What this formula explains
Shunts are used in cathodic protection circuits to measure current by reading the small voltage drop
across a known resistance. These formulas are used to determine current from a measured shunt voltage,
calculate shunt resistance from a stamped rating, and select a shunt that provides useful field
resolution for bond currents, galvanic anodes, and rectifier outputs.
Reference formulas
1) Ohm’s law for shunts (resistance method)
Use when the shunt resistance \(R_{shunt}\) is known.
\[
I = \frac{V_{shunt}}{R_{shunt}}
\]
\(V_{shunt}\) = measured
voltage across shunt (V)
\(R_{shunt}\) = shunt resistance (Ω)
2) Shunt rating method (amps per mV)
Use when the shunt is stamped with a rating such as “100 A / 50 mV”.
\[
I = V_{mV} \,\times\, \left(\frac{I_{rated}}{V_{rated(mV)}}\right)
\]
\(V_{mV}\) = measured drop across shunt (mV)
\(I_{rated}\) = shunt current rating (A)
\(V_{rated}\) = shunt voltage rating (mV)
Problem 1
You are reading a shunt rated at 50 A / 50 mV. You measure a voltage drop of
12 mV across the shunt.
What is the current flowing through the circuit?
Reveal solution
\[
\begin{array}{c}
\text{Factor}
=
\frac{I_{rated}}{V_{rated}}
=
\frac{50}{50}
=
1.0\ \text{A/mV} \\[10pt]
I
=
V_{mV} \,\times\, \text{Factor} \\[10pt]
I
=
12 \,\times\, 1.0
=
12\ \text{A}
\end{array}
\]
Answer: 12.0 Amps
Problem 2
A shunt has a resistance of 0.01 Ω. The voltage measured across it is
0.045 V (45 mV).
Calculate the current.
Reveal solution
\[
\begin{array}{c}
I
=
\frac{V}{R} \\[10pt]
I
=
\frac{0.045}{0.01}
=
4.5\ \text{A}
\end{array}
\]
Answer: 4.5 Amps
Problem 3
You have a shunt rated at 25 A / 50 mV.
What is the internal resistance of this shunt?
Reveal solution
\[
\begin{array}{c}
50\ \text{mV}
=
0.050\ \text{V} \\[10pt]
R
=
\frac{V}{I} \\[10pt]
R
=
\frac{0.050}{25}
=
0.002\ \Omega
\end{array}
\]
Answer: 0.002 Ω
Problem 4
You are measuring a resistance bond between two pipelines. The bond contains a shunt with a resistance
of 0.05 Ω. The measured voltage drop is 125 mV.
Calculate the magnitude of the current flowing through the bond.
Reveal solution
\[
\begin{array}{c}
125\ \text{mV}
=
0.125\ \text{V} \\[10pt]
I
=
\frac{V}{R} \\[10pt]
I
=
\frac{0.125}{0.05}
=
2.5\ \text{A}
\end{array}
\]
Answer: 2.5 Amps
Problem 5
A rectifier ammeter is broken. You measure the voltage drop across the panel shunt stamped
100 A / 50 mV. The multimeter reads 22 mV.
What is the rectifier output current?
Reveal solution
\[
\begin{array}{c}
\text{Factor}
=
\frac{100}{50}
=
2.0\ \text{A/mV} \\[10pt]
I
=
22 \,\times\, 2.0
=
44\ \text{A}
\end{array}
\]
Answer: 44 Amps
Problem 6
You expect a single magnesium anode to output approximately 5 mA. To obtain a reading
of at least 5 mV on your digital voltmeter, what is the minimum resistance shunt you
should install?
Reveal solution
\[
\begin{array}{c}
V
=
5\ \text{mV}
=
0.005\ \text{V} \\[10pt]
I
=
5\ \text{mA}
=
0.005\ \text{A} \\[10pt]
R
=
\frac{V}{I}
=
\frac{0.005}{0.005}
=
1.0\ \Omega
\end{array}
\]
Answer: 1.0 Ω
Problem 7
A rectifier has a maximum output of 80 A. You want to install a standard
50 mV style shunt.
- A) 50 A / 50 mV
- B) 100 A / 50 mV
- C) 25 A / 50 mV
Which shunt rating is required?
Reveal solution
\[
\begin{array}{c}
\text{Required shunt rating} \ge 80\ \text{A} \\[10pt]
50\ \text{A}: \text{ too low} \\[10pt]
25\ \text{A}: \text{ too low} \\[10pt]
100\ \text{A}: \text{ acceptable} \\[10pt]
R_{shunt}
=
\frac{0.05}{100}
=
0.0005\ \Omega \\[10pt]
V \text{ at } 80\ \text{A}
=
80 \,\times\, 0.0005
=
0.040\ \text{V}
=
40\ \text{mV}
\end{array}
\]
Answer: 100 A / 50 mV
Problem 8
A 0.01 Ω shunt and a 0.1 Ω shunt are both reading 10 mV.
Which shunt is passing more current, and how much more?
Reveal solution
\[
\begin{array}{c}
V
=
10\ \text{mV}
=
0.010\ \text{V} \\[10pt]
I_{0.01}
=
\frac{0.010}{0.01}
=
1.0\ \text{A} \\[10pt]
I_{0.1}
=
\frac{0.010}{0.1}
=
0.10\ \text{A} \\[10pt]
\frac{1.0}{0.10}
=
10
\end{array}
\]
Answer: The 0.01 Ω shunt passes 10× more current.
Problem 9
You are designing a bond station. You want the technician to read 1 mV for every 1 A
to reduce calculation errors.
What shunt resistance is required?
Reveal solution
\[
\begin{array}{c}
V
=
1\ \text{mV}
=
0.001\ \text{V} \\[10pt]
I
=
1\ \text{A} \\[10pt]
R
=
\frac{V}{I}
=
\frac{0.001}{1}
=
0.001\ \Omega
\end{array}
\]
Answer: 0.001 Ω
Problem 10
You encounter a specialized shunt marked \(R_S = 0.001\ \Omega\). You measure
3.5 mV.
What is the current?
Reveal solution
\[
\begin{array}{c}
3.5\ \text{mV}
=
0.0035\ \text{V} \\[10pt]
I
=
\frac{V}{R}
=
\frac{0.0035}{0.001}
=
3.5\ \text{A}
\end{array}
\]
If \(R = 0.001\ \Omega\), then current in amps equals the reading in millivolts.
Answer: 3.5 Amps
