Simple 1:1 divider: A voltage divider uses two equal 68 kΩ resistors across a 24 V source. What is the nominal output voltage measured at the midpoint (no load)?

Introduction / Context:
Equal-value voltage dividers are common for halving a supply, establishing bias points, or providing reference levels. With two identical resistors in series across a source, the midpoint sits at exactly half the supply under no-load conditions, assuming ideal components.

Given Data / Assumptions:

• R_top = 68 kΩ, R_bottom = 68 kΩ, connected in series.
• Source voltage V_in = 24 V (DC).
• Output is at the junction (midpoint) with no load attached.

Concept / Approach:
For a two-resistor divider: V_out = V_in * R_bottom / (R_top + R_bottom). With equal values, R_bottom = R_top, the ratio is 1/2. Therefore, V_out equals half of V_in. No load is critical here; any finite load would reduce V_out below this ideal value unless the divider is buffered.

Step-by-Step Solution:

Verification / Alternative check:
Symmetry argument: Identical resistors share equal voltage. Each drops 12 V, placing the midpoint exactly at 12 V relative to ground. This aligns with KVL and proportional division concepts.

Why Other Options Are Wrong:

• 24 V: That would be the full source at the top node, not the midpoint.
• 0 V: That is the bottom node reference, not the midpoint.
• 6 V: Would require unequal resistor values (e.g., a 1:3 ratio), which is not given.

Common Pitfalls:

• Forgetting that the stated output is no-load; a load would alter the midpoint voltage.

Final Answer:
12 V