Identifying the branch with maximum current: A 470 Ω || 1.2 kΩ parallel network is in series with a second network formed by three 3 kΩ resistors in parallel. A 200 V source feeds the series combination. Which single resistor carries the greatest current?

Introduction / Context:
This mixed series–parallel problem asks you to reason about branch currents. Within any parallel group, the branch with smaller resistance carries larger current for the same block voltage. To compare across two parallel blocks in series, find the voltage division between blocks using their equivalent resistances; then compare branch currents inside each block.

Given Data / Assumptions:

• Block A: 470 Ω in parallel with 1.2 kΩ.
• Block B: three equal 3 kΩ resistors in parallel.
• Blocks A and B are connected in series to a 200 V source.
• Ideal components; DC operation.

Concept / Approach:
First compute each block’s equivalent resistance to find how the 200 V divides between blocks. Then, within each parallel block, branch current I_branch = V_block / R_branch. The highest current will be in the smallest-resistance branch that also sits across the larger block voltage.

Step-by-Step Solution:

Verification / Alternative check:
Sanity check: In any given parallel pair at the same voltage, the lower resistance must draw more current. Even after voltage division, the computed numbers confirm the 470 Ω branch dominates.

Why Other Options Are Wrong:

• 1.2 kΩ: Higher resistance in the same parallel block; therefore less current.
• 3 kΩ: Each 3 kΩ branch carries less current than the 470 Ω branch at their respective block voltages.
• '470 Ω or 1.2 kΩ': Ambiguous; only 470 Ω is correct.

Common Pitfalls:

• Comparing resistances without accounting for block voltage division.
• Confusing series and parallel roles when simplifying the network.

Final Answer:
470 Ω