Series vs parallel resistor combinations: Worked examples

This page is a shooting gallery: we list every kind of series/parallel problem that can be thrown at you, then knock each one down with a full worked solution. If a cell in the matrix below never shows up in your homework, you were still made to see it here.

Everything rests on three facts from the parent note: Ohms Law ( $V = I R$ ), KVL (series voltages add), and KCL (parallel currents add). Nothing new is assumed — every symbol below is re-earned as we use it.

The scenario matrix

Every problem you can meet lives in one of these cells. The last column names the example that hits it.

Cell	What makes it tricky	Hit by
Pure series	just add	Ex 1
Pure parallel (unequal)	must invert at the end	Ex 2
Parallel, equal resistors	shortcut $R / n$	Ex 3
Degenerate: a short ( $0 Ω$ ) in parallel	limiting value, $R_{e q} \to 0$	Ex 4
Degenerate: an open ( $\infty Ω$ ) branch	branch vanishes, ignore it	Ex 5
Mixed ladder (innermost first)	nested collapse order	Ex 6
Real-world word problem	translate words → topology	Ex 7
Exam twist: find the unknown $R$	work the formula backwards	Ex 8
Limiting behaviour: one $R \to \infty$ in series/parallel	asymptotic reasoning	Ex 9

Let a resistor be a component that resists current; its size $R$ is in ohms ( $Ω$ ). $R_{e q}$ = the single resistor that behaves the same seen from two terminals.

Ex 1 — Pure series

Worked example Three in a row

$R_{1} = 100 Ω$ , $R_{2} = 220 Ω$ , $R_{3} = 330 Ω$ end-to-end, driven by $V = 12 V$ . Find $R_{e q}$ , the current $I$ , and the drop across $R_{3}$ .

Forecast: guess whether $R_{e q}$ is bigger or smaller than $330$ , and whether $R_{3}$ takes the most or least voltage.

Step 1 — Same current? All three lie on one path with no branch, so the same $I$ flows through each ⇒ series. Why this step? Series is defined by shared current; that lets voltages add via KVL.

Step 2 — Add. $R_{e q} = 100 + 220 + 330 = 650 Ω$ . Why this step? Series ⇒ resistances add.

Step 3 — Total current by Ohm. $I = \frac{V}{R _{e q}} = \frac{12}{650} = 0.01846 A$ . Why this step? The source sees one resistor $R_{e q}$ ; Ohm gives the current it pushes.

Step 4 — Drop across $R_{3}$ . $V_{3} = I R_{3} = 0.01846 \times 330 = 6.09 V$ . Why this step? Same $I$ through $R_{3}$ ; Ohm again. (This is the Voltage Divider in action.)

Verify: $R_{e q} = 650 > 330$ ✓ (series is biggest). Drops sum: $I (100 + 220 + 330) = 0.01846 \times 650 = 12 V$ ✓. Biggest $R$ ate most volts ✓.

Ex 2 — Pure parallel, unequal

Worked example Two branches between the same nodes

$R_{1} = 60 Ω$ and $R_{2} = 40 Ω$ share two nodes at $V = 6 V$ . Find $R_{e q}$ and each branch current.

Forecast: which branch carries more current — the $40 Ω$ or the $60 Ω$ ? Where will $R_{e q}$ land?

Step 1 — Same voltage? Both sit across the same two nodes ⇒ same $V$ ⇒ parallel. Why this step? Parallel is defined by shared voltage; KCL then adds the currents.

Step 2 — Product over sum. $R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{60 \times 40}{100} = 24 Ω$ . Why this step? For exactly two resistors this shortcut equals $\frac{1}{R _{1}} + \frac{1}{R _{2}}$ inverted.

Step 3 — Branch currents. $I_{1} = \frac{6}{60} = 0.1 A$ , $I_{2} = \frac{6}{40} = 0.15 A$ . Why this step? Each branch feels the full $6 V$ ; Ohm per branch. (The Current Divider.)

Verify: $R_{e q} = 24 < 40$ , below the smallest ✓. Total $I = 0.25 A$ ; check $V / R_{e q} = 6/24 = 0.25 A$ ✓. The smaller resistor ( $40$ ) hogs more current ✓.

Ex 3 — Parallel, all equal

Ex 4 — Degenerate: a dead short in parallel

Ex 5 — Degenerate: an open branch

Ex 6 — Mixed ladder, innermost first

Worked example Nested collapse

$R_{1} = 8 Ω$ in series with the parallel pair $R_{2} ∥ R_{3}$ , where $R_{2} = 12 Ω$ , $R_{3} = 6 Ω$ . Driven by $V = 20 V$ . Find $R_{e q}$ and total current.

Forecast: which do you simplify first — the series part or the parallel part?

Step 1 — Innermost = the parallel pair. $R_{23} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4 Ω$ . Why this step? Like nested parentheses, collapse the deepest cluster first. $R_{2}, R_{3}$ share two nodes ⇒ parallel.

Step 2 — Now series. $R_{1}$ carries the same current as the collapsed $R_{23}$ ⇒ series: $R_{e q} = 8 + 4 = 12 Ω$ . Why this step? Once $R_{23}$ is one block, it and $R_{1}$ lie end-to-end.

Step 3 — Total current. $I = \frac{20}{12} = 1.667 A$ .

Verify: $R_{23} = 4 < 6$ (below smallest) ✓; $R_{e q} = 12 > 8$ (series grew it) ✓. This is exactly Equivalent Resistance and Network Reduction.

Ex 7 — Real-world word problem

Worked example The workshop heater

A heater has two identical $20 Ω$ coils. On HIGH they're wired in parallel; on LOW in series. The wall gives $240 V$ . Find the power on each setting. ( $P = V^{2} / R_{e q}$ ; see Power Dissipation in Resistors.)

Forecast: which setting is actually hotter — series or parallel?

Step 1 — HIGH (parallel). $R_{e q} = \frac{20}{2} = 10 Ω$ . Why this step? Two equal in parallel ⇒ $R / n$ . Lower resistance.

Step 2 — Power HIGH. $P = \frac{V ^{2}}{R _{e q}} = \frac{24 0 ^{2}}{10} = \frac{57600}{10} = 5760 W$ . Why this step? Same wall voltage; lower $R$ ⇒ more current ⇒ more power ( $P = V^{2} / R$ ).

Step 3 — LOW (series). $R_{e q} = 20 + 20 = 40 Ω$ , $P = \frac{24 0 ^{2}}{40} = 1440 W$ . Why this step? Series is biggest $R$ ⇒ least power at fixed voltage.

Verify: ratio $5760/1440 = 4$ — parallel is $4 \times$ hotter, matching $R$ dropping by $4 \times$ ( $40 \to 10$ ) ✓. Units: $V^{2} /Ω = W$ ✓. So "parallel = HIGH" is correct.

Ex 8 — Exam twist: solve backwards for an unknown

Ex 9 — Limiting behaviour

Recall

Recall Which cell is this?

Q: " $0 Ω$ across a resistor" and " $\infty Ω$ in series" both do the same drastic thing — what? A ::: Both drive $R_{e q}$ to an extreme by removing the resistor's effect — a parallel short forces $R_{e q} = 0$ ; a series open forces $R_{e q} = \infty$ . Same intuition, mirrored by the same-voltage / same-current duality.

Recall Backwards problems

Q: To find an unknown $R_{2}$ in a parallel target, what's the safe procedure? A ::: Work in conductances: $\frac{1}{R _{2}} = \frac{1}{R _{e q}} - \frac{1}{R _{1}}$ , then invert last.

Connections

Ohms Law — every step used $V = I R$ .
Kirchhoffs Voltage Law & Kirchhoffs Current Law — why series adds voltages / parallel adds currents.
Voltage Divider (Ex 1) & Current Divider (Ex 2).
Equivalent Resistance and Network Reduction — the ladder collapse of Ex 6.
Conductance — the language that made Ex 4, 5, 8 clean.
Power Dissipation in Resistors — the heater in Ex 7.