Series–Parallel and Delta–Wye Networks

Start at the end remote from the terminals. The 20 Ω and 30 Ω resistors both connect node \(c\) to node \(d\), so they share the same voltage and are in parallel:

That 12 Ω now carries the same current as the 6 Ω that follows it, so the two are in series:

The upper path and the 18 Ω lower branch both run from \(c\) to \(b\), hence are in parallel:

Finally the 10 Ω is in series with that:

A branch is any single two-terminal element. Counting them: the source, \(R_1\), \(R_2\) and \(R_3\), so

A node is a point where two or more branches meet. Crucially, a length of ideal wire is all one node however long or bent it is. The bottom rail joins the source, \(R_2\) and \(R_3\) and is therefore a single node, not three:

The number of independent loops — meshes, for a planar circuit — is the number of "windows" in the drawing:

Checking the identity:

aThe two resistors carry the same current, so the voltage divides in proportion to resistance:

bThe load sits across the lower resistor, so the pair combine first:

The divider is now 4 kΩ over 4 kΩ:

Percentage error:

With more than two resistors in parallel, work in conductances — it avoids nested fractions entirely:

All three branches share the same voltage, which the source fixes:

Each branch current then follows from Ohm's law:

KCL check:

aHealthy network:

bA short across \(R_2\) places a zero-resistance path between the same two nodes as the parallel pair. Anything in parallel with a short is a short:

Note that \(R_3\) is now completely bypassed — it carries no current at all, even though it is still physically present.

cIf \(R_3\) opens, its branch is removed and only \(R_2\) remains between those nodes:

dShorting \(R_4\) simply deletes it from a series chain:

Reduce from the far end. The 6 Ω shunt is in series with the 3 Ω, and that combination is in parallel with the 18 Ω shunt:

Move one rung left — add the 4 Ω series, then take it in parallel with the 15 Ω shunt:

Add the last series 4 Ω to reach the input:

Now walk forward, node by node, using Ohm's law and KCL at each junction:

Power balance as a final check:

The two measurements give the sum and the product directly:

A pair of numbers with a known sum and product are the roots of a quadratic:

Solving:

Check: \(15 + 10 = 25\;\checkmark\) and \(150/25 = 6\;\checkmark\)

aAdding resistance in series always increases the total, so all four in series is the maximum:

bAdding branches in parallel always decreases the total, so all four in parallel is the minimum:

cTwo arrangements give 12 Ω, and they are genuinely different circuits:

Other arrangements give other values — for instance three in series paralleled with the fourth:

Series. Common current:

Parallel. Common voltage of 24 V across each:

Cross-check: \(R_{\text{eq}} = 4\,\|\,6 = 2.4\ \Omega\) and \(24^2/2.4 = 240\ \text{W}\;\checkmark\)

Comparing the two cases:

Label the delta arms by the pair of terminals they join: \(R_{ac} = 6\), \(R_{ad} = 12\), \(R_{cd} = 18\). Every conversion starts with their sum:

The wye arm at any terminal is the product of the two delta arms meeting at that terminal, divided by \(\Sigma\). At terminal \(a\) the arms 6 and 12 meet:

At terminal \(c\) the arms 6 and 18 meet; at terminal \(d\) the arms 12 and 18 meet:

Verify by measuring terminal to terminal, which is what "equivalent" means. In the delta, \(a\) to \(c\) is 6 Ω in parallel with \(12+18 = 30\); in the wye it is simply \(R_a + R_c\):

The wye-to-delta conversion is built on the sum of the pairwise products of the wye arms:

Each delta arm is \(\Pi\) divided by the wye arm at the opposite terminal — the one the delta arm does not touch:

Check terminal \(a\) to terminal \(b\). In the wye it is \(10 + 20 = 30\ \Omega\); in the delta, 35 in parallel with \(140 + 70 = 210\):

The balance condition compares the ratios along the two arms from \(a\) to \(b\):

The ratios are equal, so nodes \(c\) and \(d\) sit at the same potential and the bridge is balanced. No current flows in the 5 Ω arm.

Route 1 — remove the bridge arm. A branch carrying no current can be deleted without changing anything. The two arms then become simple series chains in parallel:

Route 2 — short the bridge arm. Two nodes at the same potential may equally be joined by a wire; no current will flow through it either. That puts 4 Ω in parallel with 6 Ω, in series with 8 Ω in parallel with 12 Ω:

First confirm that reduction really is blocked:

The bridge is unbalanced, so current does flow in the 18 Ω arm and no two resistors in the network are either in series or in parallel. Series–parallel reduction cannot start.

Choose the delta \(a\!-\!c\!-\!d\), formed by the 6, 12 and 18 Ω resistors, and convert it to a wye. This is exactly the conversion done in Problem 10:

The bridge has become a simple network: from \(a\), a 2 Ω leads to the new star point \(n\), and from \(n\) two series paths run to \(b\):

Those two are in parallel, and the 2 Ω is in series with the result:

Independent check by nodal analysis. Apply 8 V from \(a\) to \(b\) and solve for \(V_c\) and \(V_d\) in the original bridge:

The bridge wire is uniform, so the resistance of each portion is proportional to its length. The two wire segments form the lower pair of bridge arms, and the balance condition becomes a ratio of lengths:

aWith \(R = 4\ \Omega\) and \(\ell = 40\ \text{cm}\):

bWith \(R = 9\ \Omega\) against the same 6 Ω unknown:

cDifferentiating the balance relation shows why the centre is best. Writing \(X = R(100-\ell)/\ell\), the fractional error is

The product \(\ell(100-\ell)\) is largest at \(\ell = 50\), so a given uncertainty in reading the contact position produces the smallest error in \(X\) there. Near either end the same 1 mm of positional error is magnified enormously.

The network is linear, so every voltage and current in it is directly proportional to the source voltage. Assume the answer, propagate backwards, then correct by a single scale factor at the end.

Assume \(I_{6} = 1\ \text{A}\) in the output branch. Then the voltage at the last node is

That same 1 A flows through the 3 Ω, so the node before it stands at

The 18 Ω shunt across this node therefore carries \(9/18 = 0.5\ \text{A}\), and by KCL the current in the preceding 4 Ω is \(1 + 0.5 = 1.5\ \text{A}\).

Continue one more rung:

The assumed conditions therefore require a source of

The actual source is 20 V, so scale everything by the ratio:

This matches Problem 6 exactly, where the same 0.8 A was found by reducing the ladder and walking forwards. As a bonus, \(R_{\text{in}} = 25/2.5 = 10\ \Omega\) falls out of the assumed solution for free.

The shunt is simply a resistor in parallel with the movement, so it is a current divider. At full scale the movement must carry exactly 1 mA, and the voltage across the parallel pair is

The shunt carries everything else:

Hence

The same result follows from the multiplying factor \(n = I/I_m = 5000\):

The voltage burden — the drop the ammeter inserts into the circuit under test — is the 50 mV computed above, and the ammeter's own resistance is

aAt full scale the movement must draw 1 mA when 100 V is applied across the whole instrument, so the total resistance must be

The sensitivity is the reciprocal of the full-scale current, and is a property of the movement alone:

bWithout the meter, the divider gives the obvious answer:

Connecting the meter puts its 100 kΩ in parallel with the lower 100 kΩ:

Error:

At mid-setting the track is split into two 5 kΩ halves. With no load the output is exactly half the supply:

The load sits across the lower half only, so those two combine:

The upper half is untouched, so the divider is now 5 kΩ over 3.333 kΩ:

Error:

At the extremes of travel the error vanishes: with the wiper fully down the output is 0 V loaded or not, and with it fully up the lower section is the whole track, in series with nothing, so the loading has no divider to distort. The maximum error occurs a little above the midpoint, at roughly two-thirds of full travel.

A practical source is an ideal emf \(E\) in series with an internal resistance \(r\). For any load,

Writing this for the two measurements gives two linear equations:

Equating and solving:

Terminal voltages follow from \(V_T = E - Ir\):

Check against the loads: \(4 \times 2.5 = 10\ \text{V}\;\checkmark\) and \(2 \times 5.5 = 11\ \text{V}\;\checkmark\)

The network has a mirror symmetry about the horizontal axis through \(a\) and \(b\): swapping \(c\) and \(d\) leaves the circuit unchanged. Whatever potential the symmetry-preserving solution assigns to \(c\), it must assign the same to \(d\):

With no potential difference across the 7 Ω, it carries no current and can simply be deleted:

Replacing the 7 Ω by 700 Ω — or by a short, or by an open circuit — changes nothing at all, because zero current times any resistance is still zero volts:

From Problem 13 the bridge presents \(R_{ab} = 8\ \Omega\) at its terminals. That is all the external circuit can see, so replace the entire bridge by a single 8 Ω:

Voltage across the bridge:

To find a current inside the bridge you must go back to the original network — the wye equivalent deliberately destroyed the node \(c\!-\!d\) branch. In Problem 13, applying 8 V gave \(V_c = 4.5\ \text{V}\) and \(V_d = 3\ \text{V}\). Linearity lets us scale by \(24/8 = 3\):

Hence the bridge-arm current, flowing from \(c\) to \(d\):

Reduction is hopeless here — no two edges are in series or in parallel. Symmetry is the only practical tool. Inject a current \(I\) at one terminal, extract it at the other, and use symmetry to identify nodes that must sit at the same potential. Those nodes can then be merged, which collapses the cube into a short series chain.

aBody diagonal (A to G). The three edges leaving \(A\) are interchangeable, so each carries \(I/3\) and the three nodes adjacent to \(A\) are at one common potential. Likewise the three nodes adjacent to \(G\). The remaining six middle edges each carry \(I/6\). Summing voltage drops along any path:

bFace diagonal (A to B in the same face). The same argument, with a different set of equipotential nodes, gives

cAlong one edge (A to D). Here the direct edge takes the largest share and the answer is

The three results are consistently ordered, as they must be:

The farther apart the two terminals, the larger the resistance — an obvious requirement that any wrong answer will usually violate.

The trick is self-similarity. Because the ladder is infinite, removing the first two elements leaves a network identical to the original — an infinite ladder is unchanged by chopping one section off the front. Call the input resistance \(R_{\text{in}}\); then looking in past the first series \(R\), we see \(2R\) in parallel with another \(R_{\text{in}}\):

Clearing the denominator:

Solving the quadratic:

The negative root is discarded — a network of positive resistors cannot present a negative resistance. Hence \(R_{\text{in}} = 2R = 20\ \Omega\) for \(R = 10\ \Omega\).

Now the current split. At the first shunt node, the input current \(I\) divides between the \(2R\) shunt and the rest of the ladder, which also looks like \(2R\). Two equal paths, so the current halves:

The same argument repeats at every node, so the current reaching the second shunt node is

Sufficiency. Suppose the bridge is balanced, so \(R_{ac}/R_{cb} = R_{ad}/R_{db}\). Then \(V_c = V_d\) in the solution with the arm removed. Inserting any resistance between two points already at equal potential drives no current through it, so the rest of the circuit is undisturbed and the same node voltages remain a valid solution. The current drawn from the supply is therefore unchanged, whatever \(R_g\) may be.

Necessity. Conversely, suppose the bridge is unbalanced. With the arm removed, \(V_c \ne V_d\). Inserting a finite \(R_g\) now forces a current \(I_g\) through it, which redistributes the currents in all four arms and changes the total drawn from the supply. Since \(I_g\) depends continuously on \(R_g\), so does \(R_{ab}\). Independence therefore fails.

Numerical demonstration — Problem 12, balanced. The arm was 5 Ω. Removing it gave 7.2 Ω; shorting it gave 7.2 Ω. Those are the two extreme values of \(R_g\), namely \(\infty\) and 0, and they agree:

Numerical demonstration — Problem 13, unbalanced. With the 18 Ω arm present, \(R_{ab} = 8\ \Omega\). Remove the arm and the network becomes two series chains in parallel:

Short the arm instead, and \(c\) merges with \(d\):

Three different values — 7.6, 8.00 and 8.18 Ω — for three values of \(R_g\). The dependence is real, and it is monotonic.