Worked examples: two-branch divider and a household overload

One clean maths problem to lock in the formulas, one real-world problem where a parallel topology creates the very failure mode engineers have to plan for.

Example 1 — two resistors in parallel

A 12 V supply feeds a 4 Ω and a 12 Ω resistor in parallel. Find the total resistance, the source current, and each branch current.

1. Given. $V = 12\,\text{V}$, $R_1 = 4\,\Omega$, $R_2 = 12\,\Omega$.
2. Total resistance (two-in-parallel shortcut).
   $$R_{total} = \frac{R_1 R_2}{R_1 + R_2} = \frac{4 \cdot 12}{4 + 12} = \frac{48}{16} = 3\,\Omega$$
   
   Note: 3 Ω is smaller than the smallest branch (4 Ω). Always.
3. Source current. $I_{source} = V / R_{total} = 12 / 3 = 4\,\text{A}$.
4. Branch currents (V constant, use Ohm's law each branch). $I_1 = 12 / 4 = 3\,\text{A}$, $I_2 = 12 / 12 = 1\,\text{A}$.
5. Check. $I_1 + I_2 = 3 + 1 = 4\,\text{A}$, which matches the source current. KCL holds.

The smaller 4 Ω resistor hogs 3/4 of the current; the 12 Ω gets the remaining 1/4. Same pattern every time: the smaller resistor carries the bigger share of the current.

Example 2 — a household overload

A 230 V circuit feeds three appliances in parallel: a 2 300 W kettle (≈ 23 Ω), a 1 000 W toaster (≈ 52.9 Ω), and a 1 500 W heater (≈ 35.3 Ω). The breaker is rated 16 A. Does it hold when all three run at once?

1. Branch currents direct from P = V · I.
   Kettle: $I_k = 2300 / 230 = 10\,\text{A}$.
   Toaster: $I_t = 1000 / 230 \approx 4.35\,\text{A}$.
   Heater: $I_h = 1500 / 230 \approx 6.52\,\text{A}$.
2. Total (parallel → branches add). $I_{total} = 10 + 4.35 + 6.52 \approx 20.9\,\text{A}$.
3. Verdict. 20.9 A > 16 A → the breaker trips. You can run the kettle plus either the toaster or the heater, but not all three at once.