Anode groundbed resistance is the resistance associated with current discharge from an anode or anode groundbed into the surrounding earth.
In cathodic protection work, anode groundbed resistance affects how much current can be delivered for a given driving voltage. Lower resistance generally allows more current for the same voltage. Higher resistance generally limits current for the same voltage.
This page uses basic voltage, current, and resistance relationships only. It does not teach detailed anode-bed design formulas, spacing rules, backfill assumptions, or performance guarantees.
A learner-safe relationship for anode groundbed resistance review is:
R = E ÷ I
Where:
The relationship can also be rearranged:
I = E ÷ R
E = I × R
Anode groundbed resistance may be used when reviewing:
This page does not prescribe anode spacing.
This page does not prescribe backfill assumptions.
This page does not guarantee anode-bed performance.
This page does not prove CP protection.
| Solve for | Formula | Use when you know |
|---|---|---|
| Resistance | R = E ÷ I | voltage and current |
| Current | I = E ÷ R | voltage and resistance |
| Voltage | E = I × R | current and resistance |
| Symbol | Meaning | Common unit | Unit symbol | CP field meaning |
|---|---|---|---|---|
R | Resistance | ohm | Ω | Opposition to current discharge from an anode or groundbed |
E | Voltage | volt | V | Driving force available for current output |
I | Current | ampere | A | Current delivered through the circuit |
Use matching units:
Use R = E ÷ I when voltage and current are known.
Use I = E ÷ R when voltage and resistance are known.
Use E = I × R when current and resistance are known.
For learner calculations, use voltage and current values from the same operating condition.
Do not treat this simple relationship as a full anode-bed design method. Actual anode-bed performance depends on more than this formula alone.
A groundbed circuit has 50 V applied and 10 A flowing. What resistance does that represent?
Known values:
E = 50 VI = 10 AR = ?Use: R = E ÷ I
Substitute: R = 50 V ÷ 10 A
Calculate: R = 5 Ω
Answer: The resistance is 5 Ω.
Field meaning: This is the resistance represented by the voltage and current in this basic calculation.
A rectifier applies 40 V to an anode groundbed circuit with 8 Ω resistance. What current does that represent?
Known values:
E = 40 VR = 8 ΩI = ?Use: I = E ÷ R
Substitute: I = 40 V ÷ 8 Ω
Calculate: I = 5 A
Answer: The current is 5 A.
Field meaning: If voltage stays the same, higher resistance would reduce current.
A target current in a learner calculation is 6 A. The resistance is 7 Ω. What voltage does that represent?
Known values:
I = 6 AR = 7 ΩE = ?Use: E = I × R
Substitute: E = 6 A × 7 Ω
Calculate: E = 42 V
Answer: The voltage is 42 V.
Field meaning: This is the voltage represented by that current and resistance. It is not a complete rectifier or anode-bed design.
A rectifier applies 60 V. What current is represented by 6 Ω resistance? What current is represented by 12 Ω resistance?
Case 1: I = 60 V ÷ 6 Ω
I = 10 A
Case 2: I = 60 V ÷ 12 Ω
I = 5 A
Answer: At 6 Ω, the current is 10 A. At 12 Ω, the current is 5 A.
Field meaning: With the same voltage, doubling resistance cuts the current in half in this basic relationship.
ohm-cm or ohm-ft.| Problem | Answer | Calculation / explanation |
|---|---|---|
| 1 | 6 Ω | R = E ÷ I = 36 V ÷ 6 A = 6 Ω |
| 2 | 5 Ω | R = 60 V ÷ 12 A = 5 Ω |
| 3 | 6 A | I = E ÷ R = 30 V ÷ 5 Ω = 6 A |
| 4 | 4 A | I = 48 V ÷ 12 Ω = 4 A |
| 5 | 42 V | E = I × R = 7 A × 6 Ω = 42 V |
| 6 | Current is cut in half | With the same voltage, doubling resistance cuts current in half |
| 7 | No | This page teaches a basic voltage/current/resistance relationship only |
| 8 | No | Anode groundbed resistance supports review. CP protection must be evaluated with proper measurements and criteria. |