Power in a parallel network: A 5 kΩ resistor and a 25 kΩ resistor are connected in parallel across a 10 V DC supply. What is the total power dissipated by the two-resistor network?

Introduction / Context:Evaluating power in parallel circuits is a staple skill for electronics. Here, we use voltage known across each parallel branch to compute total power cleanly using conductances or individual branch powers.

Given Data / Assumptions:

• R1 = 5 kΩ, R2 = 25 kΩ in parallel.
• Supply voltage V = 10 V across both resistors.
• Ideal DC conditions.

Concept / Approach:Total power for parallel resistors can be found via P_total = V^2 * (1/R1 + 1/R2). Since voltage is the same on each branch, this approach is direct and avoids finding branch currents separately.

Step-by-Step Solution:Compute equivalent conductance: 1/R1 + 1/R2 = 1/5000 + 1/25000 = 0.0002 + 0.00004 = 0.00024 S.Compute total power: P_total = V^2 * (sum conductance) = 10^2 * 0.00024 = 100 * 0.00024 = 0.024 W.Convert to milliwatts: 0.024 W = 24 mW.

Verification / Alternative check:Branch powers: P1 = V^2/R1 = 100/5000 = 0.02 W (20 mW); P2 = 100/25000 = 0.004 W (4 mW). Sum = 24 mW. Confirms the result.

Why Other Options Are Wrong:2.4 mW and 3.3 mW are too small; 33 mW is too large relative to the calculated conductance.

Common Pitfalls:Using series formulas by mistake or forgetting that both branches see the full 10 V in parallel.

Final Answer:24 mW