In a series circuit, the voltage drop across a particular resistor is directly proportional to which quantity?

Introduction / Context:
Voltage division in a series network is governed by Ohm's law. Designers often choose resistor ratios to obtain specific node voltages. Identifying the correct proportionality helps when scaling reference voltages and setting bias points.

Given Data / Assumptions:

• All elements are resistors in series.
• Same current flows through each resistor.
• Resistors are linear and operate within ratings.

Concept / Approach:
For a given series current I, the drop on any resistor Ri is Vi = I * Ri. Because I is common to all series elements, Vi is directly proportional to Ri alone, not the total resistance or on-time. The wattage rating limits safe power dissipation but does not set the instantaneous voltage for a given I.

Step-by-Step Solution:
Let I = V_source / ΣR.Voltage across resistor Ri: Vi = I * Ri.Holding I fixed, increasing Ri increases Vi linearly; decreasing Ri reduces Vi linearly.Therefore, Vi ∝ Ri, demonstrating direct proportionality to its own resistance.

Verification / Alternative check:
Example: Series chain with I = 2 mA, Ri options 1 kΩ, 2 kΩ, 3 kΩ yield Vi values 2 V, 4 V, 6 V respectively. The ratios of Vi match the ratios of Ri, verifying proportionality.

Why Other Options Are Wrong:

• Total resistance: Affects I, but Vi for a given resistor scales specifically with its own Ri.
• Wattage rating: A specification limit; it does not define Vi unless power constraints are exceeded.
• On-time: Duration does not change instantaneous DC voltage drop.

Common Pitfalls:

• Assuming wattage rating determines voltage; it only constrains P = V * I = I^2 * R.
• Mixing series and parallel rules; in parallel, voltage is common and currents divide.

Final Answer:
its own resistance