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AMPP CP2 Technician practice questions

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Question 1Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 4 Ω resistor is connected across a 12 V DC source. What current flows through the resistor?

  1. 3 A Correct answer.
  2. 0.33 A
  3. 48 A
  4. 16 A

Explanation

Ohm's law states that current equals voltage divided by resistance (I = V/R). Dividing 12 V by 4 Ω gives 3 A. The resistance limits the current, so the answer must be smaller than the voltage value.

Question 2Electrical Fundamentals and Circuit Theory for Cathodic Protection

Two resistors, 10 Ω and 15 Ω, are connected in series across a DC source. What is the total resistance of the combination?

  1. 25 Ω Correct answer.
  2. 1.5 Ω
  3. 6 Ω
  4. 5 Ω

Explanation

In a series circuit the same current passes through every element and the resistances simply add. Therefore the total is 10 Ω + 15 Ω = 25 Ω.

Question 3Electrical Fundamentals and Circuit Theory for Cathodic Protection

Two 8 Ω resistors are connected in parallel. What is the equivalent resistance of the pair?

  1. 8 Ω
  2. 16 Ω
  3. 4 Ω Correct answer.
  4. 0.25 Ω

Explanation

For two equal resistors in parallel the equivalent resistance is half of one resistor's value, because the conductance doubles. Formally, 1/R = 1/8 + 1/8 = 0.25 S, so R = 4 Ω.

Question 4Electrical Fundamentals and Circuit Theory for Cathodic Protection

A current of 0.5 A flows through a 20 Ω resistor. What power is dissipated in the resistor?

  1. 40 W
  2. 5 W Correct answer.
  3. 10 W
  4. 2.5 W

Explanation

Resistive power can be calculated as P = I²R when current and resistance are known. Squaring 0.5 A gives 0.25 A², and multiplying by 20 Ω gives 5 W.

Question 5Electrical Fundamentals and Circuit Theory for Cathodic Protection

At a junction in a DC circuit, 5 A arrives from one branch and 2 A leaves along a second branch. What current leaves along the third and only remaining branch?

  1. 7 A
  2. 3 A Correct answer.
  3. 2.5 A
  4. 10 A

Explanation

Kirchhoff's current law requires that the total current entering a junction equals the total current leaving it. The balance is 5 A − 2 A = 3 A leaving on the remaining branch.

Question 6Electrical Fundamentals and Circuit Theory for Cathodic Protection

How is an ammeter connected when measuring the current through a resistor?

  1. Across the DC source terminals
  2. Between the circuit and earth
  3. In series with the resistor Correct answer.
  4. In parallel with the resistor

Explanation

An ammeter must carry the current being measured, so it is placed in series so that the circuit current flows through it. Its low internal resistance then adds only a small burden to the circuit.

Question 7Electrical Fundamentals and Circuit Theory for Cathodic Protection

A conductor has a resistance of 4 Ω. What is its conductance?

  1. 16 S
  2. 0.25 S Correct answer.
  3. 4 S
  4. 0.5 S

Explanation

Conductance is the reciprocal of resistance, G = 1/R, and is expressed in siemens. For 4 Ω, G = 1/4 = 0.25 S.

Question 8Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 24 V DC source is connected across a series combination of a 6 Ω and an 18 Ω resistor. What is the voltage across the 18 Ω resistor?

  1. 12 V
  2. 18 V Correct answer.
  3. 6 V
  4. 24 V

Explanation

The series current is 24 V / 24 Ω = 1 A, and the drop across the 18 Ω resistor is 1 A × 18 Ω = 18 V. Equivalently, the voltage divider splits the source in proportion to resistance: 24 V × 18/24 = 18 V.

Question 9Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 12 V battery and a 6 V battery are connected in series aiding in a loop that also contains a 3 Ω resistor. What current flows in the loop?

  1. 6 A Correct answer.
  2. 18 A
  3. 4 A
  4. 2 A

Explanation

Series-aiding sources add, giving a net driving voltage of 18 V around the loop. Dividing by the 3 Ω resistance gives 18 V / 3 Ω = 6 A.

Question 10Electrical Fundamentals and Circuit Theory for Cathodic Protection

A conductor is 50 m long, has a cross-sectional area of 1.0 × 10⁻⁶ m² and a resistivity of 1.7 × 10⁻⁸ Ω·m. What is its resistance?

  1. 8.5 Ω
  2. 0.085 Ω
  3. 0.85 Ω Correct answer.
  4. 0.34 Ω

Explanation

Resistance follows R = ρL/A. Multiplying resistivity by length gives 8.5 × 10⁻⁷ Ω·m², and dividing by 1.0 × 10⁻⁶ m² gives 0.85 Ω.

Question 11Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is the equivalent resistance of the pair?

  1. 2 Ω Correct answer.
  2. 18 Ω
  3. 4.5 Ω
  4. 9 Ω

Explanation

For two resistors in parallel, the equivalent resistance is the product divided by the sum: (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The result must always be smaller than the smallest branch resistance.

Question 12Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 120 V DC source is connected across a series combination of a 10 Ω and a 30 Ω resistor. What power is dissipated in the 30 Ω resistor?

  1. 270 W Correct answer.
  2. 120 W
  3. 30 W
  4. 90 W

Explanation

The series current is 120 V / 40 Ω = 3 A. Power in the 30 Ω resistor is I²R = 9 A² × 30 Ω = 270 W.

Question 13Electrical Fundamentals and Circuit Theory for Cathodic Protection

Why is a high input impedance desirable in a voltmeter used for structure-to-electrolyte potential measurements?

  1. It can withstand much higher DC voltages
  2. It can measure AC as well as DC potentials
  3. It permits much longer test leads to be used
  4. It draws negligible current Correct answer.

Explanation

A voltmeter is placed in parallel with the circuit under test, so any current it draws passes through the same path being measured and depresses the reading. High input impedance keeps that current negligible so the measured potential reflects the undisturbed circuit.

Question 14Electrical Fundamentals and Circuit Theory for Cathodic Protection

A current shunt of 0.001 Ω carries 10 A. What voltage appears across the shunt?

  1. 0.1 mV
  2. 10 mV Correct answer.
  3. 1 V
  4. 100 mV

Explanation

The shunt is a known low resistance, so Ohm's law gives V = IR = 10 A × 0.001 Ω = 0.010 V, which is 10 mV. This small voltage is then read by a millivoltmeter calibrated in amperes.

Question 15Electrical Fundamentals and Circuit Theory for Cathodic Protection

Which component is used in AC mitigation to pass alternating current while blocking direct current?

  1. A capacitor Correct answer.
  2. A diode
  3. An inductor
  4. A resistor

Explanation

A capacitor presents a low reactance to alternating current but an open circuit to steady direct current once charged. This property is used in decoupling devices that provide an AC path to earth while preserving the DC cathodic protection circuit.

Question 16Electrical Fundamentals and Circuit Theory for Cathodic Protection

In a series circuit the source voltage is 48 V and two of the three voltage drops are 12 V and 18 V. What is the third voltage drop?

  1. 12 V
  2. 18 V Correct answer.
  3. 66 V
  4. 30 V

Explanation

Kirchhoff's voltage law requires that the voltage drops around a closed loop sum to the applied source voltage. Subtracting the known drops from 48 V leaves 48 − 12 − 18 = 18 V.

Question 17Electrical Fundamentals and Circuit Theory for Cathodic Protection

How does the resistance of a copper conductor change as its temperature rises?

  1. It decreases
  2. It increases Correct answer.
  3. It stays the same
  4. It falls to zero

Explanation

Copper is a metallic conductor with a positive temperature coefficient, so increased thermal vibration of the lattice scatters electrons more effectively. The result is a measurable rise in resistance with temperature, which matters when correcting cable resistance to a reference temperature.

Question 18Electrical Fundamentals and Circuit Theory for Cathodic Protection

Two parallel bonds of 0.1 Ω and 0.3 Ω carry a combined current of 8 A. What current flows in the 0.1 Ω bond?

  1. 2 A
  2. 6 A Correct answer.
  3. 8 A
  4. 4 A

Explanation

Parallel branches share current in inverse proportion to their resistance. The equivalent resistance is 0.075 Ω, so the common voltage is 0.6 V, and 0.6 V / 0.1 Ω = 6 A in the lower-resistance bond.

Question 19Electrical Fundamentals and Circuit Theory for Cathodic Protection

A circuit has 3 Ω of resistance in series with 4 Ω of capacitive reactance. What is the magnitude of the impedance?

  1. 5 Ω Correct answer.
  2. 1 Ω
  3. 7 Ω
  4. 12 Ω

Explanation

Resistance and reactance are ninety degrees out of phase, so they combine vectorially: Z = √(R² + X²). Here Z = √(9 + 16) = √25 = 5 Ω.

Question 20Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 24 V DC source supplies a 6 Ω resistor in series with a parallel pair of 12 Ω and 6 Ω resistors. What total current flows from the source?

  1. 6 A
  2. 1.6 A
  3. 4 A
  4. 2.4 A Correct answer.

Explanation

The parallel pair reduces to (12 × 6)/(12 + 6) = 4 Ω, which adds to the 6 Ω series resistor to give 10 Ω. The source current is therefore 24 V / 10 Ω = 2.4 A.

Question 21Electrical Fundamentals and Circuit Theory for Cathodic Protection

A source has an open-circuit voltage of 2.0 V and a short-circuit current of 0.5 A. It is connected to a 4 Ω load. What current flows in the load?

  1. 0.25 A Correct answer.
  2. 0.5 A
  3. 0.125 A
  4. 0.4 A

Explanation

The internal (Thevenin) resistance is the open-circuit voltage divided by the short-circuit current, 2.0 V / 0.5 A = 4 Ω. With a 4 Ω load the total is 8 Ω, so the load current is 2.0 V / 8 Ω = 0.25 A.

Question 22Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 12 V source with 1 Ω of internal resistance is connected to a 5 Ω load. What power is dissipated in the load?

  1. 28.8 W
  2. 12 W
  3. 24 W
  4. 20 W Correct answer.

Explanation

The total loop resistance is 1 Ω + 5 Ω = 6 Ω, so the current is 12 V / 6 Ω = 2 A. Load power is I²R = 4 A² × 5 Ω = 20 W.

Question 23Electrical Fundamentals and Circuit Theory for Cathodic Protection

Which condition increases the AC voltage induced on a buried pipeline by a nearby high-voltage power line?

  1. A lower current in the power line conductors
  2. A shorter length of parallelism
  3. A greater separation between the pipe and the line
  4. A higher current in the power line conductors Correct answer.

Explanation

Inductive (electromagnetic) coupling depends on the magnetic field linking the pipeline, which grows with the current carried by the power line and with the length over which the two run parallel. Increasing line current therefore raises the induced AC voltage on the pipeline.

Question 24Electrical Fundamentals and Circuit Theory for Cathodic Protection

A 265 µF capacitor is connected across a 60 Hz AC supply. What is its capacitive reactance?

  1. About 10 Ω Correct answer.
  2. About 100 Ω
  3. About 1 Ω
  4. About 0.1 Ω

Explanation

Capacitive reactance is Xc = 1/(2πfC). With f = 60 Hz and C = 265 × 10⁻⁶ F, the denominator is about 0.0999, giving a reactance of approximately 10 Ω.

Question 25Cathodic Protection Theory and Protection Criteria

Cathodic protection reduces corrosion by making the protected structure the ___ of an electrochemical cell.

  1. cathode Correct answer.
  2. electrolyte
  3. reference electrode
  4. anode

Explanation

Cathodic protection supplies electrons to the structure so that it becomes the cathode, suppressing anodic metal dissolution. The structure is deliberately made the cathodic half of the corrosion cell.

Question 26Cathodic Protection Theory and Protection Criteria

The four essential elements of a corrosion cell are an anode, a cathode, an electrolyte, and a ___.

  1. rectifier
  2. coating
  3. metallic path Correct answer.
  4. reference electrode

Explanation

A corrosion cell requires an electronic path connecting the anode and cathode so that electrons can flow between them. This metallic path completes the circuit along with the electrolyte.

Question 27Cathodic Protection Theory and Protection Criteria

Which reference electrode is most commonly used for pipe-to-soil potential measurements in soil?

  1. Copper/copper sulfate Correct answer.
  2. Saturated calomel
  3. Silver/silver chloride
  4. Standard hydrogen

Explanation

The copper/copper sulfate electrode is the standard field reference for soil potential measurements and the -850 mV criterion is expressed versus CSE. It is rugged and practical for burial or direct soil contact.

Question 28Cathodic Protection Theory and Protection Criteria

The -850 mV CSE polarized potential criterion for steel is verified by measuring the potential:

  1. with the CP current continuously applied
  2. with a zinc reference electrode only
  3. 24 hours after de-energizing the rectifier
  4. immediately after interrupting all CP current Correct answer.

Explanation

The polarized potential excludes IR drop, so all CP current sources are interrupted and the potential is read immediately before the structure depolarizes. This instant-off reading represents the true polarized potential.

Question 29Cathodic Protection Theory and Protection Criteria

Which metal is commonly used as a sacrificial anode for buried steel pipelines?

  1. Mixed metal oxide
  2. Copper
  3. Graphite
  4. Magnesium Correct answer.

Explanation

Magnesium has a sufficiently negative potential and is widely used as a galvanic anode for buried steel. It corrodes preferentially and supplies protective current to the steel.

Question 30Cathodic Protection Theory and Protection Criteria

In a galvanic CP system, the anode material must have a more ___ potential than the structure being protected.

  1. alternating
  2. negative Correct answer.
  3. positive
  4. equal

Explanation

A galvanic anode must be electrochemically active relative to the structure, meaning its potential is more negative. This potential difference provides the driving voltage for protective current.

Question 31Cathodic Protection Theory and Protection Criteria

Which material is typical of an impressed current anode?

  1. Zinc
  2. Magnesium
  3. Aluminum
  4. Mixed metal oxide Correct answer.

Explanation

Mixed metal oxide anodes are inert, long-life materials used with a rectifier in impressed current CP systems. They are not consumed as sacrificial anodes.

Question 32Cathodic Protection Theory and Protection Criteria

Soil resistivity for CP design is commonly measured using which field method?

  1. Wenner four-pin method Correct answer.
  2. Half-cell potential survey
  3. Pearson survey
  4. Close interval survey

Explanation

The Wenner four-pin method passes current through the soil and measures voltage drop to calculate resistivity. Soil resistivity directly affects anode output and current distribution.

Question 33Cathodic Protection Theory and Protection Criteria

A pipe-to-soil potential measured with CP current applied is -780 mV CSE, and the IR drop is estimated at 90 mV. What is the approximate polarized (instant-off) potential?

  1. -600 mV CSE
  2. -780 mV CSE
  3. -690 mV CSE Correct answer.
  4. -870 mV CSE

Explanation

IR drop makes an applied-current reading more negative than the true polarized potential. Subtracting the 90 mV IR drop from -780 mV gives approximately -690 mV CSE.

Question 34Cathodic Protection Theory and Protection Criteria

The 100 mV polarization criterion requires a potential shift of at least 100 mV in which direction from the native potential?

  1. Alternating
  2. Negative Correct answer.
  3. Zero net
  4. Positive

Explanation

Cathodic protection polarizes the structure in the negative direction, so the 100 mV criterion is based on a negative shift from the native potential. A positive shift would indicate anodic behavior rather than protection.

Question 35Cathodic Protection Theory and Protection Criteria

Which reference electrode is preferred for potential measurements in seawater?

  1. Silver/silver chloride Correct answer.
  2. Saturated calomel
  3. Copper/copper sulfate
  4. Zinc

Explanation

The silver/silver chloride electrode is stable and well suited to chloride-rich environments such as seawater. It is commonly used for marine cathodic protection measurements.

Question 36Cathodic Protection Theory and Protection Criteria

Anode bed resistance increases when which soil property increases?

  1. Soil conductivity
  2. Soil moisture
  3. Anode surface area
  4. Soil resistivity Correct answer.

Explanation

Anode bed resistance is proportional to electrolyte resistivity, so higher soil resistivity increases resistance and reduces anode current output. Moisture and conductivity have the opposite effect.

Question 37Cathodic Protection Theory and Protection Criteria

CP current interrupters are used during potential surveys primarily to:

  1. increase rectifier output
  2. measure native potential only
  3. eliminate IR drop from the measured potential Correct answer.
  4. test anode bed resistance

Explanation

Interrupting the CP current allows the potential to be read without the IR drop caused by current flowing through the soil. The resulting instant-off reading is used to assess the polarized potential.

Question 38Cathodic Protection Theory and Protection Criteria

In a neutral soil, the principal cathodic reaction in a CP system is:

  1. Oxygen reduction Correct answer.
  2. Sulfate reduction
  3. Hydrogen oxidation
  4. Iron oxidation

Explanation

In neutral and alkaline soils, dissolved oxygen is reduced at the cathode, producing hydroxide ions. This cathodic reaction supports the protective current in aerated environments.

Question 39Cathodic Protection Theory and Protection Criteria

Stray current corrosion occurs when:

  1. the structure is electrically isolated
  2. AC voltage is induced in the coating
  3. the coating has high dielectric strength
  4. direct current from an external source enters and leaves the structure Correct answer.

Explanation

Stray direct current can enter a structure at one location and discharge at another, causing concentrated corrosion at the discharge point. The external current source may be another CP system, a DC transit system, or welding equipment.

Question 40Cathodic Protection Theory and Protection Criteria

Which pipe-to-soil potential indicates possible overprotection of a high-strength steel pipeline?

  1. -1250 mV CSE Correct answer.
  2. -950 mV CSE
  3. -600 mV CSE
  4. -850 mV CSE

Explanation

Potentials more negative than about -1200 mV CSE can generate hydrogen at the steel surface and increase the risk of hydrogen embrittlement in high-strength steels. The -1250 mV value is therefore a concern for overprotection.

Question 41Cathodic Protection Theory and Protection Criteria

A high coating conductance value for a coated pipeline generally indicates:

  1. a low current requirement
  2. that CP is unnecessary
  3. coating deterioration or many holidays Correct answer.
  4. a sound, high-resistance coating

Explanation

Coating conductance is inversely related to coating resistance, so a high value means the coating is allowing more current to pass. This typically reflects deterioration, holidays, or a generally leaky coating.

Question 42Cathodic Protection Theory and Protection Criteria

The current required to protect a bare steel structure is calculated by multiplying the required current density by the:

  1. rectifier voltage
  2. soil resistivity
  3. surface area Correct answer.
  4. anode mass

Explanation

Current requirement equals current density multiplied by the surface area exposed to the electrolyte. For a coated structure, only the bare or damaged area is normally used in the calculation.

Question 43Cathodic Protection Theory and Protection Criteria

Which protection criterion is generally applied to aluminum structures to avoid amphoteric corrosion?

  1. 100 mV polarization shift Correct answer.
  2. -850 mV CSE polarized
  3. -1200 mV CSE polarized
  4. -600 mV CSE polarized

Explanation

Aluminum can suffer amphoteric corrosion if polarized too negatively, so the 100 mV polarization shift criterion is generally used instead of a fixed -1200 mV target. This limits the risk of alkaline attack at the metal surface.

Question 44Cathodic Protection Theory and Protection Criteria

A close interval potential survey is primarily used to:

  1. determine anode consumption
  2. locate coating holidays and assess CP levels along a pipeline Correct answer.
  3. measure soil resistivity
  4. measure coating thickness

Explanation

A close interval survey records pipe-to-soil potentials at short spacing along a pipeline, revealing areas of low protection and likely coating damage. It is a CP assessment tool, not a resistivity or coating-thickness test.

Question 45Cathodic Protection Theory and Protection Criteria

A 100 mV polarization decay test shows an instant-off potential of -0.92 V CSE and a potential after 4 hours of -0.80 V CSE. Does the structure meet the 100 mV criterion?

  1. Cannot be determined without soil resistivity
  2. Yes, because the instant-off potential is more negative than -0.85 V
  3. Yes, because the decay is 120 mV Correct answer.
  4. No, because the decay is only 80 mV

Explanation

The polarization decay is the difference between the instant-off potential and the later potential: 0.92 V minus 0.80 V equals 0.12 V, or 120 mV. A decay of at least 100 mV satisfies the criterion.

Question 46Cathodic Protection Theory and Protection Criteria

A pipeline has a native potential of -0.55 V CSE and a polarized potential of -0.70 V CSE. Which criterion is satisfied?

  1. 100 mV polarization criterion Correct answer.
  2. Both criteria
  3. -850 mV CSE polarized criterion
  4. Neither criterion

Explanation

The polarization shift is 0.70 V minus 0.55 V, or 150 mV, which exceeds the 100 mV criterion. The polarized potential of -0.70 V CSE is not more negative than -0.85 V CSE, so the -850 mV criterion is not met.

Question 47Cathodic Protection Theory and Protection Criteria

A potential of -0.850 V versus a copper/copper sulfate electrode is approximately what potential versus a saturated calomel electrode? (CSE = +316 mV vs SHE; SCE = +241 mV vs SHE)

  1. -0.925 V
  2. -0.550 V
  3. -0.775 V Correct answer.
  4. -0.850 V

Explanation

Converting to the standard hydrogen scale gives -0.850 V + 0.316 V = -0.534 V vs SHE. Subtracting the SCE offset of 0.241 V gives approximately -0.775 V vs SCE.

Question 48Cathodic Protection Theory and Protection Criteria

A coated steel tank has 1,000 m² of surface area, 5% coating breakdown, and requires 0.02 A/m² of bare steel. What is the approximate current requirement?

  1. 1.0 A Correct answer.
  2. 0.1 A
  3. 0.5 A
  4. 2.0 A

Explanation

The bare area is 1,000 m² multiplied by 0.05, or 50 m². Multiplying 50 m² by 0.02 A/m² gives 1.0 A of required protective current.

Question 49Cathodic Protection Theory and Protection Criteria

A metallic short between a cathodically protected pipeline and an unprotected grounded structure will generally:

  1. drain CP current and cause underprotection Correct answer.
  2. increase coating resistance
  3. reduce soil resistivity
  4. improve CP protection on the pipeline

Explanation

A metallic short provides an unintended path for CP current to leave the protected pipeline and enter the grounded structure. This current drain can reduce the pipeline's polarization and create underprotection or interference.

Question 50Cathodic Protection Theory and Protection Criteria

Cathodic protection controls corrosion of a buried steel structure primarily by which mechanism?

  1. Supplying electrons to suppress anodic dissolution Correct answer.
  2. Driving the structure potential in the noble direction
  3. Removing dissolved oxygen from the surrounding soil
  4. Applying a conductive barrier coating to the metal

Explanation

Cathodic protection makes the structure the cathodic site of a corrosion cell by feeding it electrons from an anode, so the anodic iron-dissolution reaction is stifled. The structure potential is driven in the negative (active) direction, not the noble direction. Oxygen removal and barrier coatings may reduce current demand but are not the protective mechanism itself.