Worked examples — Compute equivalent resistance in mixed networks

The scenario matrix

Before solving anything, let's list every distinct case-class a two-terminal resistor problem can throw at you. Each worked example below is tagged with the cell it covers.

Two symbols we will lean on, stated in plain words first:

Ex 1 — Case A + B: series feeding a parallel pair

Forecast: $R_2\parallel R_3$ must come out below $6\Omega$ (the smaller branch). Then we add $8$. Guess: somewhere around $12\Omega$.

Look at figure s01: terminal $A$ on the left connects through the blue $R_1$ to node $C$; from $C$ two arcs — the yellow $R_2$ and the green $R_3$ — both fan out to terminal $B$ on the right. See the red arrow: it flags that both yellow and green touch the same pair of nodes ($C$ and $B$), which is the visual signature of "parallel".

1. Collapse the parallel pair $R_2\parallel R_3$. $R_{23}=\frac{R_2{R}_3}{R_2+R_3}=\frac{12\cdot 6}{12+6}=\frac{72}{18}=4\Omega$ Why this step? $R_2$ and $R_3$ touch both the same two nodes ($C$ and $B$) → same voltage → parallel. Product-over-sum is legal because there are exactly two.
2. Series-add $R_1$. $R_{AB}=R_1+R_{23}=8+4=12\Omega$ Why this step? After collapsing, node $C$ now joins only $R_1$ and the $4\Omega$ block, nothing else → same current → series.

Ex 2 — Case C: three in parallel (no product-over-sum!)

Forecast: three branches → answer must drop below the smallest ($2\Omega$). Guess: about $1\Omega$.

1. Add conductances $G=1/R$, because product-over-sum only works for two. $G_{AB}=G_1+G_2+G_3=\frac{1}{6}+\frac{1}{3}+\frac{1}{2}$ Why this step? Same voltage across all three → currents add (KCL) → the ease-of-flow numbers $G$ add. This is exactly the parallel rule from Conductance and Admittance.
2. Common denominator (6). $G_{AB}=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}=\frac{6}{6}=1{\Omega }^{-1}$
3. Flip $G$ back to $R$ — the classic trap is stopping at $G$. $R_{AB}=\frac{1}{G_{AB}}=\frac{1}{1}=1\Omega$ Why this step? Step 2 gave the conductance $G_{AB}$, not the resistance. Invert to recover $R_{AB}$.

Ex 3 — Case D: series first, then parallel

Forecast: the series pair is $10\Omega$, parallel with another $10\Omega$ → two equal branches → must halve to $5\Omega$.

1. Series inner group first. $R_{12}=R_1+R_2=5+5=10\Omega$ Why this step? $R_1$–$R_2$ form one path carrying the same current before rejoining $R_3$'s far node. Collapse the inner group before combining across terminals.
2. Parallel with $R_3$. $R_{eq}=\frac{R_{12}{R}_3}{R_{12}+R_3}=\frac{10\cdot 10}{10+10}=\frac{100}{20}=5\Omega$ Why this step? The collapsed $10\Omega$ block and $R_3$ now share both end nodes ($A$ and $B$) → same voltage across each → parallel. And since it is now exactly two resistors, product-over-sum is legal again.

Ex 4 — Case E: a three-rung ladder

Forecast: the C→D→B branch is $4+8=12\Omega$, parallel with $6\Omega$ → below $6$. Add $4$. Guess: around $8\Omega$.

Look at figure s02: $A$ feeds blue $R_1$ into node $C$. From $C$ the yellow $R_2$ takes the direct top road to $B$; the lower road goes green $R_3$ to node $D$, then red $R_4$ on to $B$. Follow the white arrow pointing at $D$ — it marks that $R_3,R_4$ share $D$ alone, so you must collapse that far branch first ($R_3+R_4=12\Omega$) before anything else lines up as parallel.

1. Series in the far branch. Start at the end farthest from the terminals. $R_{34}=R_3+R_4=4+8=12\Omega$ Why this step? $R_3,R_4$ share node $D$ which touches nothing else → same current → series.
2. Parallel at node $C$. $R_C=\frac{R_2{R}_{34}}{R_2+R_{34}}=\frac{6\cdot 12}{6+12}=\frac{72}{18}=4\Omega$ Why this step? Now $R_2$ and the $12\Omega$ block both span $C$ to $B$ → same voltage → parallel.
3. Series with $R_1$. $R_{AB}=R_1+R_C=4+4=8\Omega$

Ex 5 — Case F: a short circuit steals the current

Forecast: a zero-resistance wire is a free path — current abandons $R_1$ entirely. So the parallel block collapses to $0\Omega$, and only $R_2$ survives.

Look at figure s03: between $A$ and node $M$ two paths sit side by side — the blue $R_1$ arc on top and a flat yellow bare wire (labelled $0$) below; after $M$ the green $R_2$ runs to $B$. The red arrow shows current diving into the yellow wire and skipping $R_1$ entirely — that's why the parallel block reads $0\Omega$.

1. Parallel of $R_1$ with the short ($R_s=0$). $R_{1s}=\frac{R_1\cdot 0}{R_1+0}=\frac{0}{10}=0\Omega$ Why this step? Same voltage across both — but the short has $0\Omega$, so all current takes it. Formally, product-over-sum with one factor $0$ gives $0$. The short wins: it dominates anything in parallel.
2. Series with $R_2$. $R_{AB}=0+R_2=7\Omega$

Ex 6 — Case G: an open circuit carries nothing

Forecast: a broken branch means no current can flow through the $R_1$ path at all — so it might as well not exist. Only $R_2$ remains → $20\Omega$.

1. Series with the open. $R_{1o}=R_1+R_o=10+\infty =\infty$ Why this step? An open in series blocks the whole path — infinite resistance dominates any finite series partner. That branch is dead.
2. Parallel of a dead ($\infty$) branch with $R_2$. $G_{AB}=\frac{1}{R_{1o}}+\frac{1}{R_2}=0+\frac{1}{20}\Rightarrow R_{AB}=20\Omega$ Why this step? A branch of infinite resistance contributes zero conductance ($G\to 0$) — the "$1/\infty =0$" is shorthand for the limit as $R_{1o}\to \infty$. It adds nothing in parallel, so it simply drops out.

Ex 7 — Case H: symmetry / limiting value with $n$ equal resistors

Forecast: every extra branch is another door → resistance keeps falling, heading toward $0$ as $n$ grows without bound.

1. General parallel of $n$ equal branches (conductances add). $G_{AB}=\underset{n\text{ terms}}{\underbrace{\frac{1}{R}+\cdots +\frac{1}{R}}}=\frac{n}{R}\Rightarrow R_{AB}=\frac{R}{n}$ Why this step? Same voltage → conductances add; $n$ identical conductances give $n/R$.
2. Plug $n=4$, $R=12$. $R_{AB}=\frac{12}{4}=3\Omega$
3. Limit $n\to \infty$. ${\lim }_{n\to \infty }\frac{R}{n}=0\Omega$ Why this step? Infinitely many parallel doors = a short circuit. This is the limiting/degenerate boundary, meeting Case F from the other side.

Ex 8 — Case I: real-world word problem

Forecast: the two big $100\Omega$ resistors in parallel halve to $50\Omega$; the two small $2\Omega$ ones add on. Guess $\approx 54\Omega$.

1. Translate objects to a network. $R_1$ and $R_4$ are lone in-line resistors → series ends. $R_2,R_3$ share both nodes → parallel middle. Why this step? Word problems are solved by drawing first: identify which components share nodes.
2. Collapse the parallel middle. $R_{23}=\frac{100\cdot 100}{100+100}=\frac{10000}{200}=50\Omega$
3. Series-add the two in-line resistors. $R_{AB}=R_1+R_{23}+R_4=2+50+2=54\Omega$

Ex 9 — Case J: the exam twist — a balanced bridge

Forecast: this looks like it needs a Delta-Wye Transformation — no pair is cleanly series or parallel because $R_5$ bridges the middle. But look for symmetry first.

Look at figure s04: a diamond — $A$ on the left, $B$ on the right, node $C$ at the top and $D$ at the bottom. Blue $R_1$ and green $R_3$ leave $A$; yellow $R_2$ and red $R_4$ arrive at $B$; the dashed white $R_5$ bridges $C$ to $D$ across the middle. The red arrow marks that $C$ and $D$ end up at equal voltage, so the dashed bridge carries no current and can be erased.

1. Name the node voltages. Put terminal $A$ at some voltage $V_A$ and terminal $B$ at $V_B$. Node $C$ sits between $R_1$ and $R_2$ on the top path; node $D$ sits between $R_3$ and $R_4$ on the bottom path. We want to know $V_C$ vs $V_D$. Why this step? The bridge resistor $R_5$ carries current only if $V_C\ne V_D$ (Ohm's law: $I_5=(V_C-V_D)/R_5$). So the whole trick hinges on comparing those two node voltages.
2. Apply the Voltage Divider to each side. On the top path, $R_1$ and $R_2$ form a divider between $A$ and $B$, so by Voltage Divider the fraction of the $A$-to-$B$ drop that appears across $R_2$ (i.e. from $C$ down to $B$) is: $\frac{V_C-V_B}{V_A-V_B}=\frac{R_2}{R_1+R_2}=\frac{6}{3+6}=\frac{2}{3}.$ On the bottom path the divider $R_3$–$R_4$ gives $D$'s share: $\frac{V_D-V_B}{V_A-V_B}=\frac{R_4}{R_3+R_4}=\frac{6}{3+6}=\frac{2}{3}.$ Why this step? A series pair splits the terminal voltage in proportion to its resistances — that is exactly what the voltage-divider rule says. Computing each middle node's voltage as a fraction is the concrete "why" behind the balance test.
3. Compare — the balance condition. Both fractions equal $\frac{2}{3}$, so $V_C=V_D$. Equivalently, the balance test $\frac{R_1}{R_2}=\frac{R_3}{R_4}$ holds: $\frac{3}{6}=\frac{3}{6}$. $I_5=\frac{V_C-V_D}{R_5}=\frac{0}{100}=0.$ Why this step? Equal node voltages mean zero voltage across $R_5$, so by Ohm's law zero current flows through it — no matter how big or small $R_5$ is.
4. Delete the dead bridge and collapse. With no current through $R_5$, remove it (treat as an open — Case G). Now each side is plain series: $R_{\text{top}}=R_1+R_2=3+6=9\Omega ,R_{\text{bot}}=R_3+R_4=3+6=9\Omega .$ These two $9\Omega$ paths share both terminals $A,B$ → parallel: $R_{AB}=\frac{9\cdot 9}{9+9}=\frac{81}{18}=4.5\Omega .$

Active recall

Connections

• 1.2.02 Compute equivalent resistance in mixed networks (Hinglish) — the parent topic, Hinglish edition
• Ohm's Law — zero voltage across a resistor ⇒ zero current (used in Ex 9)
• Kirchhoff's Voltage Law — series drops add
• Kirchhoff's Current Law — parallel currents add (Ex 2)
• Conductance and Admittance — the $G=1/R$ that sums in parallel
• Voltage Divider — the ratio test in the balanced bridge
• Current Divider — how current splits between parallel branches
• Delta-Wye Transformation — needed only for an unbalanced bridge

Concept Map