Worked examples: two reductions from the inside out

Two networks, two reductions. Both follow the same inside-out method: find a purely-series or purely-parallel group, collapse it, repeat.

Example 1 — resistor in series with a parallel pair

A 24 V source drives a 4 Ω resistor (R₁) in series with a parallel combination of a 6 Ω (R₂) and a 12 Ω (R₃) resistor. Find the total resistance, the source current, and the current through R₂.

1. Collapse the parallel pair first.
   $$R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \cdot 12}{6 + 12} = 4\,\Omega$$
2. Add the series resistor. $R_{total} = R_1 + R_{23} = 4 + 4 = 8\,\Omega$.
3. Source current. $I_{source} = V / R_{total} = 24 / 8 = 3\,\text{A}$.
   This 3 A flows through R₁ and into the parallel block.
4. Voltage across the parallel block. $V_{23} = I_{source} \cdot R_{23} = 3 \cdot 4 = 12\,\text{V}$.
   This 12 V sits across both R₂ and R₃.
5. Branch current through R₂. $I_2 = V_{23} / R_2 = 12 / 6 = 2\,\text{A}$.

Sanity check: the other branch current $I_3 = 12 / 12 = 1\,\text{A}$. $I_2 + I_3 = 3\,\text{A} = I_{source}$ — KCL holds at the node, the arithmetic survives.

Example 2 — parallel chunk with a series tail

A 12 V source feeds a 10 Ω and a 40 Ω in parallel; that parallel pair sits in series with two more resistors, 3 Ω and 1 Ω. Find the total resistance and the power dissipated by the 40 Ω resistor.

1. Collapse the parallel pair.
   $$R_{\parallel} = \frac{10 \cdot 40}{10 + 40} = \frac{400}{50} = 8\,\Omega$$
2. Add the two series resistors. $R_{total} = 8 + 3 + 1 = 12\,\Omega$.
3. Source current. $I_{source} = 12 / 12 = 1\,\text{A}$.
4. Voltage across the parallel pair. $V_{\parallel} = I_{source} \cdot R_{\parallel} = 1 \cdot 8 = 8\,\text{V}$.
5. Current through the 40 Ω branch. $I_{40} = 8 / 40 = 0.2\,\text{A}$.
6. Power in the 40 Ω. $P_{40} = I^2 R = (0.2)^2 \cdot 40 = 1.6\,\text{W}$.