12 V across a series–parallel network: what is the total current?

The question

A 12 V battery is connected to R₁ = 6 Ω in series with a parallel combination of R₂ = 12 Ω and R₃ = 4 Ω. Find the total current drawn from the battery.

What to notice first

Identify the topology before touching Ohm's law. R₂ and R₃ share both of their nodes, so they are in parallel and must be collapsed into a single equivalent resistance first. Only then does the circuit become one loop, where a single current flows.

Working it through

Collapse the parallel pair. For exactly two resistors the product-over-sum form is quickest:

Sanity-check that number before continuing: a parallel combination is always SMALLER than its smallest member, and 3 Ω is indeed below 4 Ω.

Now the circuit is one loop, so the remaining resistances simply add:

Apply Ohm's law to the whole loop:

The answer

That 1.33 A flows through R₁ and then splits between R₂ and R₃ in inverse proportion to their resistances — 0.33 A through the $12\Omega$ and 1.00 A through the $4\Omega$, which add back to the total.

The mistake this one catches

Prefer to watch it? The lesson above walks through every line.