Parallel resistors and total power: Two resistors of 2 kΩ and 1 kΩ are connected in parallel. The measured total current into the parallel network is IT = 3 mA. What is the total power dissipated by the parallel combination?

Introduction / Context:This problem tests core DC circuit analysis for parallel resistors: combining equivalent resistance, using the given total current, and computing total power. Such steps appear routinely in electronics troubleshooting and power budgeting.

Given Data / Assumptions:

• R1 = 2 kΩ, R2 = 1 kΩ, connected in parallel.
• Total current into the parallel pair: IT = 3 mA.
• Use ideal components and steady-state DC.

Concept / Approach:For a parallel pair, compute equivalent resistance Req, then use P_total = IT^2 * Req. Alternatively, find the common voltage V = IT * Req and compute P_total = V^2 / Req; both are equivalent and consistent with power conservation.

Step-by-Step Solution:Compute Req = (R1 * R2) / (R1 + R2) = (2000 * 1000) / (2000 + 1000) = 2,000,000 / 3000 = 666.666… Ω.Compute power using total current: P_total = IT^2 * Req = (0.003)^2 * 666.666… = 9e-6 * 666.666… ≈ 0.006 W.Convert to milliwatts: 0.006 W = 6 mW.

Verification / Alternative check:Find branch currents: I1 = V/2000, I2 = V/1000 with V = IT * Req = 0.003 * 666.666… ≈ 2 V. Then I1 ≈ 1 mA, I2 ≈ 2 mA, and P_total = V * IT = 2 V * 3 mA = 6 mW. Matches the result.

Why Other Options Are Wrong:6 µW and 36 µW: off by factors of 1000 due to unit mistakes.

Common Pitfalls:Confusing series vs parallel formulas, or using P = I^2 * R with a branch current but the total resistance, which mixes totals and branches incorrectly.

Final Answer:6 mW