Current sharing in equal parallel branches: A 470 Ω ‖ 1.5 kΩ network is in series with a parallel group of five 1 kΩ resistors across a 50 V source. What percentage of the total load current flows through any single 1 kΩ branch?

Introduction / Context:
In circuits combining series and parallel sections, currents in equal-value parallel branches divide uniformly. This question checks whether you can identify the current through one branch of an equal-parallel group that sits in series with another subnetwork.

Given Data / Assumptions:

• First block: 470 Ω in parallel with 1.5 kΩ.
• Second block: five 1 kΩ resistors in parallel (equal values).
• Blocks are in series; source = 50 V.
• We seek the fraction of current through one 1 kΩ branch relative to the total load current entering the five-branch block.

Concept / Approach:

In a set of N identical resistors in parallel, the incoming current splits equally: each branch carries 1/N of the block current. Since these five 1 kΩ resistors are identical, each one conducts 1/5 of the current through that parallel group. The series part of the network does not change this equal division within the parallel group.

Step-by-Step Solution:

Verification / Alternative check:

You can compute the equivalent resistance of the five 1 kΩ in parallel (Req = 200 Ω) and confirm that currents into equal arms must be identical by symmetry; the exact magnitude is unnecessary to find the percentage split.

Why Other Options Are Wrong:

25% corresponds to four equal branches, not five. 50% would be two branches. 100% would mean only one branch exists, which is false here.

Common Pitfalls:

Trying to include the series block's value in the division calculation; forgetting that equal-value parallel branches share current equally regardless of other series elements.

Final Answer:

20%