Voltage division rule — in a pure series resistive circuit carrying a common current, larger resistance values develop larger voltage drops across them. Evaluate this statement.

Introduction / Context:
Voltage division in series circuits is a standard tool for biasing, sensing, and signal scaling. It relies on the fact that the same current flows through all series elements, so each element’s voltage drop is proportional to its resistance. This question checks understanding of that proportionality.

Given Data / Assumptions:

• Purely resistive series network.
• Common current I flows through all resistors.
• Ideal conditions (no parasitics).

Concept / Approach:
Ohm’s law says V_i = I * R_i. Since I is common to all series resistors, the element with the larger R_i will have the larger V_i. The voltage-divider expression quantifies this: V_k = (R_k / ΣR) * V_source. Hence, larger R_k yields a larger fraction of V_source, provided all resistances are positive and finite.

Step-by-Step Solution:

Verification / Alternative check:
Example: Two resistors 1 kΩ and 3 kΩ in series on 8 V give drops 2 V and 6 V respectively, confirming that the larger resistor gets the larger drop in proportion to its value.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting internal source resistance or measurement loading; confusing series with parallel where voltage across branches is equal rather than divided.

Final Answer:
Correct