Shunt Current for Cathodic Protection

A shunt is a known low resistance used to calculate current from a measured voltage drop.

In cathodic protection work, shunts are often used with rectifiers, bonds, or test stations so current can be estimated from a small voltage reading. The basic calculation is an Ohm's Law application.

This page explains the basic formula only. It does not teach detailed shunt calibration, meter connection procedure, equipment maintenance, or electrical safety procedure.

Quick Definition

Shunt current is calculated from the measured voltage drop across a known shunt resistance:

I = E ÷ R

Where:

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

Because shunt voltage readings are often small, millivolts must usually be converted to volts before using the basic formula.

When Shunt Current Is Used in CP Work

Shunt current calculations appear in CP work when current is estimated from a known resistance and a measured voltage drop.

  • estimating rectifier output current across a shunt;
  • estimating bond current;
  • checking current in a test station circuit where a known shunt is installed;
  • comparing present current with prior readings;
  • supporting basic troubleshooting.

A shunt-current calculation depends on knowing the correct shunt resistance and measuring the voltage drop correctly.

Formula Reference

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

Variable and Unit Table

Symbol Meaning Common unit Unit symbol CP field meaning
I Current ampere A Current flowing through the shunt
E Voltage drop volt V Measured voltage drop across the shunt
R Shunt resistance ohm Ω Known resistance value of the shunt

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

The most common learner mistake is forgetting to convert millivolts to volts.

50 mV = 0.050 V

I = E ÷ R

Do not guess the shunt resistance. Use the known shunt value from the equipment label, drawing, test station record, or other verified source.

Worked Example 1 — Calculate Current from Shunt Voltage

A shunt has a resistance of 0.010 ohms. The measured voltage drop across the shunt is 0.050 volts. What current is flowing?

Known values:

  • E = 0.050 V
  • R = 0.010 Ω
  • I = ?

I = E ÷ R

I = 0.050 V ÷ 0.010 Ω

I = 5 A

Answer: The current is 5 A.

Field meaning: This calculation estimates current through the shunt. It depends on the shunt resistance being correct.

Worked Example 2 — Convert Millivolts First

A 0.010-ohm shunt has a measured voltage drop of 25 mV. What current is flowing?

Known values:

  • E = 25 mV
  • R = 0.010 Ω
  • I = ?

25 mV = 0.025 V

I = E ÷ R

I = 0.025 V ÷ 0.010 Ω

I = 2.5 A

Answer: The current is 2.5 A.

Field meaning: Millivolt readings are common with shunts. Convert millivolts to volts before using the basic formula.

Worked Example 3 — Calculate Expected Voltage Drop

A current of 8 amps flows through a 0.005-ohm shunt. What voltage drop should appear across the shunt?

Known values:

  • I = 8 A
  • R = 0.005 Ω
  • E = ?

E = I × R

E = 8 A × 0.005 Ω

E = 0.040 V

0.040 V = 40 mV

Answer: The expected voltage drop is 0.040 V, or 40 mV.

Field meaning: A small shunt resistance can produce a small voltage drop that represents a larger current.

Worked Example 4 — Calculate Shunt Resistance

A current of 10 amps produces a voltage drop of 0.050 volts across a shunt. What is the shunt resistance?

Known values:

  • E = 0.050 V
  • I = 10 A
  • R = ?

R = E ÷ I

R = 0.050 V ÷ 10 A

R = 0.005 Ω

Answer: The shunt resistance is 0.005 Ω.

Field meaning: This shows the relationship between current, voltage drop, and shunt resistance. In field work, the shunt resistance should normally be verified from a reliable source rather than guessed.

Common Mistakes

Forgetting to convert millivolts to volts

25 mV is 0.025 V, not 25 V.

Using the wrong shunt resistance

Current calculation depends directly on the resistance value.

Confusing shunt voltage with structure-to-soil potential

Shunt voltage is measured across a known resistance. It is not a reference-electrode potential.

Assuming the calculation proves CP protection

Shunt current is circuit information. It does not replace structure-to-electrolyte measurements.

Rounding too aggressively

Small shunt values can make rounding errors important.

Mixing milliamps and amps

Keep current units consistent.

Treating this page as a calibration procedure

This page explains the basic calculation only.

Ignoring field condition

Loose connections, wrong test points, or bad readings can make the calculation misleading.

Practice Problems

  1. A 0.010-ohm shunt has a voltage drop of 0.030 V. What current is flowing?
  2. A 0.005-ohm shunt has a voltage drop of 0.025 V. What current is flowing?
  3. A 0.010-ohm shunt has a measured voltage drop of 40 mV. What current is flowing?
  4. A 6-amp current flows through a 0.010-ohm shunt. What voltage drop is expected?
  5. A 12-amp current flows through a 0.005-ohm shunt. What voltage drop is expected in volts and millivolts?
  6. A voltage drop of 0.060 V is measured while 3 A flows. What resistance does that represent?
  7. Convert 15 mV to volts.
  8. Does a shunt-current calculation by itself prove that a structure satisfies CP protection criteria?

Practice Problem Answer Key

Problem Answer Calculation / explanation
13 AI = E ÷ R = 0.030 V ÷ 0.010 Ω = 3 A
25 AI = E ÷ R = 0.025 V ÷ 0.005 Ω = 5 A
34 A40 mV = 0.040 V; I = 0.040 V ÷ 0.010 Ω = 4 A
40.060 VE = I × R = 6 A × 0.010 Ω = 0.060 V
50.060 V, or 60 mVE = 12 A × 0.005 Ω = 0.060 V = 60 mV
60.020 ΩR = E ÷ I = 0.060 V ÷ 3 A = 0.020 Ω
70.015 V15 mV ÷ 1000 = 0.015 V
8NoShunt current is circuit information. Protection must be evaluated with proper field measurements and criteria.

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