Groundbed Resistance for Cathodic Protection

Groundbed resistance is opposition to current discharge from a groundbed into the earth.

In an impressed-current CP system, the rectifier must drive current through the circuit. Groundbed resistance is one part of that circuit. If resistance is higher, a given voltage will produce less current. If resistance is lower, the same voltage can produce more current.

This page uses simple Ohm’s Law relationships only. It does not teach detailed groundbed design equations or prescribe acceptable design values.

Quick Definition

A basic relationship for groundbed circuit review is:

R = E ÷ I

Where:

  • R is resistance.
  • E is voltage.
  • I is current.

The same relationship can be rearranged:

I = E ÷ R

E = I × R

This page uses these relationships for basic learning. It does not guarantee groundbed performance.

When Groundbed Resistance Is Used in CP Work

Groundbed resistance may be used when reviewing:

  • rectifier output voltage and current;
  • how resistance affects current output;
  • changes in current for a given voltage;
  • basic impressed-current CP circuit behavior;
  • whether further troubleshooting or design review may be needed.

This page does not define acceptable groundbed resistance.

This page does not teach detailed groundbed design.

This page does not prove CP protection.

Formula Reference

Solve forFormulaUse when you know
ResistanceR = E ÷ Ivoltage and current
CurrentI = E ÷ Rvoltage and resistance
VoltageE = I × Rcurrent and resistance

Variable and Unit Table

SymbolMeaningCommon unitUnit symbolCP field meaning
RResistanceohmΩOpposition to current flow
EVoltagevoltVDriving force available in the circuit
ICurrentampereACurrent flowing in the circuit

Use matching units:

  • volts divided by amps gives ohms;
  • volts divided by ohms gives amps;
  • amps multiplied by ohms gives volts.

How to Use the Formula

Use R = E ÷ I when voltage and current are known.

Use I = E ÷ R when voltage and resistance are known.

Use E = I × R when current and resistance are known.

For learner calculations, use values from the same operating condition. Do not mix a voltage from one adjustment with a current from another adjustment.

This formula helps explain circuit behavior. It does not identify every cause of high or low current output.

Worked Examples

Worked Example 1 — Calculate Resistance

A rectifier output is 24 V and the measured current is 6 A. What resistance does that represent?

Known values:

  • E = 24 V
  • I = 6 A
  • R = ?

Use: R = E ÷ I

Substitute: R = 24 V ÷ 6 A

Calculate: R = 4 Ω

Answer: The resistance is 4 Ω.

Field meaning: This is the resistance represented by the voltage and current in this basic circuit calculation.

Worked Example 2 — Calculate Current from Voltage and Resistance

A rectifier applies 30 V to a circuit with 5 Ω resistance. What current does that represent?

Known values:

  • E = 30 V
  • R = 5 Ω
  • I = ?

Use: I = E ÷ R

Substitute: I = 30 V ÷ 5 Ω

Calculate: I = 6 A

Answer: The current is 6 A.

Field meaning: For the same voltage, lower resistance allows more current and higher resistance allows less current.

Worked Example 3 — Calculate Voltage Needed

A circuit current is 8 A and resistance is 3 Ω. What voltage does that represent?

Known values:

  • I = 8 A
  • R = 3 Ω
  • E = ?

Use: E = I × R

Substitute: E = 8 A × 3 Ω

Calculate: E = 24 V

Answer: The voltage is 24 V.

Field meaning: This is the voltage represented by that current and resistance in a basic Ohm’s Law calculation.

Worked Example 4 — Compare Two Resistance Values

A rectifier applies 20 V. What current is expected with 4 Ω resistance? What current is expected with 10 Ω resistance?

Case 1: I = 20 V ÷ 4 Ω

I = 5 A

Case 2: I = 20 V ÷ 10 Ω

I = 2 A

Answer: At 4 Ω, the current is 5 A. At 10 Ω, the current is 2 A.

Field meaning: With the same voltage, higher resistance results in lower current.

Common Mistakes

  1. Assuming resistance alone proves performance. Resistance is important, but CP performance still requires field measurements.
  2. Mixing voltage and current from different conditions. Use values from the same operating state.
  3. Confusing resistance with soil resistivity. Resistance is in ohms. Soil resistivity is commonly in ohm-cm or ohm-ft.
  4. Treating a learner calculation as a design rule. This page does not prescribe acceptable groundbed resistance values.
  5. Ignoring other circuit resistance. A real CP circuit includes more than one possible source of resistance.
  6. Assuming high resistance identifies only one cause. More review may be needed to understand the cause.
  7. Using milliamps as amps without converting. Convert units before using Ohm’s Law.
  8. Treating current output as proof of protection. Current output does not by itself prove protection.

Practice Problems

  1. A rectifier output is 12 V and current is 3 A. What resistance does that represent?
  2. A rectifier output is 48 V and current is 8 A. What resistance does that represent?
  3. A voltage of 24 V is applied to 6 Ω resistance. What current does that represent?
  4. A voltage of 36 V is applied to 9 Ω resistance. What current does that represent?
  5. A current of 5 A flows through 4 Ω resistance. What voltage does that represent?
  6. If voltage stays the same and resistance increases, what happens to current?
  7. Is groundbed resistance the same as soil resistivity?
  8. Does groundbed resistance by itself prove that a structure satisfies CP protection criteria?

Practice Problem Answer Key

ProblemAnswerCalculation / explanation
14 ΩR = E ÷ I = 12 V ÷ 3 A = 4 Ω
26 ΩR = 48 V ÷ 8 A = 6 Ω
34 AI = E ÷ R = 24 V ÷ 6 Ω = 4 A
44 AI = 36 V ÷ 9 Ω = 4 A
520 VE = I × R = 5 A × 4 Ω = 20 V
6Current decreasesWith the same voltage, higher resistance means lower current
7NoGroundbed resistance is measured in ohms; soil resistivity is commonly expressed as ohm-cm or ohm-ft
8NoGroundbed resistance supports review. CP protection must be evaluated with proper measurements and criteria.

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