Loaded voltage divider — consider attaching different loads to the same divider output. Which statement is more accurate: connecting a 6 kΩ load will cause a smaller drop in the divider's output voltage than connecting a 5 kΩ load (since 6 kΩ loads the source less)?

Introduction / Context:A voltage divider provides a reference voltage but is sensitive to loading. When an external load is connected, the output falls due to the divider's Thevenin resistance. This question compares two loads (6 kΩ versus 5 kΩ) to evaluate which causes less output sag.

Given Data / Assumptions:

• Same divider and same open-circuit output V_oc for both cases.
• Load options: 6 kΩ and 5 kΩ connected to the divider output node.
• Linear, time-invariant resistors; DC or low-frequency behavior; source internal resistance may be modeled as part of the Thevenin resistance R_th.

Concept / Approach:The loaded output is V_out = V_oc * (R_L / (R_th + R_L)). For fixed R_th and V_oc, a larger R_L yields a larger fraction and therefore a higher V_out (less droop). Hence, 6 kΩ (larger than 5 kΩ) loads the divider less and produces a smaller decrease from the open-circuit value.

Step-by-Step Solution:

Verification / Alternative check:Pick R_th = 2 kΩ and V_oc = 10 V. With 6 kΩ: V_out ≈ 10 * (6000/8000) = 7.5 V. With 5 kΩ: V_out ≈ 10 * (5000/7000) ≈ 7.14 V. The 6 kΩ case has less sag, matching the rule of thumb to use a load at least 10× the source resistance.

Why Other Options Are Wrong:

• Incorrect: Contradicts the divider formula; larger R_L reduces loading.
• Only true for AC sources: The reasoning is resistive and applies to DC as well.
• Only true when source resistance is zero: If R_th were zero, both loads would see the same ideal source voltage; the statement remains consistent with the general rule for R_th > 0.

Common Pitfalls:Ignoring the source/ divider's Thevenin resistance, or assuming that higher-ohm loads always draw "more" voltage drop in the source. In reality, higher-ohm loads draw less current and thus cause less droop.

Final Answer:Correct — a 6 kΩ load produces a smaller output decrease than a 5 kΩ load on the same divider.