Series-Parallel Circuits
The circuits you have studied so far — pure series and pure parallel — are useful for learning the rules, but they are rarely what you find inside real equipment. Open any transceiver, power supply, or antenna tuner and you will see resistors, capacitors, and inductors arranged in combinations: some in series, some in parallel, mixed together in ways that may look complicated at first glance. These are called series-parallel circuits, and they are the normal state of electronics.
The good news is that you do not need any new rules to analyse them. You already have everything you need — the series rules from two lessons ago and the parallel rules from the last lesson. The method for handling series-parallel networks is simply to apply those rules systematically, one group at a time, collapsing a complex circuit into progressively simpler ones until only a single equivalent resistance remains. This process is called network reduction, and it is one of the most practical skills in electronics.
This lesson teaches you the reduction method from scratch, walks through multiple fully-worked examples of increasing complexity, and shows you how to expand back through the network to find the voltage and current at every component — the process called back-substitution. By the end, you will be able to analyse any series-parallel resistor network confidently.
- Identifying Series and Parallel Groups
- The Reduction Method — Step by Step
- Worked Example 1: One Series, One Parallel Group
- Back-Substitution: Finding Individual Values
- Worked Example 2: Two Levels of Nesting
- More Practice Examples
- Power in Series-Parallel Circuits
- When Reduction Does Not Work
- Series-Parallel Circuit Calculator
- Experiment: Build and Measure a Series-Parallel Network
- Series-Parallel Circuits in Ham Radio
A common series-parallel arrangement: R1 is in series with the supply; R2 and R3 form a parallel group. The reduction method collapses the parallel group first, then the result is a simple two-resistor series circuit.
View LargerIdentifying Series and Parallel Groups
Before you can reduce a network, you need to be able to look at it and identify which components are in series with each other and which are in parallel. This is a skill that improves with practice, but there are reliable rules to apply.
Two resistors are in series if the only path between them is a single node — a point with nothing else connected to it. Equivalently, two resistors are in series if the same current must flow through both. If you have R1 and R2 end-to-end with no junction between them (no other wires or components branching off the middle), they are in series.
Two resistors are in parallel if both terminals of each connect to exactly the same pair of nodes. If the left end of R2 and the left end of R3 both connect to node A, and the right end of R2 and the right end of R3 both connect to node B, then R2 and R3 are in parallel between nodes A and B.
The quick mental test: imagine a single electron travelling through the circuit. At a junction, does it have a choice of which path to take? If it must pass through one component and then another with no choice, those components are in series. If it can choose one component and skip another, those components are in parallel.
In a schematic, the layout usually makes this clear: series components appear as a chain; parallel components appear as separate branches side by side between the same two wires. The difficulty arises with redrawn or messy schematics where the visual arrangement does not match the topology. When in doubt, mark the nodes (give each electrically distinct point a label) and check which components share both nodes — those are in parallel.
Looking for innermost groups
In a complex network, start by looking for the innermost group — the group that has no other series-parallel structure inside it. These will always be either a simple series pair or a simple parallel pair. Reduce that innermost group first. Then look for the next innermost group in the simplified circuit, and reduce again. Keep repeating until one equivalent resistor remains.
Network reduction proceeds from the innermost group outward. Each step replaces a group with its single equivalent resistor, producing a simpler circuit until only Rtotal remains.
View LargerThe Reduction Method — Step by Step
The reduction method is a systematic procedure. Follow these steps in order:
Step 2. Find the innermost purely-series group. Replace it with a single equivalent resistor using: Req = R1 + R2 + ...
Step 3. Find the innermost purely-parallel group. Replace it with a single equivalent resistor using: 1/Req = 1/R1 + 1/R2 + ... (or product-over-sum for two resistors).
Step 4. Redraw the simplified circuit.
Step 5. Repeat steps 2–4 until a single resistor remains. That is Rtotal.
Step 6. Find total current: Itotal = Vsupply / Rtotal.
Step 7. Back-substitute through the network levels to find voltage and current at every original component.
The key discipline in this method is to work systematically — always labelling your intermediate results, always redrawing the simplified circuit, and always checking your work at the end using the two conservation laws: voltage drops around any loop must sum to zero, and currents into any node must equal currents out of that node.
Worked Example 1: One Series, One Parallel Group
This is the most common configuration. A series resistor is followed by a parallel pair. The circuit is: 12 V supply → R1 = 330 Ω (series) → parallel combination of R2 = 1 kΩ and R3 = 1.5 kΩ → return to supply.
Step 1 — Reduce the parallel pair (R2 ∥ R3):
Rparallel = (R2 × R3) / (R2 + R3) = (1000 × 1500) / (1000 + 1500)
Rparallel = 1,500,000 / 2500 = 600 Ω
Step 2 — The circuit is now simply R1 in series with Rparallel:
Rtotal = R1 + Rparallel = 330 + 600 = 930 Ω
Step 3 — Total current from supply:
Isupply = Vsupply / Rtotal = 12 / 930 = 12.90 mA
Back-substitution phase:
Step 4 — Voltage across R1 (series: carries Isupply):
VR1 = Isupply × R1 = 0.01290 × 330 = 4.26 V
Step 5 — Voltage across the parallel pair (remainder of supply voltage):
Vparallel = Vsupply − VR1 = 12 − 4.26 = 7.74 V
Verify: Vparallel = Isupply × Rparallel = 0.01290 × 600 = 7.74 V ✓
Step 6 — Branch currents in the parallel pair (both see 7.74 V):
IR2 = Vparallel / R2 = 7.74 / 1000 = 7.74 mA
IR3 = Vparallel / R3 = 7.74 / 1500 = 5.16 mA
Check 1 — Currents at the parallel junction:
IR2 + IR3 = 7.74 + 5.16 = 12.90 mA = Isupply ✓
Check 2 — Voltages around the loop:
VR1 + Vparallel = 4.26 + 7.74 = 12.00 V = Vsupply ✓
Summary table:
R1 (330 Ω): V = 4.26 V, I = 12.90 mA, P = 54.9 mW
R2 (1000 Ω): V = 7.74 V, I = 7.74 mA, P = 59.9 mW
R3 (1500 Ω): V = 7.74 V, I = 5.16 mA, P = 39.9 mW
Total power: 54.9 + 59.9 + 39.9 = 154.7 mW
Check: Vsupply × Isupply = 12 × 0.01290 = 154.8 mW ✓
Back-Substitution: Finding Individual Values
Back-substitution is the reverse of the reduction process. You built up the circuit during reduction; now you expand back down from the total to each individual component. The rules at each level are:
When expanding a series group: all components in the group carry the same current as the group itself. Use V = I × R to find the voltage across each one. The sum of these voltages must equal the voltage across the group.
When expanding a parallel group: all branches in the group see the same voltage as the group itself. Use I = V / R to find the current through each branch. The sum of these currents must equal the total current into the group.
The two checks you always do at the end — sum all voltage drops around any path (must equal supply voltage) and sum all currents into each node (in must equal out) — are the foundation of circuit analysis. They are direct applications of Kirchhoff's laws, which you will study in the next two lessons. Getting into the habit of checking both of these now will serve you well throughout your electronics career.
A practical tip: work your results into a table as you go. List each component, its resistance, voltage, current, and power. The table makes it easy to spot errors and provides a complete record of the circuit's behaviour.
Worked Example 2: Two Levels of Nesting
This example is more complex, with a parallel group nested inside a series group that is itself in parallel with another resistor. Circuit: 15 V supply → parallel combination of R1 = 2.2 kΩ and [R2 = 1 kΩ in series with R3 = 680 Ω] → return.
The inner series group is R2 + R3. This feeds into a parallel combination with R1. Reduction proceeds from the innermost group outward.
Rseries = R2 + R3 = 1000 + 680 = 1680 Ω
Now the circuit is: 15 V → R1 (2200 Ω) in parallel with Rseries (1680 Ω) → return.
Step 2 — Reduce the parallel pair (R1 ∥ Rseries):
Rtotal = (2200 × 1680) / (2200 + 1680) = 3,696,000 / 3880 = 952.6 Ω
Step 3 — Total supply current:
Isupply = 15 / 952.6 = 15.74 mA
Back-substitution:
Step 4 — Both parallel branches see 15 V (full supply, since both connect directly across supply):
IR1 = 15 / 2200 = 6.82 mA (through R1)
Ibranch = 15 / 1680 = 8.93 mA (through the R2+R3 branch)
Check: 6.82 + 8.93 = 15.75 mA ≈ 15.74 mA ✓ (rounding)
Step 5 — Expand the R2+R3 series branch (both carry Ibranch = 8.93 mA):
VR2 = Ibranch × R2 = 0.00893 × 1000 = 8.93 V
VR3 = Ibranch × R3 = 0.00893 × 680 = 6.07 V
Check: VR2 + VR3 = 8.93 + 6.07 = 15.00 V = supply ✓
Summary:
R1 (2200 Ω): V = 15 V, I = 6.82 mA, P = 102.3 mW
R2 (1000 Ω): V = 8.93 V, I = 8.93 mA, P = 79.7 mW
R3 (680 Ω): V = 6.07 V, I = 8.93 mA, P = 54.2 mW
Total: P = 102.3 + 79.7 + 54.2 = 236.2 mW
Check: 15 × 15.74 mA = 236.1 mW ✓
Notice the key observation from this example: when a parallel branch contains a series sub-chain (R2 and R3 in series), the current through both components in that sub-chain is the same — the branch current — not the total supply current. Only after the branches split does each branch carry its own independent current. This is the most common source of errors in series-parallel analysis: confusing which current flows where.
More Practice Examples
Circuit: 9 V → R1 = 100 Ω (series) → R2 = 470 Ω ∥ R3 = 330 Ω (parallel) → R4 = 150 Ω (series) → return.
Step 1: Rparallel = (470 × 330) / (470 + 330) = 155,100 / 800 = 193.9 Ω
Step 2: Rtotal = R1 + Rparallel + R4 = 100 + 193.9 + 150 = 443.9 Ω
Step 3: Isupply = 9 / 443.9 = 20.27 mA
Back-substitution:
VR1 = 0.02027 × 100 = 2.03 V
Vparallel = 0.02027 × 193.9 = 3.93 V
VR4 = 0.02027 × 150 = 3.04 V
Check: 2.03 + 3.93 + 3.04 = 9.00 V ✓
Branch currents through parallel group:
IR2 = 3.93 / 470 = 8.36 mA
IR3 = 3.93 / 330 = 11.91 mA
Check: 8.36 + 11.91 = 20.27 mA ✓
A 12 V supply feeds R1 = 560 Ω in series with a parallel combination of R2 = 1.2 kΩ and an unknown R3. The measured total current is 15 mA. Find R3.
Step 1: Find Rtotal: Rtotal = V / I = 12 / 0.015 = 800 Ω
Step 2: Find Rparallel: Rparallel = Rtotal − R1 = 800 − 560 = 240 Ω
Step 3: Find R3 from product-over-sum rearranged:
Rparallel = (R2 × R3) / (R2 + R3)
240 (R2 + R3) = R2 × R3
240 × R2 + 240 × R3 = R2 × R3
240 × 1200 = R3 × (1200 − 240) = R3 × 960
R3 = 288,000 / 960 = 300 Ω
Check: (1200 × 300) / (1200 + 300) = 360,000 / 1500 = 240 Ω ✓
Rtotal = 560 + 240 = 800 Ω; I = 12/800 = 15 mA ✓
Power in Series-Parallel Circuits
Power calculations in series-parallel circuits follow the same rules as in pure series or pure parallel circuits — just applied at the level where you know both the voltage across and the current through each component.
The total power delivered by the supply equals Vsupply × Isupply. This must equal the sum of all individual component powers. If your individual power values do not sum to the supply power, you have made an arithmetic error somewhere.
Ptotal = Vsupply × Isupply = P1 + P2 + P3 + ...
For a series resistor in a series-parallel circuit, the easiest formula is P = I² × R, because you know the current through it from the total supply current. For a resistor in a parallel group, the easiest formula is P = V² / R, because you know the voltage across the group from the back-substitution. Use whichever formula is most convenient at each step — all three give the same result.
Component power ratings matter in series-parallel circuits. A low-resistance parallel branch may carry a large fraction of total current and dissipate significant power. Always check that no component is operating above its rated wattage. The worked examples above include power calculations precisely so you can verify this.
When Reduction Does Not Work
The reduction method works for any network where you can always find a pair of components that are either purely in series or purely in parallel with each other. Most practical circuits you will encounter on the bench are reducible in this way.
However, some network topologies cannot be reduced by this method. The most famous example is the Wheatstone bridge: four resistors arranged in a diamond with a meter or component across the middle. In a balanced bridge, the middle element carries no current, which simplifies the analysis. But in an unbalanced bridge, current flows through all five elements simultaneously, and no two are in a simple series or parallel relationship with each other. To solve a non-reducible network, you need the full power of Kirchhoff's laws (covered in the next two lessons) or advanced techniques like the delta-star (Y-Δ) transformation or node voltage analysis.
For now, the key lesson is: before attempting a reduction, check that the circuit is reducible. If you cannot find any two resistors that are clearly in series or clearly in parallel, the reduction method may not apply.
Series-Parallel Circuit Solver
Series-Parallel Circuit Solver
Topology: Vsupply → R1 (series) → [R2 ∥ R3] (parallel pair) → return. Enter supply voltage, the series resistor R1 (enter 0 if no series resistor), and the two parallel resistors R2 and R3. The calculator finds Rtotal, supply current, and voltage and current for each component.
⚖ Experiment: Build and Measure a Series-Parallel Network
Build the exact topology from Worked Example 1 and verify all calculated values. This experiment confirms that voltages at intermediate nodes follow the prediction and that current splits correctly at junctions.
- 9 V battery with clip connector
- R1: 330 Ω; R2: 1 kΩ; R3: 1.5 kΩ (or nearest available values)
- Breadboard and jumper wires
- Digital multimeter
- Predict first. Use the calculator above with V = 9 V, R1 = 330 Ω, R2 = 1000 Ω, R3 = 1500 Ω. Write down: Rtotal, Isupply, VR1, Vparallel, IR2, IR3. Keep these figures to compare against measurements.
- Build the circuit. Connect one end of R1 to the battery positive terminal. Connect the other end of R1 to node A (a row on the breadboard). Connect R2 from node A to the battery negative rail. Connect R3 from node A to the battery negative rail. R2 and R3 are now in parallel between node A and the negative rail, and R1 is in series between battery positive and node A.
- Measure VR1. Set multimeter to DC volts. Touch probes across R1 (battery positive to node A). Record the reading and compare to prediction.
- Measure Vparallel. Touch probes from node A to battery negative. Record and compare. Confirm that VR1 + Vparallel = battery terminal voltage.
- Measure total supply current. Break the wire between the battery positive and R1. Insert the multimeter (set to DC mA) in series. Record Isupply.
- Measure branch currents. Break the wire to R2 and insert the meter in series. Record IR2. Restore, then break and measure IR3. Confirm IR2 + IR3 = Isupply.
- Calculate node voltage from first principles. From your measured Isupply and R1 value, compute Isupply × 330 and confirm it matches your measured VR1. This completes the loop between Ohm's Law and direct measurement.
Measured values will agree with predictions within about 5–10% (due to resistor tolerance and battery internal resistance). VR1 + Vparallel equals measured battery terminal voltage. IR2 + IR3 equals Isupply. Node A maintains a stable voltage regardless of which branch you remove, demonstrating that each branch in the parallel group operates independently of the others.
Series-Parallel Circuits in Ham Radio
Transistor voltage-divider bias network
The voltage-divider bias configuration is the most common way to stabilise a transistor amplifier's operating point, and it is a direct application of series-parallel circuit analysis. Two resistors (R1 and R2) form a voltage divider from the supply to ground. The base of the transistor connects to the midpoint of this divider, so the transistor's input impedance appears in parallel with R2. The emitter resistor Re sits in series with the transistor's emitter. The complete bias network is a series-parallel circuit: Vcc → R1 → [R2 ∥ Rin(transistor)] → Re → ground. Analysing this bias network — determining the base voltage and thus the collector current — requires exactly the series-parallel reduction technique taught in this lesson.
Attenuator pads (Pi and T networks)
When connecting a signal generator to a receiver or an amplifier output to the next stage, an attenuator pad is often inserted to reduce the signal level and ensure proper impedance matching. A T-pad has two series resistors and one shunt (parallel) resistor. A Pi-pad has two shunt resistors and one series resistor. Both are series-parallel networks. Calculating the attenuation and the input/output impedance of a pad requires reducing the network and applying back-substitution — the same procedure as in this lesson, applied to the specific impedance values of the system (typically 50 Ω in RF work).
Power supply load regulation
A power supply does not have zero internal impedance — there is always some series resistance (the output impedance of the regulator or the winding resistance of the transformer). When a load is connected, the load resistance appears in parallel with any bypass capacitors. The output voltage seen by the load is determined by the voltage divider formed by the supply's series output impedance and the parallel combination of the load and bypass capacitors. Series-parallel analysis tells you immediately how much the output voltage will drop under load — this is the concept of load regulation.
Receiver front-end input networks
The input circuit of a receiver's front end is typically a series-parallel network designed to match the antenna impedance (usually 50 Ω or 75 Ω) to the input impedance of the first amplifier stage. A series inductor in parallel with the antenna terminal's bypass capacitor, feeding into the amplifier's input resistance, forms a series-parallel combination whose total impedance at the operating frequency determines how well the antenna is matched. At RF, the inductors and capacitors have frequency-dependent reactance (studied in AC circuit theory), but the topology and the reduction method are the same.
Frequently Asked Questions
How do I know whether two resistors in a schematic are in series or in parallel?
Two resistors are in series if the only connection between their joined ends is a single wire (no other components or branches tapping off that point). The same current must flow through both. Two resistors are in parallel if both terminals of each component connect to exactly the same pair of nodes — the same entry node and the same exit node. If in doubt, trace the current path: is there a choice of route, or must the electron pass through both components in sequence? A choice means parallel; no choice means series.
Can any resistor network be reduced by the series-parallel method?
No. The method requires that you can always find at least one pair of components that are purely in series or purely in parallel. Certain topologies — most notably the Wheatstone bridge — have cross-connections that prevent this. In a Wheatstone bridge, none of the four main resistors is in a simple series or parallel relationship with any other, so reduction fails. Such networks require Kirchhoff's laws (mesh or node analysis), Thevenin or Norton equivalents, or the delta-star transformation. In practice, the circuits a ham encounters at the bench are almost always reducible.
What is the difference between Rtotal, Requivalent, and RThevenin?
Rtotal and Requivalent are used interchangeably and mean the same thing: the single resistance value that would draw the same current from the supply as the entire network. RThevenin is a related but different concept: it is the equivalent resistance seen looking into the network from one pair of terminals, with all independent voltage sources replaced by short circuits and all current sources replaced by open circuits. For a simple network driven by a single supply, RThevenin and Rtotal are often the same, but for networks with multiple sources or with loads attached to specific internal nodes, they differ. Thevenin's theorem is covered in a later lesson in this module.
Does it matter which group I reduce first?
The final Rtotal will be the same regardless of order, as long as you correctly identify what is truly in series and what is truly in parallel at each step. However, reducing the wrong pair (combining components that are not actually in series or not actually in parallel with each other) gives a wrong answer. Always work from the innermost group outward — start with the tightest-nested pair where you are confident of the relationship, reduce it, then reassess the resulting circuit before choosing the next group.
Why does the voltage at a node change when I change the load?
Because the load resistance appears in parallel with whatever is already at that node, changing the total resistance at the parallel junction. This changes the voltage division between the series elements and the parallel group. In a voltage divider circuit (R1 in series with R2), if you connect a load resistor in parallel with R2, the effective parallel resistance is R2 ∥ Rload — lower than R2 alone — which reduces the voltage across the parallel section. This is called loading the voltage divider, and it is a critical concept in impedance matching and circuit design. The lesson on voltage dividers, coming up later in this module, covers this in full detail.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.