Voltage Drop for Cathodic Protection

Voltage drop is the voltage used up as current flows through resistance.

In cathodic protection work, voltage drop matters because CP current flows through soil, electrolyte, wires, bonds, shunts, and other resistive paths. Understanding voltage drop helps learners understand circuit behavior and why some field readings can be influenced by current flow.

Voltage drop is important, but this page does not teach full CP criteria interpretation. Current interruption, instant-off readings, and protection criteria must be handled on their own measurement pages.

Quick Definition

Voltage drop is calculated with Ohm's Law:

E = I × R

Where:

  • E is voltage drop, in volts.
  • I is current, in amperes.
  • R is resistance, in ohms.

Many people also write this as:

V = I × R

This page uses E for voltage, but the calculation is the same.

When Voltage Drop Is Used in CP Work

Voltage drop appears in CP work when current flows through resistance.

  • understanding why current-on readings may include voltage-drop effects;
  • understanding voltage loss in wires, bonds, shunts, and circuit paths;
  • checking whether a calculated voltage drop is reasonable;
  • supporting basic IR-drop awareness;
  • connecting Ohm's Law to current interruption and instant-off measurement topics.

Voltage drop calculations help explain circuit behavior. They do not replace proper field measurement or CP criteria evaluation.

Formula Reference

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

Equivalent notation

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

Variable and Unit Table

Symbol Meaning Common unit Unit symbol CP field meaning
E Voltage drop volt V Voltage across a resistance while current flows
I Current ampere A Current flowing through the resistance
R Resistance ohm Ω Opposition to current flow

Use matching units before calculating:

  • voltage in volts;
  • current in amperes;
  • resistance in ohms.

Common conversions:

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

How to Use the Formula

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

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

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

Keep the circuit context clear. Voltage drop is a voltage across a resistance while current is flowing. It should not be confused with a pipe-to-soil potential measured with a reference electrode.

Worked Example 1 — Calculate Voltage Drop

A CP current of 2 amps flows through 3 ohms of resistance. What voltage drop occurs across that resistance?

Known values:

  • I = 2 A
  • R = 3 Ω
  • E = ?

E = I × R

E = 2 A × 3 Ω

E = 6 V

Answer: The voltage drop is 6 V.

Field meaning: If 2 amps flow through 3 ohms, 6 volts are dropped across that resistance. This is a circuit calculation, not a protection criterion.

Worked Example 2 — Calculate Voltage Drop with Milliamps

A small current of 250 mA flows through a 4-ohm resistance. What voltage drop occurs?

Known values:

  • I = 250 mA
  • R = 4 Ω
  • E = ?

250 mA = 0.250 A

E = I × R

E = 0.250 A × 4 Ω

E = 1 V

Answer: The voltage drop is 1 V.

Field meaning: Small currents can still produce measurable voltage drop when resistance is present. Always convert milliamps to amps before using the basic formula.

Worked Example 3 — Solve for Resistance from Voltage Drop

A measured voltage drop is 0.600 V while 0.300 A flows. What resistance does that represent?

Known values:

  • E = 0.600 V
  • I = 0.300 A
  • R = ?

R = E ÷ I

R = 0.600 V ÷ 0.300 A

R = 2 Ω

Answer: The resistance is 2 Ω.

Field meaning: This calculation can help describe the resistance in a current path. It does not identify the cause of that resistance by itself.

Worked Example 4 — Connect Voltage Drop to IR-Drop Awareness

A current of 1.2 amps flows through a 1.5-ohm path. What voltage drop is produced?

Known values:

  • I = 1.2 A
  • R = 1.5 Ω
  • E = ?

E = I × R

E = 1.2 A × 1.5 Ω

E = 1.8 V

Answer: The voltage drop is 1.8 V.

Field meaning: Voltage drop caused by current flow is one reason CP measurements may require current interruption. This page only explains the basic relationship. Detailed instant-off measurement and IR-drop correction belong on measurement pages.

Common Mistakes

Confusing voltage drop with pipe-to-soil potential

Voltage drop is across a resistance. Pipe-to-soil potential is measured with a reference electrode.

Using milliamps without converting

250 mA must be entered as 0.250 A in the basic formula.

Using millivolts without converting

850 mV is 0.850 V.

Assuming every field reading is only voltage drop

Field measurements can include several influences. Use the correct measurement method.

Treating voltage drop as criteria evaluation

Voltage drop helps explain circuit behavior. It does not prove protection.

Solving for the wrong variable

Use E = I × R for voltage drop, I = E ÷ R for current, and R = E ÷ I for resistance.

Ignoring circuit path

A calculated voltage drop applies to the resistance path used in the calculation.

Using too much precision

Field values are often approximate. Do not imply more accuracy than the measurements support.

Practice Problems

  1. A current of 3 amps flows through 2 ohms. What voltage drop occurs?
  2. A current of 0.5 amps flows through 10 ohms. What voltage drop occurs?
  3. A current of 200 mA flows through 5 ohms. What voltage drop occurs?
  4. A voltage drop of 4 V is measured across 2 ohms. What current flows?
  5. A voltage drop of 0.900 V is measured while 0.300 A flows. What resistance does that represent?
  6. Convert 125 mA to amps.
  7. Convert 50 mV to volts.
  8. Does a voltage-drop calculation by itself prove that a structure satisfies a CP protection criterion?

Practice Problem Answer Key

Problem Answer Calculation / explanation
16 VE = I × R = 3 A × 2 Ω = 6 V
25 VE = I × R = 0.5 A × 10 Ω = 5 V
31 V200 mA = 0.200 A; E = 0.200 A × 5 Ω = 1 V
42 AI = E ÷ R = 4 V ÷ 2 Ω = 2 A
53 ΩR = E ÷ I = 0.900 V ÷ 0.300 A = 3 Ω
60.125 A125 mA ÷ 1000 = 0.125 A
70.050 V50 mV ÷ 1000 = 0.050 V
8NoVoltage drop explains circuit behavior. CP protection must be evaluated with proper measurements and criteria.

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