Total resistance of a mixed network: A 470 Ω ‖ 680 Ω combination is in series with the parallel combination of four 2 kΩ resistors. What is the total equivalent resistance?

Introduction / Context:Combining series and parallel subnetworks requires careful stepwise reduction. Equivalent resistance calculations are fundamentals used in analyzing bias networks, filters, and power distribution paths. Accuracy in parallel math prevents order-of-magnitude errors.

Given Data / Assumptions:

• First block: 470 Ω in parallel with 680 Ω.
• Second block: four 2 kΩ resistors in parallel.
• Blocks are connected in series.
• Ideal resistors and DC analysis.

Concept / Approach:

Reduce each parallel subnetwork independently using R_eq = (R1 * R2) / (R1 + R2) for two elements, and R_eq = R/N for N equal elements. Then add the two results because the reduced blocks are in series.

Step-by-Step Solution:

Verification / Alternative check:

Check bounds: 470 ‖ 680 must be less than 470 Ω and greater than 0 Ω; 278 Ω satisfies this. Adding 500 Ω yields about 778 Ω, matching expectations.

Why Other Options Are Wrong:

1,650 Ω and 1,078 Ω are too high; 77.8 Ω is off by a factor of 10 (likely a decimal slip in the parallel step).

Common Pitfalls:

Arithmetic slips in reciprocal sums; forgetting that equivalent resistance of parallel elements is less than the smallest branch.

Final Answer:

778 Ω