Compute equivalent resistance in mixed networks: Visual walkthrough

We are going to answer one question: if I hide a box of resistors and only let you touch two wires sticking out, how much does that box "resist"?

Step 1 — What "resistance" even means (the picture behind $R$ )

WHAT. A resistor is a component where push and flow are proportional. The push is the voltage $V$ (measured in volts) — think of it as the height difference driving water downhill. The flow is the current $I$ (measured in amps) — how much charge passes per second. The number linking them is the resistance $R$ .

WHY this tool. We need a single fact that never changes for a resistor, so we can compare boxes. That fact is Ohm's Law:

PICTURE. The straight red line: double the current, double the voltage. The slope of that line is $R$ . That straightness is the only thing we will lean on for the whole page.

Figure — Compute equivalent resistance in mixed networks

Step 2 — Naming the wiring: nodes, series, parallel

WHAT. A node is a junction — a spot where wires meet, all at the same voltage. We label nodes with letters $A, B, C, \dots$ so we can talk about them.

WHY. You cannot say "these two are in series" without pointing at the node they share. Labels turn a vague drawing into precise statements.

PICTURE. On the left, two resistors meet at node $C$ and nothing else touches $C$ — the only way out of $R_{1}$ is straight into $R_{2}$ . That is series: forced single-file, so the same current goes through both. On the right, two resistors touch at both ends (nodes $A$ and $B$ ) — that is parallel: the same voltage sits across both.

Step 3 — Deriving the series rule from the picture

WHAT. We prove that two series resistors act like one resistor whose value is the sum.

WHY these tools. Same current flows through both (Step 2), and voltages along a single path add up — that adding-up is Kirchhoff's Voltage Law (walk from $A$ to $B$ , count every drop).

PICTURE. Watch the height fall in two steps: a drop $V_{1}$ across $R_{1}$ , then a drop $V_{2}$ across $R_{2}$ . The total fall from $A$ to $B$ is both drops stacked.

V = drop over R_{1} V_{1} + drop over R_{2} V_{2} = Ohm on R_{1} I R_{1} + Ohm on R_{2} I R_{2} = I (R_{1} + R_{2})

Compare with $V = I R_{e q}$ and the $I$ cancels:

Step 4 — Deriving the parallel rule from the picture

WHAT. We prove two parallel resistors act like one resistor whose reciprocal is the sum of reciprocals.

WHY these tools. Same voltage across both (Step 2), and the currents in the branches add up at the node — that adding-up is Kirchhoff's Current Law (charge in = charge out). Because current is what splits, it's cleaner to think in conductance $G = 1/ R$ — "how easily current flows" (see Conductance and Admittance).

PICTURE. One incoming current $I$ splits at node $A$ into $I_{1}$ and $I_{2}$ , runs down the two branches, and merges back at $B$ . More doors ⇒ more total flow for the same push.

I = branch R_{1} I_{1} + branch R_{2} I_{2} = Ohm on R_{1} \frac{V}{R _{1}} + Ohm on R_{2} \frac{V}{R _{2}} = V (\frac{1}{R _{1}} + \frac{1}{R _{2}})

Compare with $I = V / R_{e q}$ and the $V$ cancels:

Step 5 — The "product over sum" shortcut, and why it is only for two

WHAT. For exactly two resistors, we compress the reciprocal rule into one clean fraction.

WHY. Two reciprocals over a common denominator combine neatly; flipping is then a single move.

PICTURE. The two branches merge into one fraction: numerator = the two resistances multiplied, denominator = the two added.

\frac{1}{R _{p a r a l l e l}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{R _{2} + R _{1}}{R _{1} R _{2}} ⟹ R_{p a r a l l e l} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}

Numerator $R_{1} R_{2}$ — product.
Denominator $R_{1} + R_{2}$ — sum.
Only two. With a third resistor the common denominator becomes $R_{1} R_{2} R_{3}$ and this tidy two-term flip breaks — go back to summing reciprocals.

Step 6 — Degenerate cases: what happens at the extremes

WHAT. We check the rules where they could misbehave, so no scenario surprises you.

WHY. A formula you trust must survive its own edges — a wire ( $0 Ω$ ), a gap ( $\infty Ω$ ), and equal twins.

PICTURE. Three little limit cards:

Step 7 — The collapse, watched end to end (the ladder)

WHAT. We run the full algorithm on the parent's ladder: from $A$ , $R_{1} = 10 Ω$ to node $C$ ; from $C$ , $R_{2} = 10 Ω$ straight to $B$ , and $R_{3} = 10 Ω$ to $D$ then $R_{4} = 10 Ω$ to $B$ .

WHY this order. Always collapse the innermost obvious group first — here the $C \to D \to B$ chain — because you cannot simplify a node until everything hanging off it is a single resistor.

PICTURE. Three frames, left to right, each redrawn after a merge. The red element is the group being collapsed in that frame.

Frame 1 — series in the lower branch (same current through $R_{3}$ , $R_{4}$ ): $R_{34} = R_{3} + R_{4} = 10 + 10 = 20 Ω$

Frame 2 — parallel at node $C$ (the $20 Ω$ chain and $R_{2}$ share nodes $C$ and $B$ ): $R_{C} = \frac{R _{2} \cdot R _{34}}{R _{2} + R _{34}} = \frac{10 \cdot 20}{10 + 20} = \frac{200}{30} = 6.67 Ω$ Note: $6.67 < 10$ — parallel shrank it, exactly as promised. Bug alarm silent. ✓

Frame 3 — series with $R_{1}$ (single path from $A$ ): $R_{A B} = R_{1} + R_{C} = 10 + 6.67 = 16.67 Ω$

The one-picture summary

Everything above, compressed: the linear line becomes one slope; series stacks drops (sum), parallel splits current (reciprocal sum); collapse inward until one red resistor remains.

Recall Feynman: the whole walk in plain words

Picture water pipes with a tap and a drain. Every pipe obeys one straight rule: more push gives proportionally more flow — that straight line is Ohm's law, and its steepness is the resistance. Put pipes end-to-end and the water fights each one in turn, so their difficulties add — that's series. Put pipes side-by-side and the water gets extra doors, so it flows easier — you add their easinesses (one-over-resistance) and flip back, and the answer is always smaller than the tightest single pipe. To measure a whole messy plumbing box you never solve it all at once: you find the smallest end-to-end or side-by-side bunch hiding deep inside, replace it with one pipe, redraw, and repeat. The ladder showed it: the far chain $R_{3} + R_{4}$ merged to $20$ , sat beside $R_{2} = 10$ to give $6.67$ , then joined $R_{1} = 10$ in line for a final $16.67 Ω$ . One box, one number, one slope.

Connections

Ohm's Law — the straight line whose slope is $R$ ; the linearity that makes one-number boxes possible.
Kirchhoff's Voltage Law — stacked drops justify series addition (Step 3).
Kirchhoff's Current Law — split currents justify parallel addition (Step 4).
Conductance and Admittance — parallel adds conductances $1/ R$ (Step 4).
Voltage Divider and Current Divider — what the drops/splits in Steps 3–4 become when you read off a branch.
Delta-Wye Transformation — the escape hatch when no group is purely series or parallel.

Visual walkthrough — Compute equivalent resistance in mixed networks