4.2 Series & Parallel Resistance in CP Groundbeds & Piping

Series & Parallel Resistance in CP Groundbeds & Piping

Cathodic protection circuits are rarely as simple as a single resistor. They are comprised of multiple resistive elements connected together. To accurately evaluate a system, a CP1 Tester must understand how to calculate total equivalent resistance for both series and parallel circuit configurations.

Series Circuits

In a series circuit, there is only one path for the electrical current to flow. The current must pass through every component in the circuit sequentially. If the path is broken at any point, the current flow completely stops (an open circuit).

Rules for Series Circuits

1. Current (I): The current is the same at all points in the circuit. (I_total = I_1 = I_2 = I_3...)
2. Voltage (E): The total voltage is the sum of the voltage drops across each individual resistor. (E_total = E_1 + E_2 + E_3...)
3. Resistance (R): The total equivalent resistance (R_eq) is the sum of all individual resistances.

R_eq = R_1 + R_2 + R_3 + ... + R_n

Application in CP

A basic CP circuit is essentially a series circuit consisting of three main components: the cable resistance, the anode-to-earth resistance, and the structure-to-earth resistance. The current from the rectifier travels through the positive cable, into the anode, through the soil, onto the pipeline, and back through the negative cable.

Example Calculation: Calculate the total circuit resistance of a single magnesium anode installation given the following measured values:

• Positive wire resistance (R_wire): 0.1 Ω
• Anode-to-earth resistance (R_anode): 15.5 Ω
• Structure-to-earth resistance (R_structure): 2.4 Ω

R_eq = R_wire + R_anode + R_structure R_eq = 0.1 Ω + 15.5 Ω + 2.4 Ω R_eq = 18.0 Ω

Parallel Circuits

In a parallel circuit, there are multiple paths (branches) for the current to flow. If one branch is broken, current continues to flow through the remaining parallel branches. In CP, multiple anodes are almost always installed in parallel to reduce the overall groundbed resistance.

Rules for Parallel Circuits

1. Voltage (E): The voltage drop is the same across all parallel branches. (E_total = E_1 = E_2 = E_3...)
2. Current (I): The total current is the sum of the currents flowing through each individual branch. (I_total = I_1 + I_2 + I_3...)
3. Resistance (R): The total equivalent resistance is calculated using the reciprocal formula, and the total resistance is always less than the smallest individual resistance.

1 / R_eq = (1 / R_1) + (1 / R_2) + (1 / R_3) + ... + (1 / R_n)

Alternatively, for exactly two resistors in parallel, you can use the "Product over Sum" rule: R_eq = (R_1 × R_2) / (R_1 + R_2)

Application in CP Groundbeds

When a design engineer needs more current than a single anode can provide, they install multiple anodes in a groundbed. These anodes are connected via a common header cable, effectively placing them in parallel. By adding anodes in parallel, the total resistance of the groundbed decreases, allowing more current to flow for a given voltage.

Example Calculation: Two Anodes in Parallel You have a groundbed with two anodes. Anode 1 has a resistance to earth of 10 Ω. Anode 2 has a resistance to earth of 15 Ω. What is the equivalent resistance of the groundbed?

Using the Product over Sum rule: R_eq = (10 × 15) / (10 + 15) R_eq = 150 / 25 R_eq = 6.0 Ω

Notice that the total resistance (6.0 Ω) is lower than the lowest individual resistor (10 Ω).

Example Calculation: Multiple Identical Anodes If all parallel resistors have the exact same value, a shortcut can be used: simply divide the resistance of one resistor by the total number of resistors. R_eq = R / N

A deep well groundbed contains ten mixed metal oxide (MMO) anodes, each with an individual resistance to earth of 20 Ohms. What is the total equivalent resistance of the deep well? R_eq = 20 Ω / 10 anodes R_eq = 2.0 Ω

Troubleshooting Parallel Groundbeds

Understanding parallel circuits is vital for troubleshooting distributed groundbeds. If a header cable is accidentally excavated and cut, a portion of the parallel anodes will be disconnected from the circuit.

Imagine a distributed groundbed of five identical 10 Ω anodes connected in parallel. Initially, R_eq = 10 Ω / 5 = 2.0 Ω.

If the cable is cut midway, disconnecting three anodes, only two anodes remain in the circuit. The new resistance is R_eq = 10 Ω / 2 = 5.0 Ω.

Because the total circuit resistance has increased significantly (from 2.0 Ω to 5.0 Ω), the current output from the rectifier will drop correspondingly if the voltage remains constant. As a CP1 tester, observing a sudden drop in current and calculating an increased circuit resistance gives you a strong indication that anodes have been lost or a cable has failed, guiding your field troubleshooting efforts.

Piping Networks

Piping networks can also form complex parallel circuits. When a pipeline is electrically bonded to another structure, such as a casing or a foreign pipeline, they act as parallel resistors to earth. Current from the CP system will split, with a portion returning via the primary pipeline and a portion returning via the bonded structure. The amount of current flowing to each structure depends inversely on its resistance to earth; the path of least resistance will carry the most current.