EC TRACK · ELECTRICAL BASICS FOR CP

Ohm's Law and Basic Circuit Calculations

Apply Ohm's Law to calculate voltage, current, and resistance — the working formula every field tech needs.

Foundation ~10 minutes PDH/CEC eligible

You’ve opened the lid of a test station out on a midway run. Inside the box, the bond cable runs through a calibrated shunt — a small precision resistor labeled 0.1 Ω on its housing — and continues out to a buried set of magnesium anodes wired to the pipeline. The shunt is there for one reason: to let you measure the current the anode group is putting onto the pipe without breaking the circuit to do it.

You set the meter to millivolts DC. Black lead on one end of the shunt, red lead on the other. The display settles on 25 mV.

That number isn’t current. It’s the voltage drop across a known resistance. To turn it into amps, you need the formula that connects the three quantities together — the same one you’ll reach for every time you read a shunt, calculate a cable IR drop, or back out a circuit resistance from a rectifier reading. Ohm’s Law. EC-001 introduced the shapes of CP circuits. This module is about the math that runs inside them.

Quick reference, before we go further

The three quantities, one more time

You met them in EC-001. We’ll keep the introductions brief here because EC-003 unpacks each one in detail, including how it’s measured at a real CP test point.

Voltage (V, measured in volts) is electrical pressure — what pushes current through a circuit. At the rectifier output, between two test station leads, or across a shunt.

Current (I, measured in amperes) is the rate of charge flow. 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 the cable, at the anode-to-soil interface, at the structure-to-electrolyte interface, and — by design — in the calibrated shunt you just put your meter across.

The relationship is Ohm’s Law:

V = I × R

Three quantities, one equation. Know any two of them and you can solve for the third. That’s not a slogan — it’s the move you’ll make at every cabinet, every test station, every rectifier check from here forward.

The shunt — Ohm’s Law as a measurement tool

Before we work the formula, it’s worth a minute to understand what’s going on in that test station you just opened.

A shunt is a precision resistor with a known, calibrated resistance — usually a small value like 0.1 Ω, 0.01 Ω, or 0.001 Ω. It’s installed in series with the current you want to measure, so all that current has to pass through it. Putting a known resistance directly in the current path is deliberate: it turns an ammeter problem into a voltmeter problem.

Here’s why that matters. A direct ammeter reading would require breaking the circuit, putting the meter inline, and pushing the current through the meter itself — not practical in the field, and not safe at higher currents. With a shunt, the current keeps flowing on its normal path. You put a voltmeter on millivolts, place the leads across the shunt, and read the IR drop. Then you reach for Ohm’s Law to convert that voltage drop into a current.

Shunt in a test station bond. Current flows on its normal path; the voltmeter reads the IR drop the shunt makes visible.

This is Ohm’s Law put to work as a measurement technique. The shunt’s resistance is small enough that it barely affects the circuit it sits in (the anode group, the bond cable, and the soil add up to far more), but it’s large enough that the voltage drop across it is readable on a millivolt meter. The whole arrangement is built around I = V ÷ R.

You’ll see two common shunt sizes in CP work, used for two different jobs:

Test station bond shunts — typically 0.1 Ω or 0.01 Ω. They’re sized to read the smaller currents flowing in galvanic anode bonds, foreign-line bonds, or test connections — currents from a few milliamps up to several amps.
Rectifier shunts — typically calibrated as a millivolt-per-amp rating like “50 mV at 50 A” (which works out to 0.001 Ω) or “100 mV at 10 A” (which is 0.01 Ω). They’re sized for the larger currents that an impressed-current rectifier outputs — typically from a few amps up to tens of amps.

In both cases, the move is the same: read the millivolts, know the shunt size, calculate the amps.

Why we measure in millivolts. A 0.1 Ω shunt carrying 1 A drops only 100 mV across itself. A 0.001 Ω rectifier shunt carrying 25 A drops 25 mV. These are small numbers — too small to read accurately on the volts scale. Switching the meter to millivolts gets you the resolution you need to read the current the shunt was built to measure.

Ohm’s Law — three forms, one relationship

Three quantities. One equation. Three rearrangements. Use whichever form lines up with what you know and what you’re solving for.

V = I × R

The first form. Use it when you know current and resistance, and you want voltage. (Cable IR drop calculations live here.)

I = V ÷ R

The second form. Use it when you know voltage and resistance, and you want current. (Shunt readings live here. Most field calculations live here.)

R = V ÷ I

The third form. Use it when you know voltage and current, and you want resistance. (Diagnostic calculations live here — backing out total circuit resistance from rectifier readings.)

Some people remember the three forms with a triangle: V on top, I and R on the bottom. Cover the quantity you want to solve for, and the operation you need is what’s left.

Ohm’s Law triangle. Cover V → I × R. Cover I → V over R. Cover R → V over I.

Whether you remember it as a triangle, a saying, or just as “the formula” — the math is the same. Three quantities related by multiplication and division, with the unit relationships baked in.

Unit consistency — the silent failure mode

The math of Ohm’s Law is simple. The mistakes are almost always about units.

The base SI units — volts, amperes, ohms — work cleanly together: 1 V = 1 A × 1 Ω. But CP work happens across a wide dynamic range. Voltages run from microvolts at sensitive measurements to tens of volts at the rectifier. Currents run from milliamps at corrosion-monitoring shunts to tens of amps at impressed-current rectifiers. Resistances run from milliohms in calibrated shunts to thousands of ohms in coating quality. Mix the unit prefixes carelessly and the math gives you an answer that’s off by a factor of a thousand or a million — and looks plausible enough to act on.

The standard prefixes:

Quantity	Base unit	Common prefixes you’ll meet in CP
Voltage	volt (V)	`1 V = 1,000 mV = 1,000,000 μV`
Current	ampere (A)	`1 A = 1,000 mA = 1,000,000 μA`
Resistance	ohm (Ω)	`1 Ω = 1,000 mΩ` · `1 kΩ = 1,000 Ω`

Two strategies work for keeping the units honest:

Convert everything to base units before you calculate. Volts, amps, ohms. The formula falls out clean every time, and the answer is in base units. Convert back at the end if you want millivolts or milliamps.
Use a consistent prefix throughout. If the voltage is in millivolts and the resistance is in ohms, the current comes out in milliamps automatically — because the prefix on the voltage carries through the division. Same logic if voltage is in volts and resistance is in kilohms — current comes out in milliamps.

Match your units. Ohm’s Law works in any consistent unit family — volts, amps, ohms OR millivolts, milliamps, milliohms — but it doesn’t work mixed. Before you calculate, get every number into the same family. The answer will land in matching units.

The mistake to avoid: mixing units mid-calculation. If a shunt reads 25 mV across 0.1 Ω, those units don’t match — millivolts and ohms aren’t from the same family. Convert one or the other first: 25 mV = 0.025 V, then 0.025 V ÷ 0.1 Ω = 0.25 A. Or work in millis throughout: 0.1 Ω = 100 mΩ, then 25 mV ÷ 100 mΩ = 0.25 A. Either way you get 0.25 A (= 250 mA). When an answer feels three orders of magnitude wrong, it’s almost always a unit mismatch.

Worked example A — the test station shunt

Back to the bond cable in the test station from the hook. The shunt is calibrated at 0.1 Ω, the meter is reading 25 mV across it, and we want the current the magnesium anode group is putting onto the pipeline.

Find the bond current — 0.1 Ω shunt, 25 mV reading

What we know: V = 25 mV across the shunt; R_shunt = 0.1 Ω. What we want: I.

The form of Ohm’s Law that solves for current:

I = V ÷ R

Convert everything to base units (volts, amps, ohms) so the units match before you calculate. The voltage reading is in millivolts — convert it: 25 mV = 0.025 V. The resistance is already in ohms, no conversion needed.

Now the math works cleanly:

I = 0.025 V ÷ 0.1 Ω = 0.25 A

The result is 0.25 amps. If you’d rather express that at a smaller scale: 0.25 A = 250 mA. Both are the same current — just different units for the same answer.

Field meaning: a quarter amp of bond current from a small mag anode group is well within the range you’d expect from a freshly-installed group on a moderately resistive pipeline. If the same shunt had read 5 mV instead — that’s only 50 mA — you’d be looking at a system that’s underperforming, anodes that are partially consumed, or soil drier than the design assumed. The number doesn’t tell you all that on its own. The number plus the design intent does.

Worked example B — the rectifier shunt

Same calculation, different scale. You’re at the rectifier cabinet of an impressed-current system. The output shunt is rated “50 mV = 25 A” — a standard CP rectifier rating that tells you the shunt drops 50 millivolts when the current is at the rated 25 amps. CP rectifier shunts almost always use this format: 50 mV at the rated full-load current, with the rated current varying by rectifier size (10 A, 25 A, 50 A, 100 A, and up). From that rating, you can back out the shunt’s resistance — it’s the same Ohm’s Law, applied to the calibration spec.

Find the rectifier shunt’s resistance from its rating — 50 mV = 25 A

What we know: at the rated current of 25 A, the shunt drops 50 mV. What we want: R_shunt.

R = V ÷ I = 50 mV ÷ 25 A

Convert the millivolts to volts so the units match: 50 mV = 0.050 V.

0.050 V ÷ 25 A = 0.002 Ω

The shunt is a 0.002 Ω resistor — sometimes also written as 2 mΩ. The mV-at-rated-A label is just a friendlier way of expressing the same physical thing.

Now read the live current. Your meter, on millivolts DC, reads 40 mV across that shunt.

Find the rectifier output current — 50 mV = 25 A shunt, 40 mV reading

What we know: V = 40 mV; R_shunt = 0.002 Ω. What we want: I.

I = V ÷ R

Convert the millivolts to volts: 40 mV = 0.040 V. The resistance is already in ohms.

I = 0.040 V ÷ 0.002 Ω = 20 A

That’s 20 amps coming out of the rectifier — 80% of the shunt’s rated 25 A load. Within spec, with headroom.

A useful habit: at any shunt with a known rating, the ratio of measured-to-rated voltage is the ratio of measured-to-rated current. Here, 40 mV ÷ 50 mV = 0.8, and 0.8 × 25 A = 20 A. Same answer, faster.

Field meaning: 20 A out of a shunt rated at 25 A is a comfortable working point — the system has headroom. If the same shunt were reading 48 mV (96% of rated), you’d be at 24 A — at the edge. Push past the shunt rating and you start cooking transformer windings. The same Ohm’s Law that turns the millivolts into amps also tells you whether the system is operating inside its design envelope.

Worked example C — solving for voltage drop

Move the unknown around. Sometimes you know the current and the resistance, and you want to know how much voltage the path is consuming. Cable IR drop is the textbook case — and a constant concern in CP because every volt the cable eats is a volt that isn’t doing work in the soil.

How much voltage does a 750-foot header cable drop at 18 A?

The cable resistance, given the gauge and length from a standard sizing table, comes out to R_cable = 0.4 Ω total. The rectifier is pushing I = 18 A down the line.

The form of Ohm’s Law that solves for voltage:

V = I × R

Plug in:

V_drop = 18 A × 0.4 Ω = 7.2 V

Seven and two tenths volts — that’s what the cable consumes. If the rectifier is set to 24 V output, the junction box at the far end of that cable sees 24 − 7.2 = 16.8 V, and only that 16.8 V is left to push current across the anode-to-soil interface.

Field meaning: the cable’s 7.2 V isn’t a defect — it’s the price of pushing 18 A through 0.4 Ω of conductor. But it’s a price you can choose to pay differently: a heavier-gauge cable would have lower resistance, which means lower IR drop, which means more volts available where they actually do work. Cable sizing is a voltage-budget decision dressed up as a hardware decision, and Ohm’s Law is the calculation under both.

Worked example D — solving for resistance

The third rearrangement. You know the voltage and the current, and you want to back out the resistance the circuit is presenting. This one is diagnostic. It’s how you compare what a system is doing today against what the design said it should be doing.

Total circuit resistance from rectifier readings — 22 V output, 12 A current

You’re at the cabinet with your multimeter. Voltage at the output terminals reads 22 V. Your shunt reading works out to 12 A of output current. What total resistance does that imply for the whole circuit — cable, junction, anode array, soil, and structure return, all summed?

The form that solves for resistance:

R = V ÷ I

Plug in, base units throughout:

R_total = 22 V ÷ 12 A ≈ 1.83 Ω

That’s the resistance the rectifier is “seeing” across its output terminals. The whole circuit looks like 1.83 Ω from the rectifier’s point of view.

Field meaning depends on the design. If the original design put R_total at 1.5 Ω, today’s system is running 0.33 Ω higher than designed. That extra third of an ohm has to be coming from somewhere: anode consumption (parallel-array R_parallel rises as anodes deplete), a connection that’s loosened up (series R_cable climbs as a lug oxidizes), drier soil increasing the anode-to-soil resistance, or some combination. Ohm’s Law turned three meter readings — V, I, and the design R — into a question worth asking. That’s what diagnostic calculation is for.

Cross-checking with what EC-001 already taught

EC-001 named two rules that go with the shapes: Kirchhoff’s Voltage Law (in any closed loop, the source voltages equal the sum of the drops across the elements in that loop) and Kirchhoff’s Current Law (at any junction, current in equals current out). Those are properties of the circuit shape — facts about how series and parallel paths behave.

You won’t re-derive them in this module. But once your Ohm’s Law calculation has produced an answer for a multi-element circuit, KVL and KCL are exactly the right tools to check the answer. Sum the voltage drops across the series elements; they should equal the source voltage. Sum the branch currents at a parallel junction; they should equal the total current at the junction. If the two numbers don’t agree to within a small rounding error, the math broke somewhere — usually a missed unit conversion or a wrong rearrangement of the formula.

Treat KVL and KCL as the sanity check that runs after the calculation. Don’t try to re-prove them every time. EC-001 already did that work.

Multi-step circuit reduction in detail: EC-001’s series-parallel section walked the full reduction procedure end-to-end (collapse parallel → add series → solve total → back out junction voltage → solve per-branch). The Apply lesson here will give you a fresh problem to work that procedure on, with different numbers than EC-001 used.

Back at the test station

Close the lid on the math for a minute and stand back at the test station you opened in the hook. The shunt was 0.1 Ω. The meter read 25 mV. You ran I = V ÷ R = 0.025 V ÷ 0.1 Ω = 0.25 A (= 250 mA). That’s the bond current — the magnesium anodes are putting a quarter amp of protective current onto the pipeline.

That number is what you came for. But what you do with it depends on knowing what to compare it to. Two questions are worth running every time:

Is the current within the expected range for this anode group, this pipeline, this soil? A quarter amp out of three 32-pound magnesium anodes in moderately moist clay is reasonable. A quarter amp out of three 17-pound zinc anodes in a high-resistivity sand-and-gravel station is generous. Same number, different stories.
How does it compare to the last reading from this station? A station that read 300 mA two months ago and 250 mA today is trending — slowly enough that no alarm is warranted, but worth watching. A station that reads 50 mA today after reading 300 mA in the spring is something else entirely.

None of that interpretation is in the math itself. The math gives you a number you can defend. The interpretation is in how that number sits inside the system you’re responsible for — which is exactly the work of CP.

Same loop applies at the rectifier. The 280 mV across the 0.01 Ω shunt told you the system was outputting 28 A. Whether that’s the right amount of current depends on the design current the system was sized for and how the structure has been responding. Ohm’s Law gives you the number. The job gives you the meaning.

Key takeaways

Ohm’s Law connects voltage, current, and resistance: V = I × R. Three quantities, three rearrangements, used to solve for whichever one is unknown.
Shunts turn current measurement into voltage measurement. A calibrated low-resistance shunt sits in the current path; you measure the millivolts across it and apply I = V ÷ R. Test station bond shunts are typically 0.1 or 0.01 Ω; rectifier shunts are typically 0.001 or 0.01 Ω, often labeled as a millivolt-per-amp rating.
Match your units before you calculate. Ohm’s Law works in any consistent unit family — volts/amps/ohms or millivolts/milliamps/milliohms — but it doesn’t work mixed. Convert everything into one family first.
Solve for V (V = IR) when calculating voltage drop across a known resistance at a known current. Cable IR drops are the standard case.
Solve for I (I = V ÷ R) when reading a shunt or any other situation where a voltage is measured across a known resistance. Most field calculations live here.
Solve for R (R = V ÷ I) when backing out a circuit’s effective resistance from voltage and current readings. Diagnostic work — comparing today’s R against design R.
KVL and KCL (from EC-001) are cross-checks. When the calculation gets multi-step, summing voltage drops should equal source voltage; summing branch currents should equal total current. If they don’t, the math broke.

References & further reading

AMPP CP 1 Cathodic Protection Tester Course Manual — basic electricity foundations including Ohm’s Law and resistivity for CP technicians.
AUCSC Basic Course — companion treatment with utility-side examples for CP work.
Peabody’s Control of Pipeline Corrosion — A.W. Peabody. CP-specific application of Ohm’s Law with practical worked examples.
AMPP CP 2 Cathodic Protection Technician Course Manual — extended CP-1 treatment covering rectifier sizing and ground bed design.
Corrosion Basics: An Introduction — broad textbook context for Ohm’s Law in CP design.
Corrosion Engineer’s Reference Book — Robert Baboian, ed. General electrical reference covering material resistivities and conductor properties.

Apply — three problems

Work the math

Three problems. Three different unknowns. Each one opens with a setup and a question — work the math first, on paper or in your head, then click each step to compare. The point is not the arithmetic. The point is reaching for the right rearrangement of Ohm’s Law without looking it up, and keeping the units honest while you do it.

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.

Problem 1 · Solve for I

Reading a rectifier shunt at a different scale

Setup. You’re at a rectifier on a four-mile section of distribution. The internal output shunt is rated 50 mV = 50 A — a standard CP rectifier rating that fixes its resistance. Your multimeter on millivolts DC reads 34 mV across the shunt terminals.

What is the shunt’s resistance, what is the rectifier output current, and how does that compare to the rectifier’s rated capacity?

Step 1 — back out the shunt resistance from its rating

The rating tells you the shunt’s spec: at the rated 50 A, it drops 50 mV. Apply Ohm’s Law to the rating itself:

R_shunt = V_rated ÷ I_rated = 50 mV ÷ 50 A

Convert to base units to keep the answer clean: 0.050 V ÷ 50 A = 0.001 Ω

The shunt is a 1 milliohm resistor. That’s the calibrated value the manufacturer guarantees.

Step 2 — solve for the live output current

Now use the same form of Ohm’s Law on today’s reading:

I = V ÷ R_shunt

Convert to base units first so the units match: 34 mV → 0.034 V (the resistance is already in ohms).

I = 0.034 V ÷ 0.001 Ω = 34 A

The rectifier is putting 34 amps out at the moment of the reading.

Step 3 — sanity-check using the proportion of rated drop

The shunt’s rated drop is 50 mV at full scale of 50 A. Today’s drop is 34 mV. The ratio:

34 ÷ 50 = 0.68

So today’s current is 68% of the shunt’s full-scale rating: 0.68 × 50 A = 34 A. Same answer.

That ratio trick is fast in the field — once you know what fraction of the rated millivolt drop you’re seeing, multiply that fraction by the rated amps.

Step 4 — compare to rated capacity

34 A out of a rectifier whose shunt is rated for 50 A says we’re running at 68% of the shunt’s capacity. The rectifier itself is presumably sized to match — likely a 50 A, 24 V or 50 V unit. Operating at two-thirds of rated capacity is a comfortable working point: enough margin to push more current if anode consumption demands it later, no thermal stress on the transformer or output diodes today.

The rating gives you the resistance. The reading gives you the current. The ratio tells you where today’s operating point sits inside the system’s design envelope. Three calculations, all Ohm’s Law, all chained off the same shunt.

Problem 2 · Solve for R

Backing out total circuit resistance from rectifier readings

Setup. A rectifier at a remote tank site is reading 30 V at the output terminals (multimeter) and 15 A on the shunt. The original system commissioning report listed the design total circuit resistance at 1.6 Ω. (That 1.6 Ω came from cable + connections + parallel anode array + soil + structure return, all summed during commissioning.)

What total resistance is the system showing today, and what does the comparison to design tell you?

Step 1 — solve for total resistance today

The unknown is R, so use the rearrangement that solves for it:

R_total = V ÷ I = 30 V ÷ 15 A = 2.0 Ω

Today the rectifier is “seeing” 2.0 Ω across its output terminals — the lumped resistance of cable, connections, anode array, soil, and structure return.

Step 2 — compare to design

Design R_total was 1.6 Ω. Today’s R_total is 2.0 Ω. The system has gained:

ΔR = 2.0 − 1.6 = 0.4 Ω

That’s a 25% increase over the design value — meaningful. Something in the circuit is presenting more opposition to current than the original design assumed.

Step 3 — what would cause the change

The 0.4 Ω is somewhere on the loop. The candidates, in roughly the order they tend to show up in real systems:

Anode consumption. As anodes deplete, individual anode-to-soil resistances climb. Because the anode array is parallel, the array’s equivalent resistance climbs too — though more slowly than any single anode’s. (EC-001: parallel R is dominated by the lowest-resistance branches.)
Drier soil at the groundbed. Earth resistivity rises through dry seasons. The anode-to-soil contact resistance scales directly with resistivity.
A loose or oxidized connection somewhere on the series side. Cable lugs at the rectifier output, header junction box terminals, anode lead splices.
Cable damage. Less common, but a partial conductor break or insulation breach can change the effective series resistance.

The 0.4 Ω alone doesn’t tell you which of these. It tells you to go look.

Step 4 — a quick implication for protective current

If the structure was designed to receive a specific protective current (call it the design I), and the resistance has crept up by 25%, then to deliver the same I the rectifier output voltage has to climb by 25%. That’s why you’ll often find rectifiers with auto-tap or constant-current modes adjusted upward over the years — the system is compensating for slowly rising R as the groundbed ages.

Eventually the rectifier hits its voltage ceiling and can’t compensate further. That’s when groundbed replacement or supplementation becomes the conversation.

R = V ÷ I gives you a single number that summarizes the whole circuit. Comparing today’s R to design R tells you whether the system is still operating where the design intended. The investigation that follows is what the comparison is for — Ohm’s Law just sets the agenda.

Problem 3 · Multi-step

Series-parallel reduction with Ohm’s Law on every step

Setup. A 24 V rectifier feeds a small impressed-current groundbed through a header cable measuring 0.8 Ω. The groundbed is five anodes in parallel, with anode-to-soil resistances of 6 Ω, 7 Ω, 8 Ω, 10 Ω, and 12 Ω — typical of a bed where soil moisture varies along the length.

What is the rectifier’s total output current, the voltage at the junction box, and the current through each individual anode?

Step 1 — collapse the parallel anode array

For unequal parallel branches, sum the reciprocals:

1 ÷ R_parallel = 1/6 + 1/7 + 1/8 + 1/10 + 1/12

= 0.1667 + 0.1429 + 0.1250 + 0.1000 + 0.0833 = 0.6179

R_parallel = 1 ÷ 0.6179 ≈ 1.62 Ω

Less than the smallest branch (6 Ω) — exactly what the parallel rule predicts.

Step 2 — total circuit resistance

The header cable is in series with the parallel anode array:

R_total = R_cable + R_parallel = 0.8 + 1.62 = 2.42 Ω

That’s the total resistance the rectifier is pushing current against.

Step 3 — total rectifier output current (Ohm’s Law on the whole circuit)

I_total = V_source ÷ R_total = 24 V ÷ 2.42 Ω ≈ 9.92 A

Just under 10 amps total flowing out of the rectifier and through the cable.

Step 4 — voltage at the junction box (KVL applied to the cable)

The cable consumes some of the source voltage as IR drop. Subtract it:

V_jct = V_source − (I_total × R_cable)

V_jct = 24 − (9.92 × 0.8) = 24 − 7.94 ≈ 16.06 V

About 16 V is what the junction box sees — and that’s the voltage every anode branch sees individually, because all five branches connect the same two points.

Step 5 — per-anode currents (Ohm’s Law on each branch)

Each branch: I_branch = V_jct ÷ R_branch

6 Ω anode → 16.06 ÷ 6 ≈ 2.68 A
7 Ω anode → 16.06 ÷ 7 ≈ 2.29 A
8 Ω anode → 16.06 ÷ 8 ≈ 2.01 A
10 Ω anode → 16.06 ÷ 10 ≈ 1.61 A
12 Ω anode → 16.06 ÷ 12 ≈ 1.34 A

Step 6 — KCL cross-check

Branch currents should sum to total current at the junction (Kirchhoff’s Current Law from EC-001):

2.68 + 2.29 + 2.01 + 1.61 + 1.34 = 9.93 A

Total from Step 3 was 9.92 A. The two agree within a hundredth of an amp — that’s rounding error, not math error. The circuit closes.

And notice the spread: the 6 Ω anode is doing twice the work of the 12 Ω anode. Both connected to the same junction, both seeing the same 16 V, but very different soil environments — and the math predicts the imbalance from the resistance values alone. (That’s not a fault. That’s the parallel rule from EC-001.)

Six steps, four applications of Ohm’s Law, two KVL/KCL cross-checks. The same procedure works on every series-parallel CP circuit you’ll meet — collapse parallel, add series, solve total, back out junction voltage, solve branches, verify the books close. Once the procedure is wired in, the only thing changing per problem is the numbers.

Three problems, three rearrangements of the same equation — and the same shape of work behind each: identify the unknown, pick the rearrangement, keep the units consistent, run the math, sanity-check the answer. That’s the practice.

You’ll do this same kind of math on real cabinets, with real shunts, on circuits whose resistances you don’t know in advance. Ohm’s Law is what turns three meter readings into a number you can defend.

Up next: a quiz to confirm the math is wired in.