the open-circuit voltage V_oc. Therefore, the output cannot decrease by removing the load; it increases (or at worst stays the same if R_th = 0).

Step-by-Step Solution:

Verification / Alternative check:
Bench check with V_th = 15 V, R_th = 1 kΩ, R_L = 3 kΩ: V_load = 15 * (3k / 4k) = 11.25 V. Removing the load yields V_oc = 15 V, which is higher than 11.25 V, confirming the principle.

Why Other Options Are Wrong:

• Decreases: Opposite to divider behavior.
• Drops to 0 V: An open-circuit does not short the source; the node retains the open-circuit voltage.
• Undefined: It is well-defined; it is the Thevenin (open-circuit) voltage.

Common Pitfalls:
Confusing open-circuit (infinite resistance) with short-circuit (zero resistance). A short would reduce the output to near 0 V (limited by R_th), but an open makes it rise to V_th.

Final Answer:
Correct — removing the load raises the output toward the open-circuit value.