Ohm's Law for Cathodic Protection

Ohm's Law connects voltage, current, and resistance in an electrical circuit. In cathodic protection work, it helps explain rectifier output, circuit resistance, voltage drop, current flow, and basic troubleshooting.

Ohm's Law is useful, but it does not prove whether a structure is protected. CP protection still has to be evaluated with proper field measurements and the correct protection criteria.

Quick Definition

Ohm's Law says that voltage equals current multiplied by resistance.

E = I × R

Symbol Meaning Common unit
E Voltage volts
I Current amperes
R Resistance ohms

Many people also write voltage as V, so you may also see the formula written as V = I × R. This page uses E for voltage, but the calculation is the same.

When Ohm's Law Is Used in CP Work

Ohm's Law appears often in cathodic protection work.

  • understanding how rectifier voltage, current, and circuit resistance are related;
  • estimating circuit resistance when voltage and current are known;
  • understanding why higher resistance can limit current flow;
  • supporting voltage-drop and IR-drop discussions;
  • helping explain shunt-current and bond-current calculations at a basic level;
  • checking whether a calculation result is reasonable.

Ohm's Law is a circuit relationship. It helps explain electrical behavior, but it is not a CP protection criterion by itself.

Formula Reference

Use the form that matches the unknown value.

Solve for Formula Use when you know
Voltage E = I × R current and resistance
Current I = E ÷ R voltage and resistance
Resistance R = E ÷ I voltage and current

Equivalent Voltage Notation

Common notation Same meaning
E = I × R voltage equals current times resistance
V = I × R voltage equals current times resistance

Variable and Unit Table

Symbol Meaning Common unit Unit symbol CP field meaning
E Voltage volt V Electrical push across a circuit or circuit element
I Current ampere A Amount of electrical current flowing
R Resistance ohm Ω Opposition to current flow

Use matching units before calculating. For the basic formula, voltage should be in volts, current should be in amperes, and resistance should be in ohms.

Common Conversions

  • 1 A = 1000 mA
  • 1 mA = 0.001 A
  • 1 V = 1000 mV
  • 1 mV = 0.001 V

How to Choose the Right Formula

Start by identifying what the problem gives you and what it asks for.

  • If the problem gives current and resistance, solve for voltage: E = I × R.
  • If the problem gives voltage and resistance, solve for current: I = E ÷ R.
  • If the problem gives voltage and current, solve for resistance: R = E ÷ I.

A simple way to check your work is to look at the units: volts come from amps times ohms; amps come from volts divided by ohms; and ohms come from volts divided by amps.

Worked Example 1 — Solve for Current

A rectifier circuit has 24 volts applied across a circuit resistance of 6 ohms. What current would Ohm's Law predict?

Known values:

  • E = 24 V
  • R = 6 Ω
  • I = ?

I = E ÷ R

I = 24 V ÷ 6 Ω

I = 4 A

Answer: The predicted current is 4 A.

Field meaning: If 24 volts are applied across a 6-ohm circuit, Ohm's Law predicts 4 amps of current. This is a circuit calculation. It does not prove that the protected structure meets a CP criterion.

Worked Example 2 — Solve for Resistance

A rectifier is measured at 18 volts and 3 amps. What is the approximate circuit resistance?

Known values:

  • E = 18 V
  • I = 3 A
  • R = ?

R = E ÷ I

R = 18 V ÷ 3 A

R = 6 Ω

Answer: The approximate circuit resistance is 6 Ω.

Field meaning: This calculation gives a simple estimate of circuit resistance based on measured voltage and current. It is useful for understanding the circuit, but it is not the same as measuring pipe-to-soil potential or confirming protection.

Worked Example 3 — Solve for Voltage

A CP circuit needs 2.5 amps of current through 8 ohms of resistance. What voltage is needed across that resistance?

Known values:

  • I = 2.5 A
  • R = 8 Ω
  • E = ?

E = I × R

E = 2.5 A × 8 Ω

E = 20 V

Answer: The required voltage is 20 V.

Field meaning: For this simplified circuit, 20 volts would be needed to drive 2.5 amps through 8 ohms of resistance. Real CP systems can be more complex, so field readings still have to be interpreted carefully.

Worked Example 4 — Limited Voltage-Drop Support

A current of 1.5 amps flows through a resistance of 2 ohms. What voltage drop occurs across that resistance?

Known values:

  • I = 1.5 A
  • R = 2 Ω
  • E = ?

E = I × R

E = 1.5 A × 2 Ω

E = 3 V

Answer: The voltage drop is 3 V.

Field meaning: Voltage drop is one reason CP field readings can be affected by current flow through resistive paths. This page only shows the basic Ohm's Law relationship. Detailed IR-drop interpretation belongs with current interruption and field measurement topics.

Common Mistakes

Using Milliamps as Amps

Convert milliamps to amps before using Ohm's Law with volts and ohms. For example, 250 mA is 0.250 A, not 250 A.

Using Millivolts as Volts

Convert millivolts to volts before using the basic formula. For example, 850 mV is 0.850 V, not 850 V.

Solving for the Wrong Variable

Use I = E ÷ R when current is unknown. Use R = E ÷ I when resistance is unknown.

Treating Rectifier Output as Proof of Protection

Rectifier voltage and current describe the circuit. They do not replace structure-to-electrolyte potential measurements.

Confusing Voltage Drop with Pipe-to-Soil Potential

Voltage drop is an electrical effect in a current path. Pipe-to-soil potential is a field measurement made with a reference electrode.

Confusing Resistance with Soil Resistivity

Resistance is measured in ohms. Soil resistivity is a different property and uses different units.

Using Too Many Decimals

A calculator may show many digits, but field values are usually approximate. Use reasonable precision.

Forgetting Circuit Context

Ohm's Law is powerful, but CP systems include soil, coating, electrolyte, anodes, bonds, rectifiers, and field measurement conditions.

Practice Problems

  1. A circuit has 12 volts across 4 ohms. What current flows?
  2. A rectifier output is 30 volts and 5 amps. What is the approximate circuit resistance?
  3. A current of 3 amps flows through 7 ohms. What voltage is required?
  4. A bond current calculation uses a measured voltage drop of 0.050 volts across a known resistance of 0.010 ohms. What current does Ohm's Law predict?
  5. Convert 250 mA to amps.
  6. Convert 750 mV to volts.
  7. A CP circuit has higher resistance than expected. For the same applied voltage, what happens to current?
  8. A rectifier is producing voltage and current. Does that alone prove the pipeline satisfies a CP protection criterion?

Practice Problem Answer Key

Problem Answer Calculation / explanation
1 3 A I = E ÷ R = 12 V ÷ 4 Ω = 3 A
2 6 Ω R = E ÷ I = 30 V ÷ 5 A = 6 Ω
3 21 V E = I × R = 3 A × 7 Ω = 21 V
4 5 A I = E ÷ R = 0.050 V ÷ 0.010 Ω = 5 A
5 0.250 A 250 mA ÷ 1000 = 0.250 A
6 0.750 V 750 mV ÷ 1000 = 0.750 V
7 Current decreases For the same voltage, higher resistance means lower current.
8 No Rectifier output is useful circuit information, but CP protection must be evaluated with proper field measurements and criteria.

Related Learning

Related Formulas