Equal series resistors: Will two equal-value resistors in series split the source voltage equally across them (ignoring loading)? Assume an ideal DC source and no additional load connected to the midpoint.

Introduction / Context:
Voltage dividers are most intuitive when resistor values are equal. This question checks understanding of proportional division: equal resistances in series produce equal drops from a common current.

Given Data / Assumptions:

• Two resistors R and R in series across a DC source Vs.
• No midpoint loading (infinite input impedance at the tap).
• Wire resistance and source internal resistance are negligible, or included symmetrically.

Concept / Approach:

Using the voltage divider relation Vk = Vs * (Rk / ΣR), if R1 = R2 = R, then ΣR = 2R and each drop is Vs * (R / 2R) = Vs/2. Since the same current flows through both resistors, and their resistances are equal, their power dissipations are equal as well (P = I^2 * R).

Step-by-Step Solution:

Verification / Alternative check:

Example: Vs = 12 V, R1 = R2 = 1 kΩ. Each resistor drops 6 V and dissipates P = 6 V * (6 mA) = 36 mW; total is 72 mW, which equals Vs * I = 12 V * 6 mA.

Why Other Options Are Wrong:

• A current source does not change the equality for equal resistors in series with a defined current; the statement concerns a voltage source divider but still demonstrates equal drops under equal current.
• Nonzero but small wire resistance slightly perturbs symmetry; with the assumption given, equal division holds.
• AC/DC distinction is irrelevant for ideal resistors with no loading; magnitudes divide equally.

Common Pitfalls:

Forgetting the effect of loading at the midpoint: a low-impedance load in parallel with the lower resistor changes the effective resistance and skews the division. Always consider Thevenin equivalent when tapping a divider.

Final Answer:

True