How do you solve a series-parallel resistor network?

Reduce it from the inside out. Replace the innermost parallel group with its equivalent resistance, then add any series elements, and repeat until one value is left. Each step uses only the plain series sum or the parallel reciprocal formula.

Do you calculate series or parallel first?

Neither by default - you start with whichever group is innermost, meaning the one with nothing nested inside it. In a resistor in series with a parallel pair, the parallel pair is innermost, so it goes first.

How do you tell whether two resistors are in series or parallel?

They are in parallel if both ends connect to the same two nodes. They are in series if they share one node and nothing else connects to that node. If a third wire joins between them, they are neither.

Can every resistor network be reduced this way?

No. Only series-parallel networks can. A bridge circuit such as a Wheatstone bridge has no two resistors sharing both nodes, so it needs a delta-to-star conversion, mesh analysis or a simulator.

How deep can a mixed network be nested?

As deep as you like. The reduction is recursive, so a parallel bank containing series chains containing further parallel banks is solved by the same two rules applied repeatedly. The calculator above accepts arbitrary nesting and shows every step.

Mixed Series-Parallel Resistance Calculator

Mixed series-parallel resistor networks are solved by reducing them from the inside out: collapse each parallel group to a single equivalent resistance, then add what is left in series, repeating until one value remains.

How to Reduce a Series-Parallel Resistor Network

There is no single formula for a mixed network, and that is the point: any series-parallel circuit, however deep, is solved by applying the two rules you already know in the right order. The skill is the order, not the arithmetic.

The two rules being applied

R_{s e r i e s} = R_{1} + R_{2} + \dots + R_{n}

\frac{1}{R_{p a r a l l e l}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \dots + \frac{1}{R_{n}}

Working from the innermost group outwards

Find the group that has no other group inside it - the innermost parallel bank or series pair - and replace it with its single equivalent value. The network is now simpler by one level. Repeat. Every series-parallel network collapses to one number this way, and the answer does not depend on which innermost group you pick first.

For a resistor in series with a parallel pair - the most common shape by far - the two steps combine into one expression:

R_{e q} = R_{1} + \frac{R_{2} \times R_{3}}{R_{2} + R_{3}}

Collapse the parallel bank first, then add the series element.

The order matters and cannot be swapped. Adding R1 to R2 before dealing with the parallel branch gives a different, wrong answer, because R1 is not across the same two nodes as R2 and R3. Reduce the deepest group first, always.

Circuits this method cannot solve

A network is series-parallel only if it can be taken apart into groups that are wholly in series or wholly in parallel. Bridge circuits - a Wheatstone bridge with a resistor across the middle - cannot, because no two resistors in them share both nodes. Those need a delta-to-star conversion or mesh analysis, and step-by-step reduction will not reach them.

Worked example

Given

R1 = 330 Ω in series with the bank
R2 = 4.7 kΩ
R3 = 1.5 kΩ (R2 ∥ R3)

Working

Step 1 — collapse the parallel bank:
R2 ∥ R3 = (4700 × 1500) / (4700 + 1500) = 1137.1 Ω
Step 2 — add the series resistor:
Req = 330 Ω + 1137.1 Ω

Answer1.4671 kΩ

Reading a Network Before You Calculate It

Most mistakes in mixed networks happen before any arithmetic, in deciding what is actually in parallel with what. Two resistors are in parallel only if both of their ends connect to the same pair of nodes. Two are in series only if they share a node that nothing else connects to.

That second condition is the one people miss. A resistor with a third wire joining its midpoint is not in series with its neighbour any more, because current can leave at the junction. Trace the nodes, not the drawing - a schematic laid out neatly can hide a parallel pair, and a messy one can suggest a series chain that is not there.

The schematic above is built from the network you enter, so it is a check on your reading of the circuit as well as a calculator. If the drawing does not look like the circuit in front of you, the arithmetic will not match it either.

Where Mixed Networks Come From

Loaded voltage dividers

A divider with anything connected to its output is a mixed network: the lower leg is in parallel with the load, and that combination is in series with the upper leg. This is the single most common series-parallel circuit in real electronics, and calculating it as an unloaded divider is the single most common reason a measured output is lower than expected.

Trimming a value that is already fitted

Adding a resistor across one part of an existing chain creates a parallel bank inside a series path. Working out what to add is exactly what the find-missing-value mode is for.

Attenuators and pads

T and pi attenuators are series-parallel by construction, and their input resistance depends on what is connected to the far end - which makes reducing them from the load backwards the only sensible way to work.

Checking a Mixed Result

A mixed answer has bounds you can verify by inspection, and they catch most errors without redoing the sum:

The total must be larger than any series element on its own, since series only adds.
Every parallel bank must have collapsed to less than its smallest member.
Shorting out a parallel bank should leave just the series elements - a useful limit to sanity-check against.
Removing one branch of a parallel bank must raise the total, never lower it.

The step-by-step derivation above the theory shows each reduction as it happens, so an intermediate value that looks wrong can be caught at the step that produced it rather than at the end.

Common mistakes

Adding the series resistor before collapsing the parallel bank. The order is not interchangeable and the wrong order gives a wrong answer with no warning.
Treating two resistors as parallel when only one end is common. Both ends have to join the same two nodes.
Treating two resistors as in series when a third wire joins at the node between them.
Trying to reduce a bridge network step by step. It is not series-parallel and needs a delta-star conversion or mesh analysis instead.
Ignoring the load on a divider, which turns a two-resistor calculation into a three-resistor mixed one.

Frequently asked questions

How do you solve a series-parallel resistor network?: Reduce it from the inside out. Replace the innermost parallel group with its equivalent resistance, then add any series elements, and repeat until one value is left. Each step uses only the plain series sum or the parallel reciprocal formula.
Do you calculate series or parallel first?: Neither by default - you start with whichever group is innermost, meaning the one with nothing nested inside it. In a resistor in series with a parallel pair, the parallel pair is innermost, so it goes first.
How do you tell whether two resistors are in series or parallel?: They are in parallel if both ends connect to the same two nodes. They are in series if they share one node and nothing else connects to that node. If a third wire joins between them, they are neither.
Can every resistor network be reduced this way?: No. Only series-parallel networks can. A bridge circuit such as a Wheatstone bridge has no two resistors sharing both nodes, so it needs a delta-to-star conversion, mesh analysis or a simulator.
How deep can a mixed network be nested?: As deep as you like. The reduction is recursive, so a parallel bank containing series chains containing further parallel banks is solved by the same two rules applied repeatedly. The calculator above accepts arbitrary nesting and shows every step.

Assumptions and limitations for Mixed Series-Parallel Resistance Calculator are listed on the About page. Every worked example on this site is checked against the same solver the calculator uses.