Loaded divider formula — for a Thevenin source of V_th in series with R_th feeding a 10 kΩ load, the output is V_out = V_th * (10 kΩ / (R_th + 10 kΩ)). Is this relationship the correct way to compute the loaded output?

Introduction / Context:Any practical voltage source has finite output resistance. The Thevenin model captures this as an ideal source V_th in series with R_th. When you connect a load, the output drops according to a simple divider rule. This item asks you to validate the formula for a 10 kΩ load.

Given Data / Assumptions:

• Thevenin equivalent: V_th in series with R_th.
• Load: R_L = 10 kΩ.
• Linear resistive behavior; DC or phasor magnitude for AC at one frequency.

Concept / Approach:The output node is the junction between R_th and R_L. By the voltage divider relation: V_out = V_th * (R_L / (R_th + R_L)). Substituting R_L = 10 kΩ gives the stated formula. This is exact for the Thevenin representation and is the standard way to quantify loading effects and set design margins.

Step-by-Step Solution:

Verification / Alternative check:SPICE or bench test with V_th = 5 V, R_th = 1 kΩ: V_out = 5 * (10k / 11k) ≈ 4.545 V, matching simulation and measurement results within tolerance.

Why Other Options Are Wrong:

• Incorrect: The divider formula follows directly from Ohm’s law.
• Only valid if R_th ≪ 10 kΩ: The formula holds for any values; that inequality only indicates small loading (V_out ≈ V_th).
• Only valid for AC: The relation is resistive and holds for DC as well.

Common Pitfalls:Forgetting the Thevenin resistance, or confusing series and parallel placements of the load; misapplying the parallel formula instead of the series divider.

Final Answer:Correct — V_out = V_th * (10 kΩ / (R_th + 10 kΩ)).