Calculate the effective resistance across AB :
Topic: Current, resistance and circuits
Answer
From the circuit: the 5 Ω and 4 Ω are in series with each other, that combination is in parallel with the 3 Ω, and the 8 Ω is in series with the whole arrangement.
Step 1 — the series pair: R₁ = 5 + 4 = 9 Ω
Step 2 — that 9 Ω in parallel with the 3 Ω: R₂ = (9 × 3) ÷ (9 + 3) = 27 ÷ 12 = 2.25 Ω
Step 3 — the 8 Ω in series with R₂: R = 8 + 2.25 = 10.25 Ω
Effective resistance across AB = 10.25 Ω
Read the diagram before you reach for a formula. The 5 Ω and 4 Ω sit end to end in the upper path, so they add. That upper path and the 3 Ω both run between the same two junctions, so those two are in parallel. Only the 8 Ω is in line with everything else.
Check the parallel step against the rule: 2.25 Ω is smaller than 3 Ω, the smaller of the two branches ✓. Adding all four resistors in series would give 20 Ω, which is the answer you get by not looking at the diagram at all.