Voltage division: Does any simple series circuit act as a voltage divider across its series elements? Interpret the phrase “voltage divider” for a resistive series network driven by a DC source.

Introduction / Context:
Voltage dividers are ubiquitous in sensing, biasing, and signal-level adjustment. Recognizing that any resistive series chain divides the applied voltage in proportion to element resistances is essential.

Given Data / Assumptions:

• Two or more resistors in series across a DC source.
• Load, if present, is assumed sufficiently high not to disturb the ideal divider (or we consider the unloaded case).
• Steady-state DC.

Concept / Approach:

In a series chain carrying current I, each resistor Rk develops a voltage drop Vk = I * Rk. Because I is the same through all series elements, the drops are proportional to their resistances. The total source voltage equals the sum of these drops. This proportionality is exactly what is meant by a “voltage divider.”

Step-by-Step Solution:

Verification / Alternative check:

Example: R1 = 1 kΩ, R2 = 3 kΩ across 8 V. Drops are V1 = 8 * (1/4) = 2 V and V2 = 8 * (3/4) = 6 V. Ratio matches resistances 1:3, confirming divider behavior.

Why Other Options Are Wrong:

• Equal values are not required; the divider works for any positive resistances.
• Knowing current is convenient but not a prerequisite; the divider formula directly uses ratios.
• AC/DC distinction is irrelevant for resistors; the principle holds for steady-state AC magnitudes too (ignoring reactive loading).

Common Pitfalls:

Ignoring loading: connecting a low-impedance load to the tap changes the effective lower resistance and the division ratio. Use Thevenin equivalents to model divider output under load.

Final Answer:

True