Series and parallel resistance: Worked examples — Physics deep dive

This page is the exhaustive worked-example lab for Series and parallel resistance. The parent note gave you four clean examples. Here we deliberately hunt down every case class a problem can be built from — including the weird, degenerate, and exam-trap ones — so you never meet a scenario you have not already seen solved.

Everything here rests on only two tools, both defined in the parent and its links:

Ohm's Law: $V = I R$ — voltage across a resistor equals its current times its resistance.
Kirchhoff's Laws: current in = current out at a node (KCL); voltages around a loop add to zero (KVL).

If any symbol below feels unfamiliar, it was built in the parent note — read that first.

The scenario matrix

Before working anything, let us enumerate the full space of cases. Every resistor-network problem you will ever be handed is one (or a stitch) of these cells:

#	Case class	What makes it special	Covered by
A	Pure series	one path, current shared	Ex 1
B	Pure parallel (two)	two nodes, voltage shared	Ex 2
C	Parallel, unequal branches	current splits unequally	Ex 2
D	Mixed (series + parallel block)	collapse in stages	Ex 3
E	Degenerate: short circuit ( $R \to 0$ branch)	a $0 Ω$ path steals everything	Ex 4
F	Degenerate: open circuit ( $R \to \infty$ branch)	a broken branch carries nothing	Ex 5
G	Equal resistors, $n$ of them	fast $R / n$ and $n R$ rules	Ex 6
H	Limiting behaviour: one $R$ ≫ the other	which one "wins"?	Ex 7
I	Real-world word problem (household wiring)	translate words → circuit	Ex 8
J	Exam twist: symmetric bridge / hidden equal split	spot the trick, avoid full algebra	Ex 9

Case A — Pure series

Worked example Ex 1 · Cell A — same current, resistances sum

Three resistors in one line: $R_{1} = 2 Ω$ , $R_{2} = 3 Ω$ , $R_{3} = 5 Ω$ , across a $V = 20$ V battery. Find $R_{e q}$ , the current $I$ , and each voltage drop.

Forecast: One path, so the current is the same everywhere. Guess: is $R_{e q}$ bigger or smaller than $5 Ω$ ? (Bigger — series always piles up.)

Step 1. $R_{e q} = R_{1} + R_{2} + R_{3} = 2 + 3 + 5 = 10 Ω$ . Why this step? One unbranched path → same $I$ through all → by KVL the drops add → resistances add.

Step 2. $I = V / R_{e q} = 20/10 = 2$ A. Why this step? Ohm's Law applied to the whole collapsed chain.

Step 3. Drops: $V_{1} = I R_{1} = 2 \cdot 2 = 4$ V, $V_{2} = 2 \cdot 3 = 6$ V, $V_{3} = 2 \cdot 5 = 10$ V. Why this step? Same $I$ , so each drop is just $I R_{k}$ .

Verify: $V_{1} + V_{2} + V_{3} = 4 + 6 + 10 = 20$ V $= V$ ✓ (KVL). And $10 Ω > 5 Ω$ , as forecast. Units: $A \cdot Ω = V$ ✓.

Cases B & C — Pure parallel, unequal split

Worked example Ex 2 · Cells B + C — same voltage, currents split unequally

$R_{1} = 4 Ω$ and $R_{2} = 12 Ω$ between the same two nodes, $V = 24$ V. Find $R_{e q}$ , the branch currents, and the total.

Forecast: Both feel the full $24$ V. Which branch carries more current — the $4 Ω$ or the $12 Ω$ ? (The smaller one, $4 Ω$ : less obstacle → more flow.)

Step 1. Product over sum: $R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{4 \cdot 12}{4 + 12} = \frac{48}{16} = 3 Ω$ . Why this step? Same two nodes → same $V$ → conductances add. Look at the figure: both green paths bridge the identical node pair (the blue dots).

Step 2. Branch currents (each sees the full $V$ ): $I_{1} = V / R_{1} = 24/4 = 6$ A, $I_{2} = V / R_{2} = 24/12 = 2$ A. Why this step? Ohm's Law per branch — the voltage is shared, the current is not.

Step 3. Total: $I = I_{1} + I_{2} = 6 + 2 = 8$ A. Why this step? KCL — the branch currents recombine.

Verify: Check via equivalent: $I = V / R_{e q} = 24/3 = 8$ A ✓. Note $I_{1} = 3 I_{2}$ exactly, and $R_{2} = 3 R_{1}$ — the smaller resistor grabbed three times the current (Cell C, the unequal split). And $R_{e q} = 3 Ω < 4 Ω$ (smaller than the smallest) ✓.

Case D — Mixed network, collapse in stages

Worked example Ex 3 · Cell D — series of a resistor with a parallel block

$R_{1} = 6 Ω$ in series with a parallel pair $R_{2} = 10 Ω$ and $R_{3} = 15 Ω$ . Supply $V = 30$ V.

Forecast: Which do you collapse first — the series part or the parallel block? (The parallel block: you must reduce it to one number before you can add it in series.)

Step 1. Parallel block: $R_{23} = \frac{R _{2} R _{3}}{R _{2} + R _{3}} = \frac{10 \cdot 15}{25} = \frac{150}{25} = 6 Ω$ . Why this step? $R_{2}, R_{3}$ share two nodes (the yellow node pair in the figure). Collapse them into a single equivalent block.

Step 2. Now it is a plain series chain: $R_{e q} = R_{1} + R_{23} = 6 + 6 = 12 Ω$ . Why this step? The same current flows through $R_{1}$ and then the whole block — one path → add.

Step 3. Battery current: $I = V / R_{e q} = 30/12 = 2.5$ A. Why this step? Ohm on the fully collapsed circuit.

Step 4. Voltage across the parallel block: $V_{23} = I \cdot R_{23} = 2.5 \cdot 6 = 15$ V. Then $I_{2} = 15/10 = 1.5$ A, $I_{3} = 15/15 = 1$ A. Why this step? The block's own voltage is shared by both its branches; each branch splits by Ohm's Law.

Verify: $I_{2} + I_{3} = 1.5 + 1 = 2.5$ A $= I$ ✓ (KCL). Drop on $R_{1} = 2.5 \cdot 6 = 15$ V, plus block $15$ V $= 30$ V ✓ (KVL).

Case E — Degenerate: a short circuit ( $R \to 0$ )

Worked example Ex 4 · Cell E — a wire shorting out a resistor

$R_{1} = 8 Ω$ in parallel with an ideal wire $R_{2} = 0 Ω$ , across $V = 12$ V.

Forecast: What is $R_{e q}$ ? Where does the current go? (Guess $R_{e q} = 0$ , and all current dodges $R_{1}$ .)

Step 1. $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{1}{8} + \frac{1}{0}$ . The term $1/0 \to \infty$ , so $1/ R_{e q} \to \infty$ , hence $R_{e q} = 0$ . Why this step? Take the parallel formula literally to its limit. Infinite conductance dominates.

Step 2. Voltage across the pair $= 0$ (it is $I \cdot R_{e q} = I \cdot 0$ ). So $I_{1} = V_{b l oc k} / R_{1} = 0/8 = 0$ A through $R_{1}$ . Why this step? Both branches share the block voltage; if the block has $0 Ω$ it drops $0$ V, so $R_{1}$ feels nothing.

Step 3. All current goes through the wire: $I_{2} = I_{t o t a l}$ (limited only by the rest of the circuit / the source). Why this step? KCL — total current must still flow, and only the wire offers a path with any voltage-independent capacity.

Verify: Sanity: $R_{e q} = 0 < 8 Ω$ (smaller than the smallest — the shortcut rule survives at the limit) ✓. Two-resistor shortcut also agrees: $\frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{8 \cdot 0}{8 + 0} = \frac{0}{8} = 0$ ✓.

Case F — Degenerate: an open circuit ( $R \to \infty$ )

Worked example Ex 5 · Cell F — a broken branch in series and in parallel

Two sub-questions with the same broken resistor $R_{2} = \infty$ (open). (a) In series: $R_{1} = 5 Ω$ in series with an open $R_{2}$ , across $V = 12$ V. (b) In parallel: $R_{1} = 5 Ω$ in parallel with an open $R_{2}$ , across $V = 12$ V.

Forecast: In which arrangement does the circuit still work — series or parallel? (Parallel: the open branch is simply ignored. Series: an open breaks the only path → no current at all.)

Step 1 (a). Series: $R_{e q} = R_{1} + R_{2} = 5 + \infty = \infty$ . Current $I = V / R_{e q} = 12/\infty = 0$ A. Why this step? One path, and it is cut. Charge cannot cross the gap → the whole line is dead.

Step 2 (b). Parallel: $\frac{1}{R _{e q}} = \frac{1}{5} + \frac{1}{\infty} = \frac{1}{5} + 0 = \frac{1}{5}$ , so $R_{e q} = 5 Ω$ . Why this step? An infinite resistor has zero conductance — it adds nothing to the parallel sum. The good branch alone survives.

Step 3 (b). Current: $I = V / R_{e q} = 12/5 = 2.4$ A, entirely through $R_{1}$ ; $I_{2} = 12/\infty = 0$ A. Why this step? Ohm on the surviving branch; the open branch carries nothing.

Verify: (a) gives $0$ A — a broken series chain is dead, matching intuition ✓. (b) gives $2.4$ A — the open branch vanished cleanly, $R_{e q}$ equals the lone resistor ✓. Contrast this with Ex 4: short in parallel = $R_{e q} \to 0$ ; open in parallel = branch ignored. Opposite extremes.

Case G — $n$ equal resistors (fast rules)

Case H — Limiting behaviour: one resistor dominates

Worked example Ex 7 · Cell H — what happens as one

$R$ swamps the other? Two in parallel: $R_{1} = 2 Ω$ fixed, and $R_{2}$ we let grow: try $R_{2} = 2$ , then $20$ , then $2000 Ω$ . Track $R_{e q}$ .

Forecast: As $R_{2} \to \infty$ , which value does $R_{e q}$ approach? (It approaches $R_{1} = 2 Ω$ — the fat resistor becomes an open, so the small one alone remains.)

Step 1. $R_{2} = 2$ : $R_{e q} = \frac{2 \cdot 2}{2 + 2} = \frac{4}{4} = 1 Ω$ (equal resistors → half). Step 2. $R_{2} = 20$ : $R_{e q} = \frac{2 \cdot 20}{22} = \frac{40}{22} \approx 1.818 Ω$ . Step 3. $R_{2} = 2000$ : $R_{e q} = \frac{2 \cdot 2000}{2002} = \frac{4000}{2002} \approx 1.998 Ω$ . Why these steps? We are watching the limit $lim_{R_{2} \to \infty} \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ . Divide top and bottom by $R_{2}$ : $\frac{R _{1}}{R _{1} / R _{2} + 1} \to \frac{R _{1}}{0 + 1} = R_{1}$ . The curve in the figure (red) flattens toward the dashed line $R_{e q} = 2$ .

Verify: The sequence $1 \to 1.818 \to 1.998$ climbs monotonically toward $2$ but never reaches it — exactly the horizontal asymptote $R_{e q} = R_{1}$ ✓. Physical reading: a very large parallel resistor is almost an open (Cell F), so it barely changes $R_{e q}$ .

Case I — Real-world word problem

Worked example Ex 8 · Cell I — household appliances on one outlet

A wall outlet supplies $V = 120$ V. You plug in a lamp of $240 Ω$ and a heater of $24 Ω$ into the same outlet (so they share the outlet's two terminals). Find each appliance's current and the total drawn from the outlet.

Forecast: Are these in series or parallel? (Parallel — both are wired across the same two outlet terminals, so both get the full $120$ V. This is why appliances work independently.) Which draws more current — lamp or heater?

Step 1. Recognise parallel: both between the same node pair → same $V = 120$ V. Why this step? Cell classification first, always — see the parent's mistake "same-looking wires."

Step 2. Lamp: $I_{l am p} = 120/240 = 0.5$ A. Heater: $I_{h e a t er} = 120/24 = 5$ A. Why this step? Ohm's Law per branch; each sees the full mains voltage.

Step 3. Total from outlet: $I = 0.5 + 5 = 5.5$ A (KCL). Step 4. Equivalent: $R_{e q} = \frac{240 \cdot 24}{240 + 24} = \frac{5760}{264} \approx 21.82 Ω$ .

Verify: $I = V / R_{e q} = 120/21.82 \approx 5.5$ A ✓. The low-resistance heater grabs $10 \times$ the lamp's current — this is why heaters trip breakers. Note $R_{e q} \approx 21.8 Ω < 24 Ω$ (below the smallest) ✓.

Case J — Exam twist: symmetry beats algebra

Worked example Ex 9 · Cell J — a balanced bridge (spot the trick)

Four resistors form a diamond between input node $P$ and output node $Q$ : top path $P \to M \to Q$ has $R_{a} = 10 Ω$ then $R_{b} = 20 Ω$ ; bottom path $P \to N \to Q$ has $R_{c} = 10 Ω$ then $R_{d} = 20 Ω$ . A fifth resistor $R_{g} = 50 Ω$ bridges $M$ to $N$ . Supply across $P$ – $Q$ is $V = 30$ V. Find the current through the bridge $R_{g}$ , then $R_{e q}$ .

Forecast: The two paths are identical ( $10$ then $20$ on both). Do you need the messy bridge algebra? (No — by symmetry $M$ and $N$ sit at the same potential, so no current crosses $R_{g}$ . It can be deleted!)

Step 1. Compare the two arms. Top: $R_{a} : R_{b} = 10 : 20$ . Bottom: $R_{c} : R_{d} = 10 : 20$ . The ratios match → balanced bridge (Wheatstone Bridge condition). Why this step? When $R_{a} / R_{b} = R_{c} / R_{d}$ , the voltage at $M$ equals the voltage at $N$ . Look at the figure: both midpoints land on the same dashed equipotential line.

Step 2. Equal potentials at $M$ , $N$ → voltage across $R_{g}$ is $0$ → $I_{g} = 0/ R_{g} = 0$ A. Why this step? Ohm's Law: no potential difference, no current. The bridge is dead weight.

Step 3. Delete $R_{g}$ . Now two independent series arms in parallel. Each arm: $10 + 20 = 30 Ω$ . Two equal $30 Ω$ arms in parallel: $R_{e q} = 30/2 = 15 Ω$ . Why this step? With the bridge removed the network is plain series-then-parallel (Cell D + Cell G).

Verify: Battery current $I = V / R_{e q} = 30/15 = 2$ A, splitting $1$ A into each arm. Potential at $M$ (from $P$ ): drop across $R_{a} = 1 \cdot 10 = 10$ V, so $M$ is $10$ V below $P$ . Same for $N$ : $1 \cdot 10 = 10$ V below $P$ . $V_{M} = V_{N}$ ✓ → indeed $I_{g} = 0$ , confirming the shortcut was legitimate.

Wrap-up: the matrix is now fully covered

Recall Which example hit which cell?

A→Ex1 · B,C→Ex2 · D→Ex3 · E (short)→Ex4 · F (open)→Ex5 · G (n equal)→Ex6 · H (limit)→Ex7 · I (word)→Ex8 · J (bridge twist)→Ex9. Every row of the scenario matrix has a worked, verified example.

Connections

Parent: Series and parallel resistance — the two formulas these examples exercise.
Ohm's Law — used in literally every step.
Kirchhoff's Laws — KCL for the current splits, KVL for the voltage checks.
Wheatstone Bridge — the balanced-bridge trick in Ex 9.
Power in Circuits — the heater in Ex 8 draws $5$ A: try $P = V I = 600$ W as a follow-up.
EMF and Internal Resistance — a real battery adds a series resistor to any of these.