Loaded voltage divider behavior: When a load resistance is attached to the output node of a DC voltage divider, the source must supply additional current. Evaluate this statement about source current increasing under load.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Voltage dividers are ubiquitous. However, the ideal unloaded divider assumption breaks once a load is connected. Understanding how loading alters currents and voltages helps in designing dividers with acceptable regulation and selecting buffer stages when needed.


Given Data / Assumptions:

  • Two-resistor divider producing an output node.
  • A finite load R_L is connected from the output node to ground (or return).
  • DC steady-state analysis; resistors are linear.


Concept / Approach:
The source current equals the sum of currents through all paths tied to the source. Without load, current is I_unloaded = V_in / (R_top + R_bottom). With load, the bottom resistor and R_L form a parallel network, decreasing the effective lower-leg resistance and increasing total divider current. Therefore, the source must deliver more current when a load is attached.


Step-by-Step Solution:

Compute R_lower_eff = R_bottom || R_L.Total divider resistance becomes R_total_loaded = R_top + R_lower_eff.Since R_lower_eff <= R_bottom, R_total_loaded <= R_top + R_bottom.For a fixed V_in, I_loaded = V_in / R_total_loaded ≥ I_unloaded.


Verification / Alternative check:
Select sample values and calculate currents with and without load; the loaded case always shows higher source current for any finite R_L.


Why Other Options Are Wrong:

  • Incorrect / depends on meter type / AC only: The rule follows from resistance reduction; instrument type and DC vs AC (for resistors) do not reverse it.
  • True only if R_L is larger than the bottom resistor: Regardless of R_L magnitude, any finite R_L in parallel reduces the effective resistance and increases current.


Common Pitfalls:
Ignoring load interaction when specifying divider resistors; forgetting that heavier loading also drops the output voltage below the ideal unloaded value.


Final Answer:
Correct

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