In short

When two resistors R_1 and R_2 are placed end-to-end so the same current flows through both, they are in series, and the combination acts as one resistor of resistance

\boxed{\;R_\text{series} \;=\; R_1 + R_2\;}.

When they are placed across the same two nodes so the same voltage sits across both, they are in parallel, and the combination acts as one resistor whose reciprocal is the sum of the reciprocals:

\boxed{\;\frac{1}{R_\text{parallel}} \;=\; \frac{1}{R_1} + \frac{1}{R_2}\;}.

Both formulas are consequences of Ohm's law and the two Kirchhoff-law statements charge is conserved (at every junction) and the potential at a point is single-valued (around every loop). For n equal resistors, series gives nR, parallel gives R/n. The voltage divider rule V_k = V \cdot R_k / R_\text{total} tells you how voltage splits across a series chain; the current divider rule I_k = I \cdot R_\text{other} / (R_1 + R_2) tells you how current splits across a parallel pair.

Once you can reduce series-parallel trees, you can also crack self-similar networks — the infinite resistor ladder that equals R(1+\sqrt{5})/2 — by replacing the infinite tail with its own symbol. The one exception is a bridge network (a Wheatstone configuration); there the diagonal resistor breaks series-parallel structure and you need Kirchhoff's laws directly — that is the next chapter.

A string of Diwali LED fairy lights is draped across the balcony of your flat. You plug it in, the whole strand lights up, and the courtyard glows like someone poured warm saffron into it. Now one bulb fails — maybe it took a small knock on the packet before you opened it. Depending on which strand you bought, one of two things happens.

Strand A (cheap, Chinese, series-wired): The whole garland goes dark. Every other bulb is fine, every wire is fine, but no current flows anywhere. You spend the next hour holding a torch to your face in the balcony, unscrewing fifty bulbs one at a time, looking for the dead one.

Strand B (slightly more expensive, parallel-wired): The failed bulb is dark. Every other bulb on the strand still shines. You shrug, unplug the strand, replace the one bulb, and the evening continues.

Same voltage supply — your 230 V wall socket — and the same LED bulbs. The only difference is the arrangement in which the bulbs are connected. That single arrangement decides whether one dead bulb brings down the whole strand or leaves it alone. Before you throw away strand A, you want to understand why — and once you do, you also understand why every socket on your flat's wall runs at a clean 230 V regardless of whether the geyser is on or the kettle is boiling, and why a cricket-stadium floodlight bank can lose two lamps without going dim.

That understanding is the content of this chapter. Ohm's law, from the previous article, gives you the rule for one resistor. This article gives you the rule for any network of resistors — how to replace fifty resistors with a single equivalent resistor, and why that equivalent is sometimes a plain sum, sometimes a reciprocal sum, and sometimes (in a ladder that never ends) an irrational number related to the golden ratio.

The two primitive arrangements — series and parallel

Every resistor network you can draw without crossings is built out of just two moves, applied again and again: stack two resistors end-to-end (series), or put two resistors side-by-side across the same pair of nodes (parallel). Almost every network you will meet in Class 12 and most of JEE Advanced reduces to a sequence of these two moves. The one exception — bridge networks — is saved for the chapters on Kirchhoff's laws and the Wheatstone bridge.

You cannot derive the series and parallel formulas from nothing. You need two physical principles, one for each arrangement:

Junction principle (conservation of charge). At every junction in a steady-state circuit, the total current flowing in equals the total current flowing out. Charge does not pile up at a point.

Loop principle (single-valued potential). The electric potential at a point is a single number. Between any two points, the potential difference is the same no matter which wires you follow.

These are the infant forms of Kirchhoff's current and voltage laws — you will meet the grown-ups in the next chapter. For resistors in series and parallel, the infants are enough.

Series — same current, voltages add

Two resistors R_1 and R_2 are in series when the only way current can flow out of R_1 is into R_2. There is no branch between them — the current leaving R_1 has nowhere else to go.

Two resistors in series with a cell and an ammeterA cell of EMF V drives current I through two resistors R1 and R2 connected end-to-end. The same current flows through both. The voltage across R1 is V1, the voltage across R2 is V2, and V1 + V2 = V. +−V R₁ R₂ I V₁ = IR₁ V₂ = IR₂ A
Two resistors end-to-end. The ammeter reads the same current $I$ no matter where you insert it in the loop, because there is no branch point between $R_1$ and $R_2$. The voltages $V_1 = IR_1$ and $V_2 = IR_2$ across the two resistors must add up to the cell EMF $V$.

Step 1. Apply the junction principle — trivially.

Since there is no junction between R_1 and R_2, the current I through R_1 is the same as the current through R_2.

Why: there is nowhere for the current to branch off. Every electron that enters R_1 must exit into R_2 — charge is conserved and there is no third wire for it to escape through.

Step 2. Apply Ohm's law to each resistor separately.

V_1 \;=\; I R_1, \qquad V_2 \;=\; I R_2.

Why: Ohm's law is a property of the individual resistor, not the circuit. Each resistor sees its own voltage drop determined by the current flowing through it.

Step 3. Apply the loop principle.

Walk around the loop starting at the cell's positive terminal. The potential rises by V at the cell, drops by V_1 at R_1, drops by V_2 at R_2, and returns to the starting point. The net change is zero:

V - V_1 - V_2 = 0 \quad\Longrightarrow\quad V = V_1 + V_2.

Why: the potential at a point is single-valued. If you walk in a closed loop and come back to the same point, your potential must be exactly where it started — the total rise must equal the total drop.

Step 4. Combine.

V = IR_1 + IR_2 = I(R_1 + R_2).

The combination behaves, from the cell's point of view, like a single resistor of resistance R_\text{series} = V/I:

\boxed{\;R_\text{series} \;=\; R_1 + R_2.\;}

Why: the cell only sees "current I at voltage V" — it cannot tell whether you have two resistors in series or one big resistor of value R_1 + R_2. This is what equivalent resistance means: the single resistor you could replace the combination with and not change anything the cell can measure.

For n resistors in series the derivation is identical, with n voltage drops instead of two:

R_\text{series} \;=\; R_1 + R_2 + \cdots + R_n.

Key consequence for your Diwali strand: if one bulb in a series string fails — the filament snaps or the LED dies open-circuit — the chain is broken. Current has nowhere to flow. Every bulb goes dark at once. This is strand A from the hook.

Parallel — same voltage, currents add

Two resistors are in parallel when both ends of R_1 are joined to both ends of R_2. They share the same two nodes. The voltage across them is the same because the nodes are the same points in space.

Two resistors in parallel with a cellA cell drives current I out of its positive terminal. At a junction the current splits into I1 through R1 and I2 through R2; the two branches rejoin at a second junction and return to the cell. Both resistors have the same voltage V across them. +−V R₁ R₂ II₁I₂ V across both
Two resistors wired across the same pair of nodes. Current $I$ leaves the cell, splits into $I_1$ and $I_2$ at the top junction, runs through the two resistors in parallel, and rejoins at the bottom junction. Because both resistors are tied to the same two nodes, the voltage across each one is the same: the cell voltage $V$.

Step 1. Apply Ohm's law to each branch.

Each resistor sees the cell voltage V directly across it.

I_1 \;=\; \frac{V}{R_1}, \qquad I_2 \;=\; \frac{V}{R_2}.

Why: the two ends of R_1 are the same two nodes as the two ends of R_2. Potential is single-valued, so the voltage difference is the same number for both resistors. Ohm's law then gives the branch currents.

Step 2. Apply the junction principle at the top node.

Current I comes in from the cell. It splits and leaves as I_1 (through R_1) and I_2 (through R_2):

I \;=\; I_1 + I_2.

Why: no charge piles up at the junction. Every coulomb coming in must leave by one of the two branches.

Step 3. Combine.

I \;=\; \frac{V}{R_1} + \frac{V}{R_2} \;=\; V\left(\frac{1}{R_1} + \frac{1}{R_2}\right).

The combination behaves like a single resistor whose resistance R_\text{parallel} satisfies V = I R_\text{parallel}, so

\boxed{\;\frac{1}{R_\text{parallel}} \;=\; \frac{1}{R_1} + \frac{1}{R_2}.\;}

Why: from the cell's point of view, the two parallel resistors draw a total current V/R_\text{parallel}. Matching this to the branch-sum formula shows that the reciprocal of the combined resistance is the sum of the reciprocals.

You can also write this as a single fraction:

R_\text{parallel} \;=\; \frac{R_1 R_2}{R_1 + R_2}.

For n resistors in parallel,

\frac{1}{R_\text{parallel}} \;=\; \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}.

For n equal resistors R in parallel, this gives R_\text{parallel} = R/n.

Key consequences. Two that will recur:

  1. The parallel combination is always smaller than the smallest resistor in the group. Add a second path and the current finds it easier to flow — total resistance goes down. A 100 Ω resistor in parallel with a 1 Ω resistor gives about 0.99 Ω, almost exactly the smaller one.

  2. If one branch opens (infinite resistance), the rest keep working. That is strand B from the hook. Every LED socket in a parallel strand is tied to the same two wires; a dead LED just raises its own branch resistance to infinity, and the remaining branches continue. Every electrical socket in your flat is wired in parallel with every other, from the same 230 V supply — that is exactly why the mixer, the geyser, and the fan can all run simultaneously without affecting each other's voltage.

An interactive: watch R_\text{eq} respond as you change one resistor

Before working through mixed networks, get a feel for the two formulas side by side. Fix R_1 = 10\ \Omega. Slide R_2 from 0.5\ \Omega up to 20\ \Omega and watch the equivalent resistance, computed both ways, as R_2 changes.

Interactive: equivalent resistance vs R2 for series and parallel with R1 = 10 Ω Two curves: the series equivalent R_s = 10 + R2 (straight line) and the parallel equivalent R_p = 10 R2 / (10 + R2) (curve approaching 10 from below). A draggable marker on the R2 axis shows both values live, together with the current drawn from a 10 V source. R₂ (Ω)equivalent resistance (Ω) 5101520 51015202530 R₁ = 10 Ω series: R₁ + R₂parallel: R₁R₂/(R₁+R₂) drag the red point along the axis
Fix $R_1 = 10\ \Omega$ and drag the red point along the $R_2$ axis. The red line shows $R_\text{series} = R_1 + R_2$ (a straight line that grows forever). The dark curve shows $R_\text{parallel} = R_1 R_2/(R_1 + R_2)$ — it stays *below the smaller* of the two resistors and asymptotes to $R_1 = 10\ \Omega$ as $R_2 \to \infty$. Notice: at $R_2 = 10$, series gives $20$ and parallel gives $5$, differing by a factor of 4.

A quick observation that the interactive makes obvious: the parallel curve is bounded above by R_1 no matter how large R_2 grows. This is the "a second path can only help" result — adding any finite resistor in parallel with R_1 can only reduce the total, never increase it.

Mixed networks — reduce one pair at a time

Most problems in Class 12 and JEE Main are mixed networks: a blend of series and parallel that reduces to a single resistor in a few steps. The method is mechanical once you see it. Find a pair of resistors that is clearly series or clearly parallel, replace that pair with a single equivalent, and continue until one resistor is left.

The only skill is seeing which pair to reduce first.

If neither pattern applies anywhere — if every pair has a branch coming off the middle — the network is irreducible by series-parallel alone. That is the Wheatstone-bridge case, and you need Kirchhoff's laws to crack it.

A mini-example: three resistors

Take R_1 = 4\ \Omega in series with the parallel combination of R_2 = 6\ \Omega and R_3 = 3\ \Omega.

Step 1. Reduce the parallel pair.

R_{23} \;=\; \frac{R_2 R_3}{R_2 + R_3} \;=\; \frac{6 \times 3}{6 + 3} \;=\; \frac{18}{9} \;=\; 2\ \Omega.

Why: R_2 and R_3 share the same two nodes — a clean parallel pair. Replace them with a single 2-Ω resistor and the rest of the circuit has no way to tell the difference.

Step 2. Reduce the remaining series pair.

R_\text{eq} \;=\; R_1 + R_{23} \;=\; 4 + 2 \;=\; 6\ \Omega.

Why: after step 1, only R_1 and the 2-Ω equivalent remain, in series. Add them.

That is the algorithm. You repeat it until one resistor is left, regardless of how large the network is.

Divider rules — reading off currents and voltages without solving again

Once you have reduced a network to its equivalent and computed the total current or voltage, you often want to know how that current or voltage is distributed inside the network. Two little formulas save you the work of re-solving Ohm's law branch by branch.

Voltage divider (for series resistors)

If a total voltage V is across a series chain R_1, R_2, \ldots, R_n, and the same current I = V/(R_1 + \cdots + R_n) flows through all of them, then the voltage across any one resistor R_k is

V_k \;=\; IR_k \;=\; V\cdot\frac{R_k}{R_1 + R_2 + \cdots + R_n}.

Why: the voltage splits in the same ratio as the resistances, because the current through all of them is the same. A big resistor in a series chain hogs the voltage; a small resistor barely drops any.

This is how a potential divider — two resistors in series tapped at the middle node — generates a smaller voltage from a bigger one. The output is V_\text{in} \cdot R_2/(R_1 + R_2).

Current divider (for two parallel resistors)

If a total current I enters a junction and splits between R_1 and R_2 in parallel, the current through each branch is given by

I_1 \;=\; I \cdot \frac{R_2}{R_1 + R_2}, \qquad I_2 \;=\; I \cdot \frac{R_1}{R_1 + R_2}.

Why: the voltage across both branches is the same, V = I R_\text{parallel} = I\,R_1 R_2/(R_1+R_2). Dividing by each branch resistance gives its current. Notice the cross-pattern — I_1 (the current through R_1) is proportional to R_2, the other resistor. A small resistor grabs more current; a large resistor grabs less.

You can verify the formula directly: I_1 + I_2 = I \cdot (R_2 + R_1)/(R_1 + R_2) = I, consistent with the junction rule.

Worked examples

Example 1: The toaster and the mixer on the same plug-point

You are at home and you plug a 1100\ \text{W} pop-up toaster (R_\text{toaster} = 48\ \Omega at operating temperature) into a kitchen extension board. Ten minutes later someone plugs a 600\ \text{W} immersion blender (R_\text{blender} = 88\ \Omega) into the next socket on the same board. The wall supply is 230\ \text{V}, and you want to know (i) what single resistor the two appliances look like together to the wall, and (ii) how much current each appliance draws and how much total current the board pulls.

Toaster and blender in parallel on a 230 V supplyA 230 V supply drives a parallel combination of a 48-ohm toaster and an 88-ohm blender through an extension board. The current splits between the two branches and rejoins to return to the supply. 230 V toaster, 48 Ω blender, 88 Ω I
Each appliance sits directly across the 230 V supply via the extension board — a classic parallel combination. The two appliances share the same two wires (live and neutral), so they share the same voltage.

Step 1. Identify the topology.

The extension board wires live, neutral, and earth from a single socket out to multiple sockets. Every appliance plugged in sits between the same two wires (live and neutral). They are in parallel.

Why: both ends of the toaster and both ends of the blender connect to the same two wires. Same two nodes → parallel. This is the whole reason home wiring is parallel: every socket sees the full 230 V regardless of what else is plugged in.

Step 2. Compute the equivalent resistance.

R_\text{eq} \;=\; \frac{R_\text{toaster}\, R_\text{blender}}{R_\text{toaster} + R_\text{blender}} \;=\; \frac{48 \times 88}{48 + 88} \;=\; \frac{4224}{136} \;\approx\; 31.06\ \Omega.

Why: the parallel formula for two resistors. As expected, the combined resistance (\approx 31\ \Omega) is less than the smaller of the two (48 Ω) — parallel combinations are always smaller than the smallest individual.

Step 3. Branch currents from Ohm's law.

Each appliance sees 230\ \text{V} across it, so:

I_\text{toaster} \;=\; \frac{230}{48} \;\approx\; 4.79\ \text{A},
I_\text{blender} \;=\; \frac{230}{88} \;\approx\; 2.61\ \text{A}.

Why: parallel means same voltage across each branch. Ohm's law for each branch gives the branch current directly — no need to worry about the other appliance.

Step 4. Total current drawn from the supply.

I_\text{total} \;=\; I_\text{toaster} + I_\text{blender} \;=\; 4.79 + 2.61 \;\approx\; 7.40\ \text{A}.

Or, equivalently, I_\text{total} = V/R_\text{eq} = 230/31.06 \approx 7.40 A. The two methods must agree, and they do.

Result. The equivalent resistance seen by the wall is about 31\ \Omega. The total current is 7.4\ \text{A}, split as 4.79\ \text{A} through the toaster and 2.61\ \text{A} through the blender.

What this shows. Every additional appliance you plug into a parallel strip increases the current drawn from the mains, even though each appliance itself runs at the same voltage. This is precisely why kitchen extension boards are rated for a maximum current (usually 6 A or 16 A) and why plugging in a 2 kW geyser alongside a 1 kW toaster will trip the breaker — the sum of currents, not the voltage, is what the wiring and breaker must handle.

Example 2: A mixed network from a CBSE board question

Find the equivalent resistance between terminals A and B of the following network: a 6\ \Omega resistor in series with the parallel combination of (i) a 3\ \Omega resistor and (ii) a series string 4\ \Omega + 8\ \Omega. A 10\ \text{V} cell is connected between A and B; find the current drawn from the cell and the current through the 8\ \Omega resistor.

Mixed series-parallel networkA 10 V cell drives a 6 ohm resistor in series with a parallel combination. The parallel combination has a 3 ohm resistor in one branch and a 4-ohm plus 8-ohm series pair in the other branch. 10 V A 6 Ω 3 Ω 4 Ω8 Ω B'B
A 6 Ω resistor in series with a parallel combination. One parallel branch is a single 3 Ω. The other is a series string 4 Ω + 8 Ω. The reduction order: collapse the 4-Ω + 8-Ω string first, then the parallel pair, then the series combination with the 6 Ω.

Step 1. Reduce the inner series pair.

The 4 Ω and 8 Ω resistors in the lower branch are in series (nothing taps off the point between them). So their combined resistance is

R_{4+8} \;=\; 4 + 8 \;=\; 12\ \Omega.

Why: collapse the innermost clearly-series or clearly-parallel pair first. Working from the inside out is always the safe order.

Step 2. Reduce the parallel pair.

Now the two branches between the two internal nodes are a 3 Ω resistor and a 12 Ω resistor — in parallel:

R_\text{par} \;=\; \frac{3 \times 12}{3 + 12} \;=\; \frac{36}{15} \;=\; 2.4\ \Omega.

Why: after step 1 the lower branch is a single effective resistor of 12 Ω. Together with the upper 3 Ω branch they span the same two nodes — clean parallel.

Step 3. Add the outer series resistor.

R_\text{eq} \;=\; 6 + 2.4 \;=\; 8.4\ \Omega.

Step 4. Total current from the cell.

I \;=\; \frac{V}{R_\text{eq}} \;=\; \frac{10}{8.4} \;\approx\; 1.19\ \text{A}.

Step 5. Use the current divider to find the 8 Ω current.

The total current I = 1.19 A hits the parallel junction and splits between the 3\ \Omega upper branch and the 12\ \Omega lower branch. The current through the 12-Ω branch (which is the same current that flows through the 4-Ω and the 8-Ω, since they are in series) is:

I_\text{lower} \;=\; I \cdot \frac{R_\text{upper}}{R_\text{upper} + R_\text{lower}} \;=\; 1.19 \cdot \frac{3}{3 + 12} \;=\; 1.19 \cdot \frac{1}{5} \;=\; 0.238\ \text{A}.

Why: the current divider. The upper (3 Ω) branch is smaller, so more of the current takes that path. The 8\ \Omega resistor lies in the lower branch and carries the whole lower-branch current — there is no junction within the lower branch for the current to split further.

Result. R_\text{eq} \approx 8.4\ \Omega, total current \approx 1.19 A, current through the 8\ \Omega resistor \approx 0.238 A.

What this shows. Mixed networks collapse by repeated application of the two primitive rules. No matter how complicated a network of n resistors appears, if every pair is either clearly series or clearly parallel somewhere in the reduction, you will eventually reduce it to one resistor. The divider rules then back-substitute to find individual branch currents.

The infinite ladder — self-similarity in action

Some networks are not finite, but they can still have a finite equivalent resistance because of a structural trick: the network looks the same when you chop off the first layer. This is self-similarity, and it turns an "infinite" problem into a one-line algebra problem.

Setup

Consider an infinite ladder: at each rung you have a resistor R going in, and a resistor R going across to the return rail. The ladder extends to infinity on the right.

Infinite resistor ladderAn infinite ladder of resistors. The top rail has a resistor R between each pair of consecutive rungs. Each rung is a resistor R connecting the top rail to the bottom rail. The ladder extends infinitely to the right, denoted by three dots. ··· ··· AB RRRRRR
An infinite ladder of identical resistors $R$. Each "bay" contributes a series $R$ on the top rail and a parallel rung $R$ from top to bottom. The ladder extends indefinitely to the right. You want the equivalent resistance seen between the two red terminals A and B on the far left.

Let R_\infty denote the equivalent resistance between A and B. You want to find R_\infty in terms of R.

The self-similarity trick. Look at the ladder after you strip off the first bay (one series R on the top rail and one rung R). What is left? The same infinite ladder as before. Its equivalent resistance is also R_\infty, because the ladder was infinite to begin with — removing one bay leaves an infinite ladder that is indistinguishable from the original.

So the full ladder is equivalent to:

which gives a finite circuit you can reduce. Write the equation:

R_\infty \;=\; R + \frac{R\cdot R_\infty}{R + R_\infty}.

Why: the first series resistor contributes R. After that, the rung R sits in parallel with the rest-of-ladder R_\infty. The parallel combination is R \cdot R_\infty / (R + R_\infty). Add these two and you have the whole ladder.

Solve the equation. Multiply both sides by R + R_\infty:

R_\infty(R + R_\infty) \;=\; R(R + R_\infty) + R\, R_\infty
R R_\infty + R_\infty^2 \;=\; R^2 + R R_\infty + R R_\infty
R_\infty^2 \;=\; R^2 + R R_\infty
R_\infty^2 - R R_\infty - R^2 \;=\; 0.

Why: standard quadratic in R_\infty. The self-similarity turned the infinite ladder into a single quadratic — far simpler than summing an infinite series.

Apply the quadratic formula with a = 1, b = -R, c = -R^2:

R_\infty \;=\; \frac{R \pm \sqrt{R^2 + 4R^2}}{2} \;=\; \frac{R(1 \pm \sqrt{5})}{2}.

Take the positive root (resistance must be positive):

\boxed{\;R_\infty \;=\; \frac{R(1 + \sqrt{5})}{2} \;\approx\; 1.618\, R.\;}

Why: the negative root would give a negative resistance. Physical resistance is positive, so we take (1 + \sqrt{5})/2, which is the golden ratio \varphi \approx 1.618 — a little surprise gift from resistor networks to number theory.

For R = 1\ \Omega, R_\infty \approx 1.618\ \Omega. Infinite resistors, finite answer. The same self-similarity trick works for any infinite network with repeating structure — the secret is always to replace the tail with a symbol.

Common confusions

If you came here to reduce networks and solve CBSE/JEE Main problems, you already have what you need. What follows is the JEE-Advanced layer: the symmetry argument that solves resistor cubes, the general "balanced bridge" shortcut, and the connection to graph theory.

Symmetry on a resistor cube

A famous JEE Advanced set-up: a cube of 12 identical resistors R, one along each edge. Find the equivalent resistance between two opposite corners (the body diagonal).

Directly reducing by series-parallel fails — every edge of the cube is in a Wheatstone-like configuration with several others. But symmetry does the work at a stroke.

Resistor cube showing equipotential verticesA cube drawn in perspective with 8 vertices. Two opposite corners are marked A and B (body diagonal). The 3 vertices adjacent to A are grouped as equipotential set X. The 3 vertices adjacent to B are grouped as equipotential set Y. Between X and Y lies the middle layer of 6 edges. AB X (3 vertices, all at same V) Y (3 vertices, all at same V)
The resistor cube between opposite corners A and B. By symmetry, the three vertices adjacent to A (set X) are at a single common potential, and the three vertices adjacent to B (set Y) are at another single common potential. This lets you merge each set into a single node and reduce the cube to a simple three-layer parallel combination.

Step 1. Identify the equipotential sets.

Connect the cell between A (body diagonal start) and B (body diagonal end). The three vertices adjacent to A are all in symmetric positions — the cube looks the same under a 120° rotation about the A-B diagonal. So those three vertices must be at the same potential. Call this set X. Similarly, the three vertices adjacent to B form set Y at a common potential.

Why: any two vertices related by a symmetry of the problem must be at the same potential. Hook up a voltmeter between them and the needle reads zero — they are interchangeable by a relabelling of the cube, and currents through them are symmetric.

Step 2. Short the equipotential vertices.

Wires at the same potential can be connected with no effect on the rest of the circuit. Merge the three X-vertices into one node, and the three Y-vertices into one node.

The cube now has three layers:

  • From A to the merged X-node: 3 edges in parallel (each resistor R). Equivalent: R/3.
  • From X-node to Y-node: 6 edges in parallel (the 6 middle-layer edges). Equivalent: R/6.
  • From Y-node to B: 3 edges in parallel. Equivalent: R/3.

Step 3. Combine the three layers in series.

R_{AB} \;=\; \frac{R}{3} + \frac{R}{6} + \frac{R}{3} \;=\; \frac{2R + R + 2R}{6} \;=\; \frac{5R}{6}.

Why: after merging equipotential vertices the cube becomes a three-layer, series-parallel network. The trick was the first step; the reduction is elementary.

For R = 1\ \Omega, the cube's body-diagonal resistance is 5/6\ \Omega \approx 0.833\ \Omega. Without symmetry, you would have to set up Kirchhoff's laws in 12 edge-currents and 8 node-equations. With symmetry, it is one line of arithmetic.

The same idea — find equipotential vertices and short them — cracks every "resistor lattice" problem. It is the most powerful tool at this level.

Connection to graph theory — the Laplacian and resistance distance

For readers who like context: the equivalent resistance between two nodes of any network equals a specific matrix quantity called the effective resistance or resistance distance. It is computed from the graph Laplacian L of the network. The resistance between nodes i and j is

R_{ij} \;=\; (e_i - e_j)^T L^+ (e_i - e_j),

where L^+ is the Moore–Penrose pseudoinverse of the Laplacian and e_i is the unit vector at node i. This formula is one line, handles any topology — including bridges and irreducible networks — and reduces to the series-parallel results for the easy cases. It is how circuit-simulation software (SPICE) handles resistor networks internally. You do not need it for JEE, but it is the grown-up language of the problem, and it answers the question "what is the right general framework?" cleanly.

A beautiful consequence: the resistance distance R_{ij} is a genuine distance on the graph — it obeys the triangle inequality R_{ij} \le R_{ik} + R_{kj} for any intermediate node k. Resistor networks secretly embed graphs in a metric geometry.

Balanced bridge shortcut — a preview

A Wheatstone bridge with ratio-balance (R_1/R_2 = R_3/R_4) is not truly series-parallel, but the galvanometer arm carries zero current, so you can either delete the diagonal resistor or short-circuit it — both give the same total resistance. This shortcut will save you many lines of algebra in the next chapter. Look for bridge configurations in every resistor problem; if the ratio condition holds, the diagonal is invisible.

Where this leads next