Introduction to Electrical Circuits
Series, parallel, and series-parallel — the shape of every CP circuit.
You’re out on a routine bi-monthly check, standing at the cabinet, and you check the output with your multimeter: 12 volts DC, 8 amps. Two months ago, on the last visit, the same dial setting was putting 14 amps. Six amps have gone somewhere in the circuit.
You check the connections. Tight. The rectifier isn’t in alarm and nothing on the cabinet hardware looks like it’s failing. Using your multimeter, you start tracing the circuit — out along the header cable to the groundbed, where six anodes sit in a row. At the junction box, voltage reads 11.1 V. By the time you do the math on what the cable alone is eating, you’ve lost almost a full volt before the current ever reached the soil. And when you look at how current splits across the six anode branches, it isn’t even — the two anodes nearest the junction box are carrying nearly twice the current of the two farthest out.
This isn’t a rectifier problem. It isn’t a soil problem. It’s a circuit problem. Understanding why it’s happening — and what to do about it — starts with knowing how the three shapes of DC circuits behave. That’s what this module is for.
Quick reference, before we go further
The vocabulary you’ll need
Every CP circuit has the same three quantities running through it. We’ll use them by name throughout this module, and we’ll keep the introductions brief here because two later modules in this rung unpack them in detail.
Voltage (V, measured in volts) is electrical pressure — the driving force that pushes current through a circuit. Read it at the rectifier output, between two test station leads, or across a shunt.
Current (I, measured in amperes) is the rate of charge flow through the circuit. In CP work, current is what does the actual job — current onto the pipe drives the protective electrochemical reaction.
Resistance (R, measured in ohms, Ω) is opposition to current flow. In a CP circuit, resistance lives in many places: in the cable, at the anode-to-soil contact, at the structure-to-electrolyte interface.
The relationship between the three is Ohm’s Law:
That formula appears repeatedly in the worked examples below. We use it. We don’t teach it from every angle in this module.
For depth on these three: EC-002 (Ohm’s Law and Basic Circuit Calculations) works the formula from every angle, including the rearrangements R = V÷I and I = V÷R, plus a full set of practice problems. EC-003 (Voltage, Current, and Resistance in DC Systems) walks through how each quantity is measured at a real CP test point. From here on, EC-001 is about the shape of the circuit these three quantities live in.
Series circuits — one path, one current
A series circuit has only one path for current to flow. Every element sits end-to-end on that single path. If the path opens anywhere, everything downstream stops.
Three rules govern series:
- Current is the same through every element. Whatever current leaves the source is the same current at every point along the path. (Kirchhoff’s Current Law, or KCL, applied to the trivial case of a single path — there’s nowhere else for the current to go.)
- Resistances add. Total resistance RT = R1 + R2 + R3 + …
- Voltage drops sum to the source voltage. If the source is putting out 20 V and the path has elements that each drop some voltage, all those drops have to add up to 20 V. (Kirchhoff’s Voltage Law, or KVL.)
Where this lives in CP: any cable run is a series element. Every length of conductor between the rectifier and the groundbed adds resistance in series, which means every length of conductor consumes some of the available voltage before the current ever reaches the soil. We call that loss IR drop — current times resistance, expressed as a voltage.
Why IR drop matters in the field: the rectifier might be putting out 24 V, but if 4 V are lost across the cable on the way to the anodes, only 20 V are doing useful work in the soil. That can be the difference between a system that meets criteria and one that’s chronically underprotected. (24 V = 4 V + 20 V — that’s KVL in numbers.)
Worked example A — a single cable run
500-foot header cable from rectifier to junction box
The cable resistance, given the gauge and length, comes out to 0.6 Ω. The rectifier is putting 15 A down the line.
The voltage drop across the cable alone:
Vdrop = I × R = 15 A × 0.6 Ω = 9 V
That’s nine volts the cable consumed before the current ever reached the soil. If the rectifier output is set at 24 V, the junction box sees 24 − 9 = 15 V — and only that 15 V is available to push current across the anode-to-soil interface.
Worked example B — a multi-segment cable chain
Same path, with a loose lug in the mix: three series elements
Main feed cable 0.5 Ω, a loose lug at the (+) header junction box terminal 0.1 Ω, anode lead wire 0.3 Ω. All of those sit on the single path from the rectifier to the soil.
Total series resistance:
RT = 0.5 + 0.1 + 0.3 = 0.9 Ω
With the same 15 A flowing through that path:
Vdrop, total = 15 A × 0.9 Ω = 13.5 V
Almost a third more loss than the single cable in example A — and that’s why cable sizing, lug torque, and connection quality keep showing up as the things that eat your voltage budget. Each one is a series element. Each one adds.
In a series circuit, the current is the same everywhere on the path. Resistance and voltage drop are what change from one element to the next. (That’s KVL — the voltage drops add up to the source.)
Parallel circuits — multiple paths, divided current
A parallel circuit has more than one path between the same two points. Current arriving at the start of those paths divides among them and rejoins at the end. Every parallel branch sees the same voltage across it because every branch connects the same two points.
Three rules govern parallel:
- Voltage is the same across every branch. All branches connect the same two points; that defines the voltage they share.
- Branch currents add to total current. Whatever splits out at the front end has to add up to whatever flowed in. IT = I1 + I2 + I3 + … (KCL applied at the junction.)
- Total resistance is always less than the smallest branch resistance. Adding more parallel paths gives current more options, which lowers overall opposition.
The formula for parallel resistance:
And the shortcut for branches of equal resistance: RT = R ÷ N (where N is the count of equal branches).
Where this lives in CP: every anode array is a parallel circuit. Each anode is a separate current path from the junction box into the soil. Adding anodes adds paths, which lowers total array resistance, which lets the rectifier push more current at the same voltage.
Why parallel anode design matters in the field: each anode in a groundbed is one more path for current, and adding paths drops the array’s resistance to the soil. Six 6-Ω anodes present just 1 Ω to the circuit. That’s how groundbeds get sized for a target current — by counting paths.
Worked example A — three equal anodes
Three anodes, each 5 Ω, wired in parallel
Using the equal-branch shortcut:
RT = 5 Ω ÷ 3 = 1.67 Ω
That’s the punch line. Three 5-ohm anodes don’t behave like 5 ohms to the circuit — they behave like 1.67 Ω. Less than a third of any single anode’s resistance, because the current now has three paths to choose from.
Worked example B — six equal anodes (a typical groundbed)
Six anodes, each 6 Ω, wired in parallel
RT = 6 Ω ÷ 6 = 1.0 Ω
Six anodes, each individually a 6-ohm path, present a 1-ohm load to the circuit as a group. The pattern is consistent: R drops as N grows. Doubling the number of equal anodes halves the array resistance.
One thing this also means: the current any individual anode carries is its share of the total. If the array sees 1 Ω and the rectifier delivers 15 A into it, each anode carries roughly 15 ÷ 6 ≈ 2.5 A — assuming the soil resistance at each anode is actually equal. (In practice it isn’t always — that’s where uneven current distribution shows up, and it’s also why we measure individual anode currents during a system check.)
In a parallel circuit, the voltage is the same across every branch. Currents and resistances are what split. (That’s KCL — the branch currents add up to the total.)
Series-parallel — how real CP circuits are wired
Real impressed-current CP circuits aren’t pure series and they aren’t pure parallel. They’re series-parallel: a series element (the cable from rectifier to junction box) feeds a parallel element (the array of anodes), which collectively returns through the soil and the structure.
A typical impressed-current CP circuit reads as series (cable) feeding parallel (anode array).
To solve a series-parallel circuit, reduce it in steps:
- Calculate the equivalent resistance of the parallel portion (the anode array).
- Add that to the series elements (cable, connections). That’s total circuit resistance.
- Apply Ohm’s Law:
IT = Vsource ÷ RT. - Calculate the voltage at the junction box:
Vjct = Vsource − (IT × Rseries). (KVL.) - Calculate individual branch currents:
Ibranch = Vjct ÷ Rbranch.
Worked example A — equal anode resistances (the textbook case)
20 V rectifier · 0.5 Ω cable · six anodes at 6 Ω each
Step 1. Parallel anode equivalent: Rparallel = 6 ÷ 6 = 1.0 Ω
Step 2. Total circuit resistance: RT = 0.5 + 1.0 = 1.5 Ω
Step 3. Total current: IT = 20 ÷ 1.5 ≈ 13.3 A
Step 4. Junction voltage: Vjct = 20 − (13.3 × 0.5) = 13.35 V
Step 5. Per-anode current: Ianode = 13.35 ÷ 6 ≈ 2.23 A
Six anodes, each carrying about 2.23 A, summing to roughly 13.3 A back through the system. That’s what the circuit should look like in steady state.
Worked example B — unequal anode resistances (the realistic case)
20 V rectifier · 0.5 Ω cable · six anodes — but R = 4, 5, 6, 6, 8, 10 Ω
Same procedure, no equal-branch shortcut.
Step 1. Parallel anode equivalent:
1 ÷ Rparallel = 1/4 + 1/5 + 1/6 + 1/6 + 1/8 + 1/10
= 0.250 + 0.200 + 0.167 + 0.167 + 0.125 + 0.100 = 1.009
Rparallel = 1 ÷ 1.009 ≈ 0.99 Ω
Step 2. Total circuit resistance: RT = 0.5 + 0.99 ≈ 1.49 Ω
Step 3. Total current: IT = 20 ÷ 1.49 ≈ 13.4 A
Step 4. Junction voltage: Vjct = 20 − (13.4 × 0.5) ≈ 13.3 V
Step 5. Per-anode currents (Vjct ÷ Reach):
- 4 Ω → 3.33 A
- 5 Ω → 2.66 A
- 6 Ω → 2.22 A
- 6 Ω → 2.22 A
- 8 Ω → 1.66 A
- 10 Ω → 1.33 A
The branch currents add to about 13.4 A — the math closes (small rounding). (KCL: the per-branch currents flowing out of the junction sum back to the total flowing in.) But notice the spread: the lowest-resistance anode is carrying 2.5× the current of the highest-resistance anode. Both are connected to the same junction, both seeing the same 13.3 V — but one anode at 4 Ω is doing roughly two and a half times the work of the one at 10 Ω.
That’s not a system that’s broken. That’s a system where the soil around each anode is different — which is normal. It’s also why uneven current distribution shows up in the field: that’s what series-parallel circuit math predicts when individual anode resistances vary.
Three shapes, three rules: add resistances in series, sum reciprocals in parallel, reduce series-parallel branch by branch. That’s the whole toolkit.
Real CP circuits are series-parallel — a cable chain feeding a parallel anode array. The math doesn’t change. We name the shape, apply the rule, compute the result.
Up next: a walk-through of the same three shapes with the diagrams in motion.
Listen — narrated walkthrough
Introduction to Electrical Circuits
Same scope as the read — series, parallel, and series-parallel — walked through visually so the shapes start to read at a glance instead of as math you have to set up. The deck is built around real CP circuits: a single cable run, an anode array, and the full impressed-current circuit you’d actually meet in the field.
Narrated by Mike Roberts · ~20 min
Listen on the drive in or while waiting for the coating to cure. Come back for the deck or the worked problems whenever you want.
Once you’ve worked through the audio or the deck, head to the Apply lesson for three problems on the same circuit shapes — and then the quiz to lock it in.
Apply — three problems
Work the math
Three problems. One per circuit shape. Each problem opens with a setup and a question. Work the math first — on paper, in your head, however you’d do it in the field — then click each step to compare. The point is not the arithmetic. The point is recognizing which shape applies and reaching for the right rule without looking it up.
How to use this lesson. Read the setup. Solve before you click. Each step reveals the working we used. Match your answer; if it doesn’t match, the steps are there to compare against.
Cable IR drop on a real header run
0.4 Ω, the lugs and connections through the header box at 0.05 Ω, and the anode lead wire at 0.3 Ω.
What total IR drop does the cable path consume, and what voltage is left at the soil interface?
Step 1 — total series resistance
In a series path, resistances add:
RT = 0.4 + 0.05 + 0.3 = 0.75 Ω
Step 2 — total IR drop
With 18 A flowing through every element on the single path:
Vdrop = I × RT = 18 × 0.75 = 13.5 V
That’s 13.5 volts the path consumes before the current ever reaches the soil.
Step 3 — voltage available at the soil
What’s left of the 24 V source after the path drops its share:
Vsoil = 24 − 13.5 = 10.5 V
Only 10.5 volts is doing work across the anode-to-soil interface. More than half the rectifier’s output went to moving current along the wire.
Series rule applied: the same current flows through every element, and resistances add. When the path has more elements, IR drop grows fast. Cable sizing and connection quality are not cosmetic — they are voltage-budget items.
An anode array with mismatched soil
5 Ω, 6 Ω, 8 Ω, and 12 Ω (the 12-ohm anode is in drier soil). The voltage at the junction box reads 12 V.
What’s the equivalent resistance of the parallel array, what current does each anode carry, and what’s the total array current?
Step 1 — sum of reciprocals
For unequal parallel branches, no shortcut. Reciprocals first:
1/RT = 1/5 + 1/6 + 1/8 + 1/12
= 0.200 + 0.167 + 0.125 + 0.0833 = 0.575
Step 2 — equivalent parallel resistance
Reciprocal of the sum:
RT = 1 ÷ 0.575 ≈ 1.74 Ω
Less than the smallest branch (5 Ω) — the parallel rule predicts that, every time.
Step 3 — current per anode
Each branch sees the full 12 V across it. Per-anode current is V ÷ Rbranch:
5 Ω anode → 12 ÷ 5 = 2.40 A6 Ω anode → 12 ÷ 6 = 2.00 A8 Ω anode → 12 ÷ 8 = 1.50 A12 Ω anode → 12 ÷ 12 = 1.00 A
Step 4 — total current and sum check
Branch currents add to total. (KCL.)
IT = 2.40 + 2.00 + 1.50 + 1.00 = 6.90 A
Cross-check using the equivalent resistance: IT = 12 ÷ 1.74 ≈ 6.90 A — the math closes.
And notice the spread: the wettest-soil anode is doing 2.4× the work of the driest-soil anode. Same junction, same voltage, very different currents — predicted entirely by the resistance differences.
Parallel rule applied: voltage is common, currents and resistances split. Equivalent resistance always lands below the smallest branch. Mismatched branch resistances always produce mismatched branch currents — and that’s normal field behavior, not a fault.
Full reduction of a real impressed-current circuit
0.6 Ω. The array has four anodes at 4 Ω, 5 Ω, 6 Ω, and 7 Ω.
What’s the total circuit current, the voltage at the junction box, and the current through each anode?
Step 1 — collapse the parallel anode array
1/Rparallel = 1/4 + 1/5 + 1/6 + 1/7
= 0.250 + 0.200 + 0.167 + 0.143 = 0.760
Rparallel = 1 ÷ 0.760 ≈ 1.32 Ω
Step 2 — add the series cable to get total circuit resistance
Cable in series with the parallel array:
RT = 0.6 + 1.32 = 1.92 Ω
Step 3 — total current from the rectifier
Ohm’s Law on the whole circuit:
IT = Vsource ÷ RT = 30 ÷ 1.92 ≈ 15.6 A
That’s how much current the rectifier is putting out into the system.
Step 4 — junction box voltage
Subtract the cable IR drop from the source voltage. (KVL.)
Vjct = Vsource − (IT × Rcable)
Vjct = 30 − (15.6 × 0.6) = 30 − 9.4 ≈ 20.6 V
The cable consumed about 9.4 volts. The junction box sees roughly 20.6 V — and that’s the voltage every anode branch sees individually.
Step 5 — per-anode currents
Each branch: Ibranch = Vjct ÷ Rbranch
4 Ω → 20.6 ÷ 4 ≈ 5.15 A5 Ω → 20.6 ÷ 5 ≈ 4.12 A6 Ω → 20.6 ÷ 6 ≈ 3.43 A7 Ω → 20.6 ÷ 7 ≈ 2.94 A
Sum: 5.15 + 4.12 + 3.43 + 2.94 ≈ 15.6 A — closes back to the total current. Math is consistent.
Series-parallel rule applied: collapse parallel first, add series, then solve total. Once total current is known, back out junction voltage with the cable IR drop, then per-branch currents from the junction voltage. Five steps. Same procedure every time.
Three problems, three shapes — and the same pattern showed up in each: name the shape, apply the rule, compute the result. That’s the work.
You’re going to do this same kind of math on real cabinets, with real meter readings, on circuits whose anode resistances you don’t know in advance. The shape rules are what let you turn three meter readings into a diagnosis.
Up next: a quiz to confirm the shapes are wired in.
Introduction to Electrical Circuits
Foundation tier · EC TRACK · ELECTRICAL BASICS FOR CP
One module done. Keep going — you'll earn the certificate when you finish this section, and the Foundation medal when you complete every section in the tier.
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