What this formula explains
When measuring pipe-to-soil potentials, the voltmeter becomes part of the electrical circuit.
If the measurement circuit resistance is high, the meter can draw current and alter the reading.
This effect is called meter loading. The formulas below allow calculation of
the true polarized potential and allow estimation of the circuit resistance
that causes the loading error.
Reference formulas
1) True potential correction (two input impedances)
\[
\begin{aligned}
K &= \frac{R_l}{R_h} \\[10pt]
E_{true}
&=
\frac{V_h \,\times\, (1 - K)}
{1 - K \,\times\, \left(\frac{V_h}{V_l}\right)}
\end{aligned}
\]
\(K\) = ratio of input resistances
\(R_l\) = lower meter input resistance
\(R_h\) = higher meter input resistance
\(V_l\) = measured
voltage using \(R_l\)
\(V_h\) = measured voltage using \(R_h\)
\(E_{true}\) = true polarized potential
2) Meter loading (voltage divider)
\[
V_m = E_{true} \,\times\, \left(\frac{R_m}{R_m + R_{ext}}\right)
\]
\(V_m\) = meter reading (loaded)
\(E_{true}\) = true polarized potential
\(R_m\) = meter input resistance
\(R_{ext}\) = external/contact resistance
3) Solve for external/contact resistance
\[
R_{ext} = R_m \,\times\, \left(\frac{E_{true}}{V_m} - 1\right)
\]
\(R_{ext}\) = external/contact resistance
\(R_m\) = meter input resistance
\(E_{true}\) = true potential
\(V_m\) = measured meter voltage
Problem 1
You suspect high contact resistance while measuring a structure potential.
You take two readings:
- 1 MΩ meter reading = −0.700 V
- 10 MΩ meter reading = −0.850 V
Calculate the true potential.
Reveal solution
\[
\begin{aligned}
K &= \frac{1}{10} = 0.1 \\[10pt]
\frac{V_h}{V_l}
&=
\frac{-0.850}{-0.700}
=
1.2143 \\[10pt]
E_{true}
&=
\frac{-0.850 \,\times\, (1 - 0.1)}
{1 - 0.1 \,\times\, 1.2143} \\[10pt]
E_{true}
&=
\frac{-0.765}{0.87857}
=
-0.870\ \text{V}
\end{aligned}
\]
Answer: −0.870 V
Problem 2
Using the results of Problem 1
(\(E_{true}=-0.870\ \text{V}\), \(V_m=-0.850\ \text{V}\)),
calculate the external resistance if \(R_m=10\ \text{M}\Omega\).
Reveal solution
\[
\begin{aligned}
R_{ext}
&=
10{,}000{,}000
\,\times\,
\left(
\frac{-0.870}{-0.850} - 1
\right) \\[10pt]
R_{ext}
&=
10{,}000{,}000
\,\times\,
0.02353 \\[10pt]
R_{ext}
&=
235{,}300\ \Omega
\approx 235\ \text{k}\Omega
\end{aligned}
\]
Answer: 235 kΩ
Problem 3
A pipeline has a true potential of −1.000 V.
External resistance is 0.5 MΩ and the meter input resistance is 10 MΩ.
What voltage will the meter display?
Reveal solution
\[
\begin{aligned}
V_m
&=
(-1.000)
\,\times\,
\frac{10}{10 + 0.5} \\[10pt]
V_m
&=
(-1.000)
\,\times\,
0.95238 \\[10pt]
V_m
&=
-0.952\ \text{V}
\end{aligned}
\]
Answer: −0.952 V
Problem 4
Two readings are taken in dry soil:
- 10 MΩ reading = −900 mV
- 20 MΩ reading = −920 mV
Determine the true potential.
Reveal solution
\[
\begin{aligned}
K &= \frac{10}{20} = 0.5 \\[10pt]
\frac{V_h}{V_l}
&=
\frac{-920}{-900}
=
1.02222 \\[10pt]
E_{true}
&=
\frac{-920 \,\times\, (1 - 0.5)}
{1 - 0.5 \,\times\, 1.02222} \\[10pt]
E_{true}
&=
\frac{-460}{0.48889}
=
-940.8\ \text{mV}
\end{aligned}
\]
Answer: −941 mV
Problem 5
Calculate \(R_{ext}\) using the results of Problem 4
(\(E_{true}=-940.8\ \text{mV}\), \(V_m=-920\ \text{mV}\), \(R_m=20\ \text{M}\Omega\)).
Reveal solution
\[
\begin{aligned}
R_{ext}
&=
20{,}000{,}000
\,\times\,
\left(
\frac{-940.8}{-920} - 1
\right) \\[10pt]
R_{ext}
&=
20{,}000{,}000
\,\times\,
0.02261 \\[10pt]
R_{ext}
&=
452{,}200\ \Omega
\approx 452\ \text{k}\Omega
\end{aligned}
\]
Answer: ≈452 kΩ
Problem 6
If the true potential is −1200 mV and the meter reads −1000 mV due to loading,
what is the percentage error magnitude?
Reveal solution
\[
\begin{aligned}
\%\text{Error}
&=
\left|
\frac{-1000 - (-1200)}{-1200}
\right|
\,\times\, 100 \\[10pt]
\%\text{Error}
&=
\left|
\frac{200}{-1200}
\right|
\,\times\, 100 \\[10pt]
\%\text{Error}
&=
16.7\%
\end{aligned}
\]
Answer: 16.7%
Problem 7
Measurements:
- 50 MΩ meter reading = −1.50 V
- 100 MΩ meter reading = −1.60 V
Calculate the true potential.
Reveal solution
\[
\begin{aligned}
K &= \frac{50}{100} = 0.5 \\[10pt]
\frac{V_h}{V_l}
&=
\frac{-1.60}{-1.50}
=
1.06667 \\[10pt]
E_{true}
&=
\frac{-1.60 \,\times\, (1 - 0.5)}
{1 - 0.5 \,\times\, 1.06667} \\[10pt]
E_{true}
&=
\frac{-0.800}{0.46667}
=
-1.714\ \text{V}
\end{aligned}
\]
Answer: −1.714 V
Problem 8
Given \(E_{true}=-821\ \text{mV}\), \(V_m=-800\ \text{mV}\), and \(R_m=10\ \text{M}\Omega\),
calculate the total circuit resistance.
Reveal solution
\[
\begin{aligned}
R_t
&=
R_m
\,\times\,
\left(
\frac{E_{true}}{V_m}
\right) \\[10pt]
R_t
&=
10
\,\times\,
\frac{-821}{-800} \\[10pt]
R_t
&=
10
\,\times\,
1.02625 \\[10pt]
R_t
&=
10.26\ \text{M}\Omega
\end{aligned}
\]
Answer: 10.26 MΩ
Problem 9
You accidentally use a meter with only 1 MΩ input impedance to measure a circuit with
1 MΩ external resistance. The true potential is −1.400 V.
What voltage will the meter read?
Reveal solution
\[
\begin{aligned}
V_m
&=
(-1.400)
\,\times\,
\frac{1}{1+1} \\[10pt]
V_m
&=
(-1.400)
\,\times\,
0.5 \\[10pt]
V_m
&=
-0.700\ \text{V}
\end{aligned}
\]
Answer: −0.700 V
Problem 10
Concrete measurements:
- 10 MΩ reading = −400 mV
- 100 MΩ reading = −450 mV
Calculate the corrected polarized potential.
Reveal solution
\[
\begin{aligned}
K &= \frac{10}{100} = 0.1 \\[10pt]
\frac{V_h}{V_l}
&=
\frac{-450}{-400}
=
1.125 \\[10pt]
E_{true}
&=
\frac{-450 \,\times\, (1 - 0.1)}
{1 - 0.1 \,\times\, 1.125} \\[10pt]
E_{true}
&=
\frac{-405}{0.8875}
=
-456.3\ \text{mV}
\end{aligned}
\]
Answer: −456 mV
