All resistors 30 Ω: R1 is in series with a parallel network of R2 and R3 (each 30 Ω). What is the total resistance seen by the source?

Introduction:
Combining series and parallel resistors correctly is a foundational skill in circuit analysis. This question checks your ability to compute a simple series–parallel equivalent when all component values are the same, a common exam and lab scenario.

Given Data / Assumptions:

• R1 = 30 Ω in series with the parallel pair R2 ∥ R3.
• R2 = 30 Ω and R3 = 30 Ω.
• Ideal conductors and a DC source; no reactive elements.

Concept / Approach:

For series elements, resistances add directly. For two parallel resistors of equal value R, the equivalent is R/2. Therefore, first reduce the parallel pair to a single equivalent, then add the series resistor to obtain the total resistance presented to the source.

Step-by-Step Solution:

Verification / Alternative check:

Using reciprocals: 1 / R_parallel = 1/30 + 1/30 = 2/30 → R_parallel = 15 Ω. Series addition then confirms 45 Ω. Any measurement with a DMM and low test current would read near 45 Ω ignoring lead resistance.

Why Other Options Are Wrong:

• 10 ohms / 20 ohms: Too small; would imply additional parallel paths not present.
• 90 ohms: Would be the sum if all three were in series, which they are not.
• 30 ohms: Ignores the added series effect of R1 plus the reduced parallel pair.

Common Pitfalls:

• Accidentally adding all three (30 + 30 + 30) instead of reducing the parallel first.
• Forgetting that equal parallel resistors halve the value.

Final Answer:

45 ohms