Power accounting — in a purely resistive series circuit, does the total power dissipated equal the sum of the powers dissipated by each resistor, or does it exceed that sum?

Introduction / Context:
Energy conservation in circuits dictates that the power delivered by sources is absorbed by elements as heat or stored energy. In steady-state resistive networks, all source power ends up as heat in the resistors. This question checks whether you understand that total power is an additive quantity across elements, not something that mysteriously exceeds the parts’ contributions.

Given Data / Assumptions:

• Purely resistive series network driven by a source.
• Steady state (no reactive storage).
• Ideal interconnects without parasitic losses.

Concept / Approach:
Use P = V*I at the source, and P_k = I^2 * R_k (or V_k^2 / R_k) at each resistor. Since the same current I flows in series, the total resistor power is Σ (I^2 * R_k) = I^2 * Σ R_k. The equivalent resistance is R_total = Σ R_k, so the power seen from the source is P_total = I^2 * R_total = I^2 * Σ R_k, exactly the same as the sum of the individual powers. There is no extra “hidden” power term in an ideal resistive loop.

Step-by-Step Solution:

Verification / Alternative check:
Measure current I and each drop V_k, compute P_k = V_k * I. Summing P_k matches the source measurement V_source * I within meter tolerance, confirming conservation of energy.

Why Other Options Are Wrong:

• Exceeds the sum: Violates conservation of energy in an ideal resistive loop.
• Less than the sum: Also violates conservation; nothing else absorbs the “missing” power.
• Only for equal resistors: Equality of values is irrelevant to the additivity of power.

Common Pitfalls:
Confusing instantaneous versus average power, or mixing AC reactive elements (which store and return energy) with purely resistive networks. In a DC or resistive AC case, powers simply add.

Final Answer:
Correct — total power equals the sum of the individual resistor powers in a series resistive circuit.