Power accounting in parallel networks: Is the total power in a parallel circuit obtained by summing the power dissipated in each individual resistor? Assume a steady-state circuit with an ideal voltage source feeding multiple resistive branches in parallel.

Introduction / Context:
Understanding how power combines in circuits is essential for thermal design, fuse sizing, and efficiency analysis. In a parallel circuit, each branch has the same applied voltage, but different currents may flow depending on branch resistances. The question asks whether total power equals the sum of individual branch powers.

Given Data / Assumptions:

• Ideal voltage source maintains the same voltage across all parallel branches.
• Elements are linear resistors (no active sources within branches).
• Steady-state operation (DC or sinusoidal AC with RMS values).

Concept / Approach:

Power conservation states that input electrical power from the source equals the total power consumed by all loads. In parallel, each resistor k dissipates Pk = V^2 / Rk when supplied by a common voltage V. The source supplies the algebraic sum of these powers, so P_total = ΣPk.

Step-by-Step Solution:

Verification / Alternative check:

Using currents: the source current I_source = ΣIk. Source power is P_source = V * I_source = V * Σ(V / Rk) which simplifies to Σ(V^2 / Rk) = ΣPk, confirming additivity.

Why Other Options Are Wrong:

• Equal resistors are unnecessary; additivity holds for any set of resistances.
• In AC, using RMS values yields the same result for pure resistors.
• Finite source resistance can be modeled as an additional element whose power simply adds to the total dissipation.

Common Pitfalls:

Confusing voltage addition (which does not occur in parallel across the same two nodes) with power addition. Also, forgetting to use RMS values in AC analysis can lead to incorrect power calculations.

Final Answer:

True