Compute equivalent resistance in mixed networks

The two atomic rules (derived, not memorized)

The algorithm (the 80/20 core)

Figure — Compute equivalent resistance in mixed networks

Worked Example 1 — one parallel inside a series

Find $R_{AB}$ : $R_1 = 6\,\Omega$ in series with ( $R_2 = 4\,\Omega \parallel R_3 = 4\,\Omega$ ).

Step 1 — collapse the parallel pair. $R_{23} = \frac{4\cdot 4}{4+4} = \frac{16}{8} = 2\,\Omega$ Why this step? $R_2$ and $R_3$ share both nodes → same voltage → parallel rule.

Step 2 — series add. $R_{AB} = R_1 + R_{23} = 6 + 2 = 8\,\Omega$ Why this step? After collapsing, $R_1$ and the $2\,\Omega$ block share a single node with nothing else → same current → series.

Worked Example 2 — nested series-then-parallel

$R_1 = 2\,\Omega$ and $R_2 = 4\,\Omega$ are in series; that combo is in parallel with $R_3 = 6\,\Omega$ .

Step 1 — series first (it's the inner group). $R_{12} = 2 + 4 = 6\,\Omega$ Why this step? $R_1$ – $R_2$ form one path carrying identical current before rejoining $R_3$ 's node.

Step 2 — parallel with $R_3$ . $R_{eq} = \frac{6 \cdot 6}{6+6} = \frac{36}{12} = 3\,\Omega$ Why this step? The $6\,\Omega$ path and $R_3$ now share both terminals → same voltage.

Worked Example 3 — a ladder (Forecast-then-Verify)

Terminals A–B. $R_1=10\,\Omega$ from A, then node C. From C: $R_2=10\,\Omega$ to B directly, and $R_3=10\,\Omega$ to node D then $R_4=10\,\Omega$ to B.

Step 1 — series in the lower branch. $R_{34} = 10 + 10 = 20\,\Omega \quad(\text{same current through }R_3,R_4)$

Step 2 — parallel at node C. $R_{C} = \frac{10\cdot 20}{10+20} = \frac{200}{30} = 6.67\,\Omega$

Step 3 — series with $R_1$ . $R_{AB} = 10 + 6.67 = 16.67\,\Omega$ Forecast (17) matched the verified answer (16.67) → structure understood. ✓

Common mistake Steel-manning the classic errors

Wrong idea 1: "They look next to each other, so they're in series." Why it feels right: physical layout looks linear. The fix: series needs a shared node with nothing else attached. If a third wire taps that node, current splits — not series.

Wrong idea 2: Adding parallel resistors directly ( $R_1+R_2$ ). Why it feels right: series addition is the first rule you learn, so you over-apply it. The fix: parallel adds conductances $1/R$ , not resistances. Result is always smaller — use that as a bug alarm.

Wrong idea 3: Forgetting to flip the reciprocal. You compute $\frac1{R_1}+\frac1{R_2}=0.5$ and write $R=0.5\,\Omega$ . The fix: that $0.5$ is $1/R_{eq}$ ; the answer is $R_{eq}=1/0.5 = 2\,\Omega$ .

Recall Feynman: explain to a 12-year-old

Imagine water pipes. Putting pipes end-to-end (series) makes one long skinny pipe — harder for water, so resistance adds up. Putting pipes side-by-side (parallel) gives water more doors to go through — easier, so total resistance drops below even the narrowest single pipe. To find the "one pipe" that acts like the whole messy plumbing, you keep merging little side-by-side and end-to-end bunches until only one pipe is left between the tap and the drain.

Active recall

Two resistors are in series when…

they share exactly one node that connects to nothing else, so they carry the same current.

Two resistors are in parallel when…

they share both end nodes, so they have the same voltage across them.

Series equivalent resistance formula

R_{eq}=R_1+R_2+\dots

(resistances add).

Parallel equivalent resistance formula

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\dots

(conductances add).

Product-over-sum rule and its restriction

R_{eq}=\frac{R_1R_2}{R_1+R_2}

, valid ONLY for exactly two resistors.

Why does series increase resistance?

Current travels one long path; voltage drops add (KVL) → same current, larger total drop.

Why does parallel decrease resistance?

Extra branches give current more paths; currents add (KCL) → same voltage, larger total current, smaller effective R.

Strategy to solve a mixed network

Label nodes, collapse innermost series/parallel groups one at a time, redraw, repeat until one resistor remains.

What if nothing is purely series or parallel?

It's a bridge network; apply a Y–Δ (delta-wye) transformation.

Common bug alarm after a parallel step

R_{eq}

came out bigger than the smallest branch, you made an error (parallel must shrink it).

Connections

Ohm's Law — the linearity that lets a box equal one resistor
Kirchhoff's Voltage Law — justifies series addition
Kirchhoff's Current Law — justifies parallel addition
Delta-Wye Transformation — for non-series-parallel bridges
Voltage Divider and Current Divider — direct consequences of these rules
Conductance and Admittance — parallel adds conductances

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, "mixed network" ka matlab hai resistors ka aisa jaal jisme kuch series me hain aur kuch parallel me. Poore circuit ko ek saath solve karne ki koshish mat karo — ye sabse badi galti hai. Instead, andar se bahar ki taraf collapse karo: sabse pehle koi chhota clearly series ya parallel group dhundo, usko ek single resistor se replace karo, phir circuit dobara draw karo. Repeat karte raho jab tak sirf ek hi resistor bach jaaye do terminals ke beech.

Do basic rule yaad rakho. Series tab hota hai jab do resistor ek hi node share karein aur us node pe koi teesra wire na juda ho — tab dono me same current behta hai, isliye resistance add ho jaati hai ( $R_1+R_2$ ). Parallel tab hota hai jab dono resistor ke dono ends same nodes pe judein — tab dono pe same voltage hota hai, aur $1/R$ (conductance) add hota hai. Isiliye parallel ka result hamesha sabse chhote branch se bhi chhota hota hai. Agar tumhara parallel answer bada aa raha hai, samajh jao galti ho gayi.

Ek zabardast trick hai product-over-sum: $R = \frac{R_1 R_2}{R_1+R_2}$ — lekin ye sirf do resistors ke liye chalta hai. Teen ya zyada ke liye wapas reciprocal wala formula use karo. Aur exam me time bachane ke liye pehle forecast karo: "parallel hai toh answer chhota hona chahiye, series hai toh bada" — phir calculate karke verify karo. Isse silly mistakes turant pakad me aa jaati hain. Agar koi circuit purely series-parallel me todaa hi na ja sake (bridge type), tab Y–Δ transformation lagana padta hai — wo alag topic hai.

Compute equivalent resistance in mixed networks

The two atomic rules (derived, not memorized)

The algorithm (the 80/20 core)

Worked Example 1 — one parallel inside a series

Worked Example 2 — nested series-then-parallel

Worked Example 3 — a ladder (Forecast-then-Verify)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Circuit Analysis Fundamentals

Connections