2A672 Pre-test Edit Code 06

The direct current (DC) card on the B809A generator set senses the condition of overtemperature. This means that the card is designed to detect and monitor the temperature of the generator set, and if it exceeds a certain threshold, it will trigger an alarm or shut down the system to prevent damage or hazards. This is an important feature as overheating can lead to equipment failure or fire, so monitoring and controlling the temperature is crucial for the safe and efficient operation of the generator set.

When the butterfly valve is closed on an A/M32A-95 gas turbine compressor, chambers 1, 3, and 4 are pressurized.

The turbine start switch on the A/M32A-95 gas turbine compressor starts the automatic starting cycle. This switch is specifically designed to initiate the starting process of the turbine. It plays a crucial role in activating the necessary mechanisms and systems required for the turbine to start and operate efficiently.

The probable cause of the trouble is that S30 is shorted. This is indicated by the fact that when the person tries to reset S23, the fault lights stay on. If S30 is shorted, it would prevent the reset of S23 and keep the fault lights illuminated.

The operating voltage of the blower motor and control relays on an A/M24T-8 load bank is 28 VDC.

A loose push-on connector on the output filter capacitor could cause one phase voltage to be lower than the other two phases. When the connector is loose, it may result in a poor electrical connection, leading to a decrease in voltage on that particular phase. This can cause an imbalance in the voltage levels between the phases.

The correct answer is "The analog board has failed." This is because an under voltage (UV) fault problem indicates a malfunction in the voltage regulation system of the frequency converter. The analog board is responsible for regulating and controlling the voltage, so if it has failed, it could result in an under voltage fault.

In a bridge rectifier circuit, there are four diodes arranged in a bridge configuration. During the first cycle of AC, the two diodes on the same side of the bridge as the positive terminal of the AC source will be forward biased, allowing current to flow through them. The other two diodes will be reverse biased and will not conduct current. Therefore, only two diodes are forward biased during the first cycle of AC.

The emitter in a unijunction transistor (UJT) always points toward the base 1 lead.

The blower component on an MA-3D air conditioner increases the ambient air temperature approximately 60 degrees Fahrenheit.

The oil tank vent line prevents overpressurization of the bearing cavity on an A/M32A-95 gas turbine compressor. This vent line allows excess pressure to be released from the bearing cavity, ensuring that the pressure does not build up to dangerous levels. By allowing the pressure to escape, the vent line helps to maintain the proper operating conditions for the compressor and prevents any damage that could be caused by overpressurization.

The value of I3 can be found by applying Kirchhoff's current law, which states that the total current entering a junction is equal to the total current leaving the junction. In this case, I1 = I2 + I3. Given that I1 = 9 amps and I2 = 5 amps, we can solve for I3, which is 4 amps. To find the total current (IT), we can simply add up the currents in the series and parallel branches, giving us IT = I1 = 9 amps. Therefore, the correct answer is I3 = 4 amps and IT = 9 amps.

The basic principle of operation for transformers is mutual induction. Mutual induction refers to the process in which a changing current in one coil induces a voltage in a nearby coil. In a transformer, this principle is used to transfer electrical energy from one circuit to another through electromagnetic induction. The primary coil, connected to a power source, creates a changing magnetic field which induces a voltage in the secondary coil, allowing for the transfer of energy.

The secondary winding of a transformer acts as a conductor and provides the path into which the voltage is induced. When an alternating current flows through the primary winding, it creates a changing magnetic field. This changing magnetic field induces a voltage in the secondary winding, which can then be used to power electrical devices or transmit electricity to a different location. The secondary winding is typically connected to the load or the electrical system that requires the transformed voltage.

A capacitor opposes any change in voltage in an AC circuit because it stores and releases electrical energy as the voltage alternates. In an AC circuit, the voltage continuously changes direction, causing the capacitor to charge and discharge repeatedly, which creates an opposition to any change in voltage. On the other hand, in a DC circuit, the voltage remains constant, so a capacitor does not oppose any change in voltage. Therefore, the correct answer is AC.

When a negative potential is connected to the cathode and a positive potential is connected to the anode of a PN junction, the diode is forward biased. In this configuration, the positive potential at the anode repels the majority charge carriers in the P-region towards the junction, while the negative potential at the cathode attracts the majority charge carriers in the N-region towards the junction. This creates a low resistance path for current to flow through the diode, allowing current to flow from the anode to the cathode.

The probable cause of the voltage not building up in the generator set is that the K23 contacts are open.

To test the load contactor relay K1, the automatic-manual voltage control switch must be placed in the AUTOMATIC position. This is because the AUTOMATIC position allows the relay to be tested using the test probes, ensuring that the relay is functioning properly. Placing the switch in any other position may not provide the correct conditions for testing the relay.

The resistance in a circuit can be calculated using the formula: Resistance = Power / Current^2. In this case, the power rating is given as 40 watts and the total current is 2 amps. Plugging these values into the formula, we get: Resistance = 40 / (2^2) = 40 / 4 = 10 ohms. Therefore, the correct answer is 10 ohms.

In a series-parallel circuit, the total resistance (RT) is calculated by adding the resistances in series and then adding the reciprocal of the resistances in parallel. In this case, the resistors R2 and R3 are in parallel, so their combined resistance is 1/(1/R2 + 1/R3) = 1/(1/8 + 1/8) = 1/(2/8) = 1/(1/4) = 4 ohms. The total resistance is then the sum of R1 and the combined resistance of R2 and R3, which is 7 + 4 = 11 ohms.

In an inductive circuit, current lags voltage by 90 degrees means that the maximum value of current occurs when the voltage is at its minimum value, and the maximum value of voltage occurs when the current is at its minimum value. This is because in an inductive circuit, the current takes time to reach its maximum value after the voltage is applied. Therefore, there is a phase difference of 90 degrees between the current and voltage waveforms.

If the arrow in a bipolar transistor is pointing away from the base, it indicates that the transistor is of NPN type. In an NPN transistor, the base is made of P-type material, while the emitter and collector are made of N-type material. The arrow represents the direction of conventional current flow, which is from the emitter to the base and then to the collector in an NPN transistor.

The heat sink on the MA-3D cools the ethylene glycol/water solution. This means that the heat sink is responsible for reducing the temperature of the ethylene glycol/water solution, which is likely used for cooling purposes in the system. The other options, such as cooling the discharge air from the blower or cooling ambient air, are not mentioned in the question and therefore cannot be assumed to be correct.

The electrostatic fields around a positive ion move outward. This is because positive ions have a deficiency of electrons, resulting in an overall positive charge. As a result, the electrostatic field lines emanate from the positive ion and extend outward into the surrounding space. This movement of the electrostatic fields outward is a characteristic behavior of positive ions.

Voltage is the correct answer because it is the force or potential difference that pushes free electrons through a conductor. It is the driving force behind the flow of electric current.

In P-type material, the majority carriers are holes, which are positively charged. Therefore, the minority carriers would be the opposite charge, which are electrons. Electrons are negatively charged and are present in lower concentrations compared to holes in P-type material.

The firing time of a unijunction transistor (UJT) is controlled by the resistance and capacitance in the circuit. The resistance determines the rate at which the capacitor charges, while the capacitance determines the time it takes for the capacitor to discharge. By adjusting the values of resistance and capacitance, the firing time of the UJT can be controlled.

The correct answer is S5, run/idle toggle switch. This switch is closed to provide potential to the exciter field while operating a B809A generator. The run/idle toggle switch controls the generator's operation mode, allowing it to switch between running and idle states. When the switch is closed, it completes the circuit and provides the necessary potential to the exciter field, enabling the generator to generate power.

The probable cause of the DC voltmeter failing to indicate during operation is that TR1 is defective.

The probable cause of the engine not operating when the MA-3D air conditioner SW1 is placed in the START position is that the override switch is not closed. The override switch needs to be closed in order for the engine to start.

Atoms having more than four electrons, but less than eight are known as insulators. Insulators are materials that do not conduct electricity easily due to the arrangement of their electrons. In these atoms, the outermost energy level is not fully filled, making them less likely to share or transfer electrons with other atoms. This incomplete outer electron shell creates a large energy gap between the valence band and the conduction band, making it difficult for electrons to move freely and conduct electricity.

The probable cause of the trouble is that the contacts C and D of the K11 relay are opened. This would prevent the alternating current (AC) contactor light from coming on when S3 is closed during operation of the A/M32A-60A.

The correct answer is Blower motor B2. According to the information provided in foldout 6, CR2 supplies 28 volts direct current (VDC) to operate the panel lights, the wind switch assembly, and the alternating current (AC) voltmeter. However, there is no mention of CR2 supplying power to Transformer PT3 or Blower motor B1. Therefore, the only correct option is Blower motor B2.

The low coolant light (DS50) illuminates, but the unit does not shut down. This suggests that the low coolant switch (S55) and the low coolant relay (S48) are both closed. When the coolant level is low, the low coolant switch should open, causing the low coolant relay to open as well and shut down the unit. However, in this case, both the switch and relay are closed, indicating a malfunction in either one of them.

The oil pressure sequencing switch is a component of the A/M32A-60A gas turbine engine lubrication system that is responsible for preventing engine operation if the oil pressure is too low. This switch acts as a safety mechanism to ensure that the engine is not operated without sufficient oil pressure, which is essential for proper lubrication and protection of the engine components. If the oil pressure is below the specified threshold, the sequencing switch will interrupt the engine operation, preventing any potential damage that could occur due to inadequate lubrication.

When an A/M32A-60A generator is started, the ignition unit de-energizes at 95 percent. This means that the ignition unit, which is responsible for providing the initial spark to start the generator, is no longer needed once the generator is running at 95 percent capacity. De-energizing the ignition unit helps to conserve energy and prevent any potential damage that could occur if the ignition unit were to remain active while the generator is running.

The probable cause of the engine cranking but not starting is that the SV1 coil is open. An open coil means that there is a break in the electrical circuit, preventing the coil from receiving power and functioning properly. Without the coil functioning, the necessary spark to ignite the fuel mixture in the engine is not being generated, resulting in the engine not starting.

The regulator card output pulse sent to the exciter field of the B809A generator is powered by 110 VDC.

The silicon-controlled rectifier (SCR) has four layers. An SCR is a type of semiconductor device that acts as an electric switch. It consists of three P-N junctions, which form four layers: P-N-P-N. The four layers allow the SCR to control the flow of electric current by blocking or allowing it to pass through.

When the emitter-base (E-B) current is increased in a transistor amplifier, it causes the transistor to enter into the active region of operation. In this region, the transistor acts as an amplifier and allows a larger current to flow through the emitter-collector (E-C) circuit. Therefore, the correct answer is that the current increases through the E-C circuit.

When K25 is energized during the operation of an A/M32A-60A, it does not cause any negative effect or disruption. Instead, the unit continues to operate normally, indicating that the energization of K25 does not have any significant impact on the functioning of the equipment.

During normal operation of the self-generating nitrogen servicing cart (SGNSC), the starter solenoid disengages due to the alternator output. The alternator is responsible for generating electrical power to charge the battery and run the electrical components of the cart. When the alternator output reaches a certain level, it signals the starter solenoid to disengage, allowing the engine to continue running without the need for the starter motor. This ensures that the starter motor is not continuously engaged and prevents damage to the starter system.

In a PNP bipolar transistor, the emitter is the most positive point. This is because the emitter is connected to the positive terminal of the power supply, while the base and collector are connected to the negative terminal. The emitter is responsible for injecting majority charge carriers (electrons in this case) into the base region, allowing current to flow through the transistor. Therefore, the emitter is considered the most positive point in the transistor.

The load rheostat is the component of the voltage regulator that compensates for losses in the power cable by sensing changes in current flow. It adjusts the resistance in the circuit to maintain a constant voltage output, compensating for any losses in the power cable by increasing or decreasing the current flow as needed. This ensures that the voltage remains stable and consistent, regardless of any losses in the power cable.

In a series-parallel circuit, the total voltage is divided between the series and parallel portions. Since the total voltage is given as 12 volts and the value of E2 in the parallel branch is given as 3 volts, the remaining voltage (12 - 3 = 9 volts) is divided between E1 and E3. Therefore, E1 is 9 volts and E3 is 3 volts.

The zener diode is connected in parallel to the load in order to protect it. When the voltage across the load exceeds the breakdown voltage of the zener diode, it starts conducting and limits the voltage to the breakdown voltage. This prevents any excessive voltage from reaching the load and protects it from damage.

The probable cause of the trouble is that K3 is de-energized and its contacts did not remain closed. This means that the circuit for charging the battery is not being completed, resulting in the battery not charging.

The components that receive potential directly from the input auto transformer are the SCR voltage regulator and phase sequence transformer.

A metal oxide varistor (MOV) is made up of two semiconductors. This is because a MOV is typically composed of a zinc oxide (ZnO) disc and a ceramic disc. The ZnO disc acts as a non-linear resistor and the ceramic disc acts as an insulator. When a voltage surge occurs, the ZnO disc conducts and absorbs the excess energy, protecting the circuit from damage. Therefore, the correct answer is 2.

The correct answer is Transformer rectifier (TR1). The transformer rectifier (TR1) component on the MA-3D air conditioner provides 24 volts alternating current (VAC) for operation of the temperature controller. This means that the TR1 converts the electrical current from the power source into the necessary voltage and type of current needed for the temperature controller to function properly.

A junction diode has only one PN junction. A PN junction is formed when a P-type semiconductor material is joined with an N-type semiconductor material. This creates a region where the free electrons from the N-type material combine with the holes from the P-type material, creating a depletion region. This depletion region acts as a barrier to the flow of current in one direction, allowing the diode to function as a one-way valve for electrical current. Therefore, a junction diode has only one PN junction.

The control circuit in a bipolar transistor is responsible for regulating the flow of current through the transistor. It controls the base current, which in turn controls the collector current. The control circuit is crucial for the proper functioning of the transistor and ensures that the desired amount of current flows through it.

During high voltage operation of the B809A generator set, the action that occurs to prevent the direct current (DC) contactors from closing is that the High Sense Relay (HSR) de-energizes.

The correct answer is to check if the 95 percent centrifugal switch (S14) is closed. This is because the question states that the generator set runs up to 80 percent and the ready to load light comes on. In normal operation, the generator set should run up to 95 percent before the ready to load light comes on. Therefore, the most likely cause of the issue is that the 95 percent centrifugal switch (S14) is not closed, which is preventing the generator set from reaching its full capacity.

The probable cause of the trouble is that K110 is open. This means that the circuit breaker K110, which controls the load selector switches, is not allowing electricity to flow through it. As a result, even though the AC voltmeter indicates that there is power, the load selector switches are unable to activate any loads because they are not receiving any electricity.

The correct answer is B7 which steps open to meter refrigerant into the evaporator. This means that when the output air temperature is too high, the temperature controller adjusts B7 to allow more refrigerant into the evaporator, which helps in cooling the air.

The correct answer is that the suction pressure is too low, so the boiling point of the refrigerant must be increased to increase the output air temperature. This means that when the suction pressure is too low, the temperature controller opens the hot gas bypass valve to increase the boiling point of the refrigerant. By doing so, the output air temperature can be increased.

The probable cause of the generator shutting down when S7 is released is that K18 is de-energizing. This means that the circuit or component controlled by K18 is losing power, causing the generator to shut down. The other options mentioned (S10 being left on, K17 energizing, and K16 de-energizing) are not directly related to the problem of the generator shutting down when S7 is released.

The probable cause of the shutdown solenoid energizing immediately after the engine run indicator illuminates is that S49 is shorted.

The probable cause of the trouble is that S53 is open.

The probable cause of the engine rotating without combustion during operation of the A/M32A-60A is a faulty ignition system.

The probable cause of the engine on light continuing to glow after shutting down the generator set is that S48 is closed. When S48 is closed, it indicates that the engine is still receiving power and is not fully shut down.

When CB2 opens on an A/M24T-8 load bank, it will cause the alternating current (AC) load selection switches (S15 through S20) to stop working. This means that the load bank will not be able to switch between different AC loads using these switches. The other options are incorrect as they do not describe the effect of CB2 opening on the load bank operation.

The oil pressure switch on the A/M32A-95 gas turbine compressor closes to complete the electrical circuit to the fuel solenoid and ignition unit. This means that when the oil pressure reaches a certain level, the switch is activated and allows the electrical current to flow, enabling the fuel solenoid and ignition unit to function properly.

When the bleed air switch is turned on, the engine speed drops to 98 percent. This indicates that the spring tension in the governor is greater than the centrifugal force, causing the flyweights to move inward. As a result, the bypass port is closed, which increases the fuel flow to the engine and brings the engine speed back up to 100 percent.

A capacitive circuit is characterized by the fact that the current flowing through it leads the applied voltage by 90 degrees. This means that the current reaches its maximum value before the voltage does. This behavior is due to the fact that a capacitor stores energy in an electric field, and as the voltage across it changes, the electric field charges or discharges, causing the current to flow. Therefore, the correct answer is that the current leads the applied voltage by 90 degrees.

The correct answer is the Air load switch. On an A/M32A-95 gas turbine compressor, the Air load switch is the component that energizes SV4.

The probable cause of the blower motor not working on the load bank is CR2.

In this question, the correct answer is TD2. When the air load switch S2 is placed in the ON position and the butterfly valve starts to open at 95 percent, you would need to check TD2. TD2 refers to the turbine discharge temperature. By checking TD2, you can monitor the temperature of the gas turbine compressor and ensure that it is within acceptable limits during the operational check.

A defective K4 relay causes both K607 and K1 to de-energize.

The probable cause of the B phase on the A/M24T-8 load bank not working is most likely due to an issue with component S56.

If the ignition control unit fails to operate on an A/M32A-60A during operation, the fuel solenoid L1 should be checked. This is because the fuel solenoid L1 is responsible for controlling the flow of fuel to the ignition system. If it is not functioning properly, it can lead to a failure of the ignition control unit. Therefore, checking the fuel solenoid L1 is necessary to ensure proper operation of the ignition system.

The probable cause of the trouble is that CR1 is malfunctioning. This is indicated by the fact that the blower motors are turning when the AC cable is plugged in, but nothing happens when the shock load switches are turned on. CR1 is responsible for controlling the flow of current to the load bank, so if it is malfunctioning, it would prevent the load bank from receiving power and activating the shock load switches.

The probable cause of the trouble is that S2 is open. This means that the switch S2, which is responsible for completing the circuit and allowing power to flow, is not functioning properly and remains in the open position. As a result, when the switch S5 is closed, the contactor light illuminates, indicating that power is being applied. However, when the switch is released, the contactor opens again because S2 is open and the circuit is interrupted.

You would start troubleshooting at K2.

A zener diode is connected in series with the resistor in a circuit to regulate voltage. When the voltage across the zener diode exceeds its breakdown voltage, it starts conducting and maintains a constant voltage drop across itself. By connecting it in series with the resistor, the zener diode ensures that the voltage across the load remains constant, even if the input voltage fluctuates.