Power in parallel branches: Two resistors are connected in parallel and dissipate 6 W and 10 W respectively. What is the total power loss in the network?

Introduction / Context:
Power bookkeeping is essential for thermal design and source sizing. In any circuit, the total power delivered by the source equals the sum of the powers dissipated in all elements (ignoring storage in ideal reactive components at steady state).

Given Data / Assumptions:

• Two parallel branches with measured or calculated dissipations: P1 = 6 W, P2 = 10 W.
• DC or steady-state AC with resistive loads so real power adds arithmetically.

Concept / Approach:
Conservation of energy requires that the source supplies the sum of branch powers. In parallel circuits, voltage across each branch is common, but power still adds linearly: P_total = Σ P_branch.

Step-by-Step Solution:
Write: P_total = P1 + P2.Substitute: P_total = 6 + 10.Compute: P_total = 16 W.

Verification / Alternative check:
If branch currents are I1 and I2 at common voltage V, then P_total = V * (I1 + I2) = V * I_total = V * I1 + V * I2 = P1 + P2, confirming additivity.

Why Other Options Are Wrong:

• 3.75 W and 4 W: do not reflect power addition; possibly confused with equivalent resistance ideas.
• 60 W: gross overestimate without basis.
• 6 W: ignores the second branch.

Common Pitfalls:
Mixing current-division or resistance calculations with power addition; power does not divide inversely like resistance.

Final Answer:
16 watts