Series–parallel reduction rule: For a network where R1 is in series with a parallel group of R2, R3, and R4, the total resistance is given by RT = R1 + (R2 || R3 || R4). Determine whether this formula is valid for that topology.

Introduction / Context:
Reducing mixed series–parallel resistor networks is a core skill in circuit analysis. Recognizing when a single element is in series with a parallel group allows a quick two-step computation of the equivalent resistance, speeding up hand calculations and sanity checks in design work.

Given Data / Assumptions:

• Topology: R2, R3, and R4 share the same two nodes (are in parallel).
• R1 is connected in series with that parallel combination.
• Linear, time-invariant resistors; frequency effects neglected.
• No dependent sources or bridging connections that would break simple series/parallel relations.

Concept / Approach:
Series elements carry the same current and their resistances add arithmetically. Parallel elements share the same voltage and their conductances add: R_parallel = 1 / (1/R2 + 1/R3 + 1/R4). Combining both rules yields RT = R1 + R_parallel for this specific topology.

Step-by-Step Solution:

Verification / Alternative check:
Pick numeric values (e.g., R2=300 Ω, R3=600 Ω, R4=600 Ω). Compute R_p=150 Ω, then RT=R1+150 Ω, confirming additivity of the series element with the parallel block.

Why Other Options Are Wrong:

• Incorrect / AC only: The rule is purely topological and applies equally in DC.
• Valid only when R2=R3=R4 or when R1 is very large: Equality or magnitude ratios are not required for the formula to hold.

Common Pitfalls:
Accidentally including a bridging element that spoils pure parallel, or adding all four resistors directly as if series. Always confirm node sharing before applying ||.

Final Answer:
Correct