Power in series networks: Are individual element powers additive to give the total power consumed from the source? Consider a DC series circuit with multiple resistive elements and an ideal voltage source.

Introduction / Context:
Power accounting is essential for thermal design and energy budgeting. This question asks whether the total power drawn equals the sum of powers dissipated by each element in a series path.

Given Data / Assumptions:

• Steady-state DC with resistive elements.
• No energy storage or conversion elsewhere (no ideal lossless transformers or active sources inside the series chain).
• Ideal source with negligible internal resistance (or internal resistance treated as an additional series element).

Concept / Approach:

Conservation of energy dictates that input electrical power from the source equals the total power dissipated as heat in resistors (and any other elements that consume power). For each element, Pk = Vk * Ik = I^2 * Rk in series (same I). The source delivers P_total = V_source * I. Since V_source = Σ Vk, it follows that P_total = Σ (Vk * I) = Σ Pk.

Step-by-Step Solution:

Verification / Alternative check:

Numerical check: with R1 = 100 Ω and R2 = 300 Ω across 20 V, I = 20 / 400 = 0.05 A. P1 = I^2 * 100 = 0.25 W, P2 = I^2 * 300 = 0.75 W. ΣPk = 1.0 W. Source power = V * I = 20 * 0.05 = 1.0 W, confirming additivity.

Why Other Options Are Wrong:

• Equality of drops or resistor values is unnecessary; power still sums.
• AC caveat: when averaged over a cycle, the same conservation holds for passive elements; DC was assumed for simplicity.

Common Pitfalls:

Mixing instantaneous versus average power in AC; however, the additive property still holds for passive components when power is computed consistently.

Final Answer:

True