Tricky Networks

You have spent the last few lessons taking networks apart. Two resistors end to end add. Two side by side combine as reciprocals. A capacitor charging through a resistor rolls off with a time constant $\tau = RC$. Every circuit so far has yielded to the same move: find a pair that is clearly in series or clearly in parallel, replace it with one resistor, repeat until one resistor is left.

Now draw a square with a diagonal wire across it, a resistor on all five branches, and ask for the resistance corner to corner. There is no series pair: every junction has three wires meeting at it, so no resistor has another following it with nowhere else to go. There is no parallel pair either: no two resistors share both of their ends. Reduction stalls on the first move, and the network has only five resistors in it.

That shape is the Wheatstone bridge, the smallest network that defeats series-parallel reduction. A cube of twelve resistors defeats it too, and so does a ladder that repeats forever. Here are the tools that take over: symmetry, the balance condition, the delta-to-Y conversion, self-similarity, and the brute-force fallback that always works.

Equal potentials, and what you may do to them

Current flows through a resistor only when there is a potential difference across it. So if two nodes sit at exactly the same potential, any resistor joining them carries no current, and a resistor carrying no current is doing nothing: you may delete it. Equally you may fuse the two nodes with a wire, since a wire forces them to a potential they already shared. Both surgeries leave every current and voltage elsewhere untouched, and one of them usually breaks the deadlock.

The question is how to spot equal potentials without solving the circuit first, and the answer is symmetry. Suppose some relabelling of the nodes maps the network onto itself (every resistor lands on an identical resistor) while leaving both terminals where they are. The relabelled circuit is the same circuit, driven the same way, so it has the same solution. Any node and its image therefore share a potential.

Take the cube: twelve identical resistors $R$, one per edge, measured between opposite corners $A$ and $G$. Rotating by $120^\circ$ about the long diagonal $AG$ maps the cube onto itself and holds both $A$ and $G$ fixed. It cycles the three corners next to $A$ among themselves, so those three share a potential, and it cycles the three corners next to $G$ among themselves, so those share a second.

Fuse each triple into one node and the twelve edges sort themselves. Three edges run from $A$ to the first tier: three $R$ in parallel, $R/3$. Six run from the first tier to the second: $R/6$. Three run from the second tier to $G$: $R/3$. Those blocks are in series, so

For $R = 6.0\ \Omega$ that is $5.0\ \Omega$, and you got it without writing a single node equation.

The same rotation tells you the currents. Feed $1.0\ \mathrm{A}$ into $A$. It has three identical ways out, so each of the three edges at $A$ carries $0.333\ \mathrm{A}$.

Rings are simpler still. Six $6.0\ \Omega$ resistors in a closed hexagon, measured across opposite vertices: two arcs of three, $18 \parallel 18 = 9.0\ \Omega$. Across adjacent vertices, one resistor against the other five, $6 \parallel 30 = 5.0\ \Omega$. And in a square $ABCD$ with a diagonal brace from $B$ to $D$, measured $A$ to $C$: the reflection swapping $B$ and $D$ fixes $A$ and $C$, so the brace carries nothing and two paths of $2R$ in parallel leave $R$.

The bridge, and when it balances

Now the five-resistor bridge. Label the input $P$ and the output $Q$. One branch goes $P \to A \to Q$ through $R_1$ and $R_2$, the other $P \to B \to Q$ through $R_3$ and $R_4$, and a fifth resistor $R_g$ joins $A$ to $B$. Historically $R_g$ was a galvanometer, a sensitive current meter, and the point of the arrangement was to get it to read zero.

Suppose no current flows through $R_g$. Then the left branch is an undisturbed voltage divider: the current through $R_1$ has nowhere to go but $R_2$. With $P$ at potential $V$ and $Q$ at zero,

The right branch is an undisturbed divider too, giving $V_B = V R_4 / (R_3 + R_4)$. For the assumption to be consistent you need $V_A = V_B$, which after cancelling $V$ and cross-multiplying reduces to

That is the balance condition: the bridge balances when the two branches divide the voltage in the same proportion. It involves neither $R_g$ nor the supply voltage, which is what made the device useful. You find an unknown resistor by adjusting a known one until the meter reads zero, and you never need to know how strong your battery is or how sensitive your meter is.

Worked example 1. Arm $PA = 100\ \Omega$, $AQ = 200\ \Omega$, $PB = 300\ \Omega$, $BQ = 600\ \Omega$, with a $50\ \Omega$ galvanometer from $A$ to $B$. Find the resistance from $P$ to $Q$.

Check balance first, always: $100/200 = 0.500$ and $300/600 = 0.500$. They match, so no current crosses the bridge and you may delete the $50\ \Omega$ resistor entirely. What is left is two simple chains in parallel:

Balancing works in reverse just as easily. If $PA = 300\ \Omega$, $AQ = 450\ \Omega$ and $PB = 120\ \Omega$, then $BQ$ must satisfy $300/450 = 120/BQ$, so $BQ = 120 \times 450 / 300 = 180\ \Omega$.

When the bridge will not balance: delta to Y

If the ratios do not match, current crosses the bridge and you cannot delete anything. You need a real transformation, and the useful one turns a triangle of resistors into a three-spoked star.

Three nodes $X$, $Y$, $Z$ joined pairwise by $R_{XY}$, $R_{YZ}$, $R_{ZX}$ form a delta. The same terminals joined through a hidden centre node by spokes $R_X$, $R_Y$, $R_Z$ form a Y. They are interchangeable if they present the same resistance between every pair of terminals, and matching those three conditions gives

with $R_Y$ and $R_Z$ by the same pattern: the spoke at a terminal is the product of the two delta arms touching that terminal, divided by the sum of all three.

For three equal resistors $R$ in a delta, every spoke is $R^2/3R = R/3$: three $30\ \Omega$ resistors in a triangle behave exactly like three $10\ \Omega$ spokes. Corner to corner an equal delta is one arm in parallel with the other two in series, $R \parallel 2R = 2R/3$, which for $9.0\ \Omega$ gives $6.0\ \Omega$; the Y picture gives $R/3 + R/3$, the same thing.

Worked example 2. The unbalanced bridge: $PA = 100\ \Omega$, $AQ = 200\ \Omega$, $PB = 300\ \Omega$, $BQ = 400\ \Omega$, bridge $AB = 50\ \Omega$. Find $R_{PQ}$.

Ratios are $0.500$ and $0.750$, so it is not balanced. Pick the delta $P$–$A$–$B$, whose arms are $PA = 100$, $PB = 300$ and $AB = 50$. Their sum is $100 + 300 + 50 = 450\ \Omega$. The spokes are

The network is series-parallel again. From the hidden centre one path runs $R_A$ then $AQ$, $11.11 + 200 = 211.11\ \Omega$; the other runs $R_B$ then $BQ$, $33.33 + 400 = 433.33\ \Omega$. In parallel,

and adding the spoke $R_P$ in series gives $66.67 + 141.95 = 208.6\ \Omega$.

The fallback that always works

Delta-to-Y gives the equivalent resistance but destroys the node you folded away, so it cannot tell you the current in the bridge arm. When a problem asks for a current or a power inside the network, fall back on node voltages: ground one terminal, name the unknown potentials, and write Kirchhoff's current law at each unknown node, current in equals current out.

Worked example 3. Same unbalanced bridge, now with a $6.0\ \mathrm{V}$ battery from $P$ to $Q$. Find the current through the $50\ \Omega$ galvanometer and the power it dissipates.

Set $V_P = 6.0\ \mathrm{V}$ and $V_Q = 0$. Two unknowns, $V_A$ and $V_B$. At $A$, what arrives through $PA$ must leave through $AQ$ and through the bridge:

Multiply through by $200$ to clear the fractions: $2(6 - V_A) = V_A + 4(V_A - V_B)$, so $12 - 2V_A = 5V_A - 4V_B$, which tidies to $7V_A - 4V_B = 12$.

At $B$:

Multiply by $1200$: $4(6 - V_B) = 3V_B + 24(V_B - V_A)$, so $24 - 4V_B = 27V_B - 24V_A$. Gather the unknowns on one side, watching the signs as each term crosses: $24V_A - 4V_B - 27V_B = -24$, that is $31V_B - 24V_A = 24$.

From the first equation $V_A = (12 + 4V_B)/7$. Substitute, then multiply through by $7$ to clear the fraction:

so $217V_B - 288 - 96V_B = 168$, and $121V_B = 456$. Therefore

Both land where a divider says they should, which is the check worth doing every time. The bridge voltage is $V_A - V_B = 0.0992\ \mathrm{V}$, so

A bridge only mildly off balance passes very little current through its middle arm. That is why the instrument is precise, and also why a strain gauge bridge needs amplification: four arms of $350.00\ \Omega$ with one strained to $350.35\ \Omega$ on a $5.00\ \mathrm{V}$ supply gives an open-circuit output of $V \Delta R / 4R = 5.00 \times 0.35 / 1400 = 1.25\ \mathrm{mV}$.

Ladders: solving a network with itself

An infinite ladder is a series resistor $R_s$ along the top rail, then a shunt resistor $R_p$ across the rails, then another series, another shunt, forever. There is nothing to reduce, because there is no end to start from.

Use the infinity. Chop off the first section; what remains is the identical infinite ladder, because removing one rung from an endless sequence of rungs leaves an endless sequence of rungs. Call the input resistance $R$. The whole thing is then $R_s$ in series with $R_p$ parallel to a second copy of $R$:

Multiply out: $R(R_p + R) = R_s(R_p + R) + R_p R$, so $R^2 - R_s R - R_s R_p = 0$, and the quadratic formula gives

The other root is negative and you throw it away: a network of resistors cannot have negative resistance.

Worked example 4. All resistors $2.0\ \Omega$. Then $R = \tfrac{1}{2}(2 + \sqrt{4 + 16}) = 1 + \sqrt{5} = 3.236\ \Omega$. With $R_s = 4.0\ \Omega$ and $R_p = 6.0\ \Omega$, $R = \tfrac{1}{2}(4 + \sqrt{16 + 96}) = \tfrac{1}{2}(4 + 10.583) = 7.292\ \Omega$.

A finite ladder is easier: start at the far end, where you can see what is there, and work backwards. Three sections of $4.0\ \Omega$ series and $6.0\ \Omega$ shunt: last shunt $6$, add the series in front for $10$, parallel with the next shunt for $6 \parallel 10 = 3.75$, add $4$ for $7.75$, parallel with $6$ for $3.382$, add $4$ for $7.382\ \Omega$: already within $1.2\%$ of the infinite answer.

For currents down a ladder you want a slightly different quantity. Let $X$ be the resistance looking into the ladder from a shunt node, so $X$ is $R_p$ in parallel with everything beyond, and the input resistance is $R_s + X$:

Worked example 5. $R_s = 1.0\ \Omega$, $R_p = 2.0\ \Omega$. The equation becomes $X(3 + X) = 2(1 + X)$, so $X^2 + X - 2 = 0$ and $X = 1.0\ \Omega$. At any shunt node the arriving current sees two ways forward: the shunt, $2.0\ \Omega$, and the rest of the ladder, $R_s + X = 2.0\ \Omega$. Equal, so the current splits in half, and the half that carries on meets the identical choice at the next node. The shunt currents run $I_0/2$, $I_0/4$, $I_0/8$: the third carries $0.125\,I_0$.

Worked example 6. Section $n$ has series $100 \cdot 2^n\ \Omega$ and shunt $200 \cdot 2^n\ \Omega$. The ladder from section 1 onward is the whole ladder with every resistor doubled, and doubling every resistor doubles the resistance. So if $R$ is the input resistance at section 0, the input resistance at section 1 is $2R$, and

Clearing: $R(100 + R) = 100(100 + R) + 200R$, so $R^2 - 200R - 10000 = 0$ and $R = 100 + 100\sqrt{2} = 241.4\ \Omega$.

Capacitor ladders work the same way, with the series and parallel rules swapped. Identical capacitors $C$, series along the rail and shunt across: the input capacitance $C_{in}$ is $C$ in series with $(C + C_{in})$, giving $1/C_{in} = 1/C + 1/(C + C_{in})$, which rearranges to $C_{in}^2 + C\,C_{in} - C^2 = 0$ and $C_{in} = C(\sqrt5 - 1)/2$. For $C = 2.0\ \mu\mathrm{F}$ that is $1.236\ \mu\mathrm{F}$.

The infinite grid

One last network, because it uses a tool nothing else here does. An infinite square grid of nodes, every nearest-neighbour pair joined by a resistor $R$. What is the resistance between two adjacent nodes?

Inject a current $I$ into node $A$ and let it drain to infinity. By the four-fold symmetry of the grid it leaves $A$ equally along four edges, so the edge from $A$ to its neighbour $B$ carries $I/4$. Separately, draw current $I$ out of node $B$ from infinity: by the same symmetry that edge again carries $I/4$, in the same direction.

Both are linear problems, so you may add them. Superposed, current $I$ enters at $A$ and leaves at $B$ with nothing at infinity: the measurement you want. Edge $AB$ now carries $I/2$, so the voltage across it, and hence between $A$ and $B$, is $IR/2$:

For $8.0\ \Omega$ resistors that is $4.0\ \Omega$. Diagonal neighbours cannot be done this way, because the potential drop between them runs through many edges rather than one. The Fourier-integral result, worth knowing rather than deriving here, is $R_{\text{diag}} = 2R/\pi$; for $10\ \Omega$ that is $6.366\ \Omega$.

Recognising which tool a problem wants

Count the resistors and look at the junctions. If every junction has exactly two wires, reduce as usual and stop reading.

Reach for symmetry when the network is named after a shape (cube, square, hexagon, octahedron, tetrahedron) or when the resistors are all identical. Ask what rigid motion of the shape leaves both terminals alone, then find the nodes it swaps. Those are your equipotentials.

Reach for the balance condition the moment you see five resistors in a bridge pattern, or a resistor bridging the midpoints of two dividers. Compute the two ratios first; the cost is two divisions. "Galvanometer reads zero" or "null" is a direct instruction to use it.

Reach for delta-to-Y when balance fails and the question asks only for equivalent resistance. Pick the triangle containing the terminal you measure from, so the fold-away node is one you do not care about.

Reach for node voltages when the question asks for a current or a power inside the network. The shortcuts give the resistance and destroy the interior; only the node equations keep it.

Reach for self-similarity at the words "infinite", "repeating", "forever", or a ladder drawn with dots trailing off the page. Identical sections give $R = R_s + R_p \parallel R$. Sections scaling by a constant factor give the same kind of equation once you relate section 1 to section 0. Sections that do not scale at all leave you a series to sum, not a fixed point to find.

One habit beats all five: check the size of your answer. If two nodes are joined by a direct $10\ \Omega$ edge, the answer cannot exceed $10\ \Omega$, because every extra path only pulls the total down.

That takes resistor networks about as far as static circuits go. The next lesson leaves the wires behind and asks what those currents do to each other: Magnetic Force.