Series of two parallel blocks with a 100 V source: The combination (6.8 kΩ ∥ 10 kΩ) is in series with (2.2 kΩ ∥ 1 kΩ). With 100 V across the entire network, which resistor(s) experience the greatest voltage drop?

Introduction / Context:This question blends series–parallel reduction with voltage division. When parallel blocks are in series, the source voltage divides according to the equivalent resistances of the blocks. Within any one parallel block, each branch shares the same block voltage. We identify which component(s) see the largest voltage by first finding how the total voltage splits between blocks.

Given Data / Assumptions:

• Block A: 6.8 kΩ in parallel with 10 kΩ.
• Block B: 2.2 kΩ in parallel with 1 kΩ.
• Blocks A and B are connected in series across 100 V.
• Ideal resistors, DC operation.

Concept / Approach:Compute the equivalent resistance of each block; the larger equivalent takes a larger portion of the total voltage (voltage division). Then, inside each parallel block, all branch resistors see the same block voltage. The branch(es) in the higher-voltage block will therefore have the greatest drop.

Step-by-Step Solution:

Verification / Alternative check:Approximate split: V_A ≈ 100 * 4.0476 / (4.0476 + 0.6875) ≈ 85.5 V; V_B ≈ 14.5 V. Each branch inside A sees ≈ 85.5 V (larger than any drop in B), confirming the selection.

Why Other Options Are Wrong:

• 6.8 kΩ (alone) or 2.2 kΩ or 2.2 kΩ and 1 kΩ: All branches in Block B see the smaller block voltage; none can exceed Block A’s drop.

Common Pitfalls:

• Comparing individual branch resistances without accounting for the block-level voltage division first.

Final Answer:6.8 kΩ and 10 kΩ