Module 3 : Electrical Fundamentals

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Module 3 : Electrical Fundamentals - Quiz

This is only a preparation for EASA Module 3 and not an official EASA exam.


Questions and Answers
  • 1. 

    A material in which there are no free charge carriers is known as

    • A.

      Conductor

    • B.

      Insulator

    • C.

      Charges

    Correct Answer
    B. Insulator
    Explanation
    An insulator is a material in which there are no free charge carriers. Unlike conductors, which have free electrons that can move easily within the material, insulators have tightly bound electrons that are not able to move freely. This lack of free charge carriers makes insulators poor conductors of electricity.

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  • 2. 

    Conventional flow is:

    • A.

      Always from negative to positive

    • B.

      In the same direction as electron movement

    • C.

      In the opposite direction to electron movement

    Correct Answer
    C. In the opposite direction to electron movement
    Explanation
    Conventional flow refers to the direction of current flow in a circuit, which is opposite to the actual movement of electrons. In reality, electrons flow from the negative terminal to the positive terminal of a battery or power source. However, conventional flow is defined as the opposite direction, from positive to negative. This convention was established before the discovery of the electron, and it simplifies circuit analysis and calculations. Therefore, the correct answer is "in the opposite direction to electron movement."

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  • 3. 

    The power factor in an AC circuit is the same as:

    • A.

      The sine of the phase angle

    • B.

      The cosine of the phase angle

    • C.

      The tangent of a phase angle

    Correct Answer
    B. The cosine of the phase angle
    Explanation
    The power factor in an AC circuit is the same as the cosine of the phase angle. This is because the power factor represents the ratio of the real power to the apparent power in the circuit. The real power is the component of power that performs useful work, while the apparent power is the product of the voltage and current magnitudes. The phase angle represents the phase difference between the voltage and current waveforms in the circuit. The cosine of the phase angle is used to calculate the power factor, as it relates the real power to the apparent power.

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  • 4. 

    A passenger cabin has 110 passenger reading lamps each rated at 10W 28V. What is the maximum load current imposed by these lamps?

    • A.

      25.5 A

    • B.

      39.3 A

    • C.

      308 A

    Correct Answer
    B. 39.3 A
    Explanation
    The maximum load current imposed by the lamps can be calculated by dividing the total power consumed by the lamps by the voltage supplied. In this case, the total power consumed by the lamps is 110 lamps * 10W = 1100W. Since the lamps are rated at 28V, the maximum load current can be calculated as 1100W / 28V = 39.3A.

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  • 5. 

    Two isolated charges have dissimilar polarities. The force between them will be:

    • A.

      A force of attraction

    • B.

      A force of repulsion

    • C.

      Zero

    Correct Answer
    A. A force of attraction
    Explanation
    Two isolated charges with dissimilar polarities will experience a force of attraction. This is because opposite charges attract each other according to the principle of electrostatics. The charges will exert an attractive force on each other, causing them to move towards each other.

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  • 6. 

    The relationship between voltage (V), current (I) and resistance (R) for a resistor is:

    • A.

      V = IR

    • B.

      V = R/I

    • C.

      V = IR * R

    Correct Answer
    A. V = IR
    Explanation
    The correct relationship between voltage (V), current (I), and resistance (R) for a resistor is V = IR. This equation, known as Ohm's Law, states that the voltage across a resistor is equal to the product of the current flowing through it and the resistance of the resistor. This relationship holds true for most linear resistive circuits and is widely used in electrical and electronic engineering to calculate unknown values in circuits.

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  • 7. 

    A potential difference of 7.5 V appears across a 15-ohm resistor. Which of the following is the current through the resistor?

    • A.

      0.25 A

    • B.

      0.5 A

    • C.

      2 A

    Correct Answer
    B. 0.5 A
    Explanation
    The potential difference across a resistor is directly proportional to the current flowing through it according to Ohm's Law (V = IR). In this case, a potential difference of 7.5 V is applied across a 15-ohm resistor. To find the current, we can rearrange Ohm's Law to solve for I: I = V/R. Plugging in the values, we get I = 7.5 V / 15 ohms = 0.5 A. Therefore, the correct answer is 0.5 A.

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  • 8. 

    Which one of the following gives the symbol and abbreviated unit for electrical charge?

    • A.

      Symbol, Q ; unit, C

    • B.

      Symbol, C ; unit, V

    • C.

      Symbol, C ; unit, F

    Correct Answer
    A. Symbol, Q ; unit, C
    Explanation
    The symbol for electrical charge is Q, and the abbreviated unit for electrical charge is C, which stands for coulomb.

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  • 9. 

    The unknown current shown in the figure will be:

    • A.

      11 A flowing towards the junction

    • B.

      17 A flowing away from the junction

    • C.

      41 A flowing away from the junction

    Correct Answer
    A. 11 A flowing towards the junction
    Explanation
    Based on the given information, the unknown current shown in the figure will be 11 A flowing towards the junction.

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  • 10. 

    An aircraft supply has an RMS value of 115 V. Which one of the following gives the approximate peak value of the supply voltage?

    • A.

      67.5 V

    • B.

      115 V

    • C.

      163 V

    Correct Answer
    C. 163 V
    Explanation
    The RMS value of an AC voltage is equal to the peak value divided by the square root of 2. To find the approximate peak value of the supply voltage, we can multiply the RMS value by the square root of 2. Therefore, 115 V multiplied by the square root of 2 is approximately equal to 163 V, which is the correct answer.

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  • 11. 

    An AC waveform has a period of 4 ms. Which one of the following gives its frequency?

    • A.

      25 Hz

    • B.

      250 Hz

    • C.

      4 kHz

    Correct Answer
    B. 250 Hz
    Explanation
    The frequency of an AC waveform is the number of complete cycles it completes in one second. To find the frequency, we can use the formula: frequency = 1 / period. In this case, the period is given as 4 ms. To convert this to seconds, we divide by 1000, giving us a period of 0.004 seconds. Plugging this into the formula, we get a frequency of 1 / 0.004 = 250 Hz. Therefore, the correct answer is 250 Hz.

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  • 12. 

    Which one of the following gives the angle between the successive phases of a three-phase supply?

    • A.

      60 degrees

    • B.

      90 degrees

    • C.

      120 degrees

    Correct Answer
    C. 120 degrees
    Explanation
    The angle between the successive phases of a three-phase supply is 120 degrees. In a three-phase system, there are three separate phases that are 120 degrees apart from each other. This phase difference allows for a more balanced and efficient distribution of power. Each phase carries its own alternating current waveform, and the 120-degree phase shift ensures that the power is evenly distributed across the three phases, reducing the overall load on the system.

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  • 13. 

    The peak value of current supplied to an aircraft TRU is 28 A. Which one of the following gives the approximate value of RMS current supplied?

    • A.

      10 A

    • B.

      14 A

    • C.

      20 A

    Correct Answer
    C. 20 A
    Explanation
    The peak value of current is given as 28 A. To find the RMS current, we can use the formula RMS = peak / √2. By substituting the given peak value, we get RMS = 28 / √2 ≈ 19.8 A. Among the given options, the closest approximate value to 19.8 A is 20 A.

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  • 14. 

    A transformer has 2400 primary turns and 600 secondary turns. If the primary is supplied from a 220 VAC supply, which one of the following gives the resultant secondary voltage?

    • A.

      55 V

    • B.

      110 V

    • C.

      880 V

    Correct Answer
    A. 55 V
    Explanation
    The secondary voltage of a transformer can be calculated using the formula: secondary voltage = (primary voltage * secondary turns) / primary turns. In this case, the primary voltage is 220 VAC, the primary turns are 2400, and the secondary turns are 600. Plugging these values into the formula, we get: secondary voltage = (220 * 600) / 2400 = 55 V. Therefore, the correct answer is 55 V.

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  • 15. 

    Identify the components shown.

    • A.

      Generators

    • B.

      Inductors

    • C.

      Transformers

    Correct Answer
    C. Transformers
    Explanation
    The components shown in the question are transformers. Transformers are electrical devices that transfer electrical energy between two or more circuits through electromagnetic induction. They consist of two or more coils of wire, known as windings, which are wound around a common magnetic core. Transformers are commonly used in electrical power distribution systems to step up or step down voltage levels, allowing for efficient transmission of electricity over long distances.

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  • 16. 

    Decreasing the current in the field coil of a DC generator will:

    • A.

      Decrease the output voltage

    • B.

      Increase the output voltage

    • C.

      Increase the output frequency

    Correct Answer
    A. Decrease the output voltage
    Explanation
    Decreasing the current in the field coil of a DC generator will result in a decrease in the magnetic field strength. Since the output voltage of a DC generator is directly proportional to the magnetic field strength, a decrease in the current will cause a decrease in the output voltage. Therefore, the correct answer is that decreasing the current in the field coil of a DC generator will decrease the output voltage.

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  • 17. 

    The rotor of an induction motor consists of:

    • A.

      A laminated iron core inside a 'squirrel cage' made from copper or aluminum

    • B.

      A series of coil windings on a laminated iron core with connections via slip rings

    • C.

      A single copper loop which rotates inside the field created by a permanent magnet

    Correct Answer
    B. A series of coil windings on a laminated iron core with connections via slip rings
    Explanation
    The correct answer is a series of coil windings on a laminated iron core with connections via slip rings. This is because an induction motor uses a rotor that consists of coil windings on a laminated iron core. These coil windings are connected via slip rings, which allow for the flow of electrical current to the rotor. This design allows for the creation of a rotating magnetic field, which interacts with the stator's magnetic field to generate torque and make the motor run.

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  • 18. 

    The slip speed of an AC induction motor is the difference between:

    • A.

      The synchronous speed and the rotor speed

    • B.

      The frequency of the supply and the rotor speed

    • C.

      The maximum speed and the minimum speed

    Correct Answer
    A. The synchronous speed and the rotor speed
    Explanation
    The slip speed of an AC induction motor is the difference between the synchronous speed and the rotor speed. The synchronous speed is the speed at which the rotating magnetic field of the stator rotates, determined by the frequency of the supply. The rotor speed is the actual speed at which the rotor of the motor is rotating. The slip speed represents the relative speed between the rotating magnetic field and the rotor, which is necessary for the motor to generate torque.

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  • 19. 

    When compared with three-phase induction motors, a single phase induction motors:

    • A.

      Are not 'inherently' self-starting

    • B.

      Have more complicated stator windings

    • C.

      Are significantly more efficient

    Correct Answer
    A. Are not 'inherently' self-starting
    Explanation
    Single-phase induction motors are not inherently self-starting because they require an external means to initiate rotation, such as a starting capacitor or centrifugal switch. This is in contrast to three-phase induction motors, which can self-start due to the rotating magnetic field produced by the three-phase power supply. While single-phase induction motors may have more complicated stator windings compared to three-phase motors, they are generally less efficient due to the presence of additional components required for starting.

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  • 20. 

    A three-phase induction motor has three pairs of poles and is operated from a 60Hz. Which one of the following gives the motor's synchronous speed?

    • A.

      1200 rpm

    • B.

      1800 rpm

    • C.

      3600 rpm

    Correct Answer
    A. 1200 rpm
    Explanation
    The synchronous speed of a three-phase induction motor is given by the formula: Synchronous speed (in RPM) = (120 * Frequency) / Number of Poles. In this case, the frequency is 60Hz and the number of poles is 3 pairs, which means 6 poles. Plugging these values into the formula, we get (120 * 60) / 6 = 1200 RPM. Therefore, the correct answer is 1200 RPM.

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  • 21. 

    In a star-connected three-phase system, the line voltage is found to be 200V. Which one of the following gives the approximate value of phase voltage?

    • A.

      67 V

    • B.

      115 V

    • C.

      346 V

    Correct Answer
    B. 115 V
    Explanation
    In a star-connected three-phase system, the line voltage is equal to the square root of 3 times the phase voltage. Therefore, to find the approximate value of the phase voltage, we divide the line voltage (200V) by the square root of 3. This calculation gives us an approximate value of 115V, which matches the second option.

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  • 22. 

    A single-phase AC generator has twelve poles and it runs at 600 rpm. Which of the following gives the output frequency of the generator?

    • A.

      50 Hz

    • B.

      60 Hz

    • C.

      120 Hz

    Correct Answer
    B. 60 Hz
    Explanation
    The output frequency of an AC generator is determined by the number of poles and the rotational speed. In this case, the generator has twelve poles and runs at 600 rpm. The formula to calculate the frequency is given by f = (p * n) / 120, where f is the frequency, p is the number of poles, and n is the rotational speed in rpm. Plugging in the values, we get f = (12 * 600) / 120 = 60 Hz. Therefore, the output frequency of the generator is 60 Hz.

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  • 23. 

    The slip rings in an AC generator provide a means of:

    • A.

      Connecting an external circuit to a rotating armature winding

    • B.

      Supporting a rotating armature without the need for bearings

    • C.

      Periodically reversing the current produced by an armature winding

    Correct Answer
    A. Connecting an external circuit to a rotating armature winding
    Explanation
    The slip rings in an AC generator function by connecting an external circuit to a rotating armature winding. As the armature rotates, the slip rings maintain electrical contact with the stationary brushes, allowing the generated current to be transferred to an external circuit. This enables the generator to provide a continuous flow of electricity to power devices connected to it.

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  • 24. 

    The commutator in a DC generator is used to:

    • A.

      Provide a means of connecting an external field current supply

    • B.

      Periodically reverse the connections to the rotating coil winding

    • C.

      Disconnect the coil winding when the induced current reaches a maximum value

    Correct Answer
    B. Periodically reverse the connections to the rotating coil winding
    Explanation
    The commutator in a DC generator is used to periodically reverse the connections to the rotating coil winding. This is necessary because the coil winding needs to continuously change the direction of the current flow in order to maintain a steady flow of electricity in one direction. By reversing the connections, the commutator ensures that the current flow in the coil changes direction every half rotation, allowing for the generation of a direct current.

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  • 25. 

    The brushes fitted to a DC motor/generator should have:

    • A.

      Low coefficient of friction and low contact resistance

    • B.

      High coefficient of friction and low contact resistance

    • C.

      Low coefficient of friction and high contact resistance

    Correct Answer
    A. Low coefficient of friction and low contact resistance
    Explanation
    The brushes fitted to a DC motor/generator should have a low coefficient of friction to minimize wear and frictional losses. This allows for smooth rotation and efficient operation of the motor/generator. Additionally, they should have a low contact resistance to ensure good electrical conductivity between the brushes and the commutator/collector, which is necessary for the transfer of electrical current.

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  • 26. 

    In a shunt-wound generator:

    • A.

      None of the armature current flows through the field

    • B.

      Some of the armature current flows through the field

    • C.

      All of the armature current flows through the field

    Correct Answer
    B. Some of the armature current flows through the field
    Explanation
    In a shunt-wound generator, some of the armature current flows through the field. This is because the field winding is connected in parallel with the armature winding. As a result, a portion of the armature current splits and passes through the field winding, creating a magnetic field. This magnetic field interacts with the armature's magnetic field, inducing voltage and generating power. However, not all of the armature current flows through the field, as some of it is used to power the load connected to the generator.

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  • 27. 

    When combined with a CSD, a brushless three-phase AC generator is often referred to as:

    • A.

      A compound generator

    • B.

      A 'frequency wild' generator

    • C.

      An IDG

    Correct Answer
    C. An IDG
    Explanation
    When combined with a CSD (Constant Speed Drive), a brushless three-phase AC generator is often referred to as an IDG (Integrated Drive Generator). The IDG is a type of generator commonly used in aircraft systems. It combines the functions of a generator and a constant speed drive into a single unit, providing a reliable and efficient power source. The IDG is designed to maintain a constant frequency output, making it suitable for powering various electrical systems in an aircraft.

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  • 28. 

    An out-of-balance condition in an AC three-phase system can be detected by means of:

    • A.

      Voltage sensors connected across each output line

    • B.

      A dedicated field coil monitoring circuit

    • C.

      A current transformer connected in the neutral line

    Correct Answer
    C. A current transformer connected in the neutral line
    Explanation
    A current transformer connected in the neutral line can detect an out-of-balance condition in an AC three-phase system. This is because the neutral line carries the return current, and any imbalance in the system will result in an unequal distribution of current among the phases. By using a current transformer in the neutral line, the current flowing through it can be measured and compared to the expected balanced values. If there is a significant difference, it indicates an out-of-balance condition in the system.

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  • 29. 

    Self-excited generators derive their field current from:

    • A.

      The current produced by armature

    • B.

      A separate field current supply

    • C.

      An external power source

    Correct Answer
    A. The current produced by armature
    Explanation
    Self-excited generators derive their field current from the current produced by the armature. In a self-excited generator, the armature current is used to create a magnetic field in the stator. This magnetic field then induces a voltage in the stator windings, which in turn generates a current in the field winding. This self-sustaining process allows the generator to continue generating power without the need for an external power source or a separate field current supply.

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  • 30. 

    In a simple cells, electrons are:

    • A.

      Removed from the positive (anode) and deposited on the negative (cathode)

    • B.

      Removed from the positive (cathode) and deposited on the negative (anode)

    • C.

      Removed from the negative (cathode) and deposited on the positive (anode)

    Correct Answer
    C. Removed from the negative (cathode) and deposited on the positive (anode)
    Explanation
    In a simple cell or battery, electrons are removed from the negative (cathode) and deposited on the positive (anode). Electrons flow from the negative terminal (cathode) to the positive terminal (anode) through an external circuit, creating an electric current.

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  • 31. 

    The energy storage capacity of a cell is determined by the:

    • A.

      Terminal voltage

    • B.

      Electrolyte specific gravity

    • C.

      Amount of material available for chemical reaction

    Correct Answer
    C. Amount of material available for chemical reaction
    Explanation
    The energy storage capacity of a cell is determined by the amount of material available for chemical reaction. This is because the chemical reactions that occur within the cell are responsible for storing and releasing energy. The more material available for these reactions, the greater the energy storage capacity of the cell. Terminal voltage and electrolyte specific gravity may impact the efficiency or performance of the cell, but they do not directly determine its energy storage capacity.

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  • 32. 

    When mixing electrolyte:

    • A.

      Acid is always added to the water

    • B.

      Water is always added to the acid

    • C.

      It is not important how water and acid are mixed

    Correct Answer
    A. Acid is always added to the water
    Explanation
    The correct answer is that acid is always added to the water. This is because adding water to acid can cause a violent reaction, resulting in splattering and potential release of heat. By adding acid to water slowly and carefully, the reaction is more controlled and the risk of splattering and heat release is minimized. This is a safety precaution to prevent accidents and ensure the proper mixing of electrolyte.

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  • 33. 

    Servicing equipment used for lead-acid batteries:

    • A.

      Can also be used for nickel-cadmium batteries

    • B.

      Must not be used for nickel cadmium batteries

    • C.

      Must be disposed of after use

    Correct Answer
    B. Must not be used for nickel cadmium batteries
    Explanation
    The correct answer is "must not be used for nickel cadmium batteries". This is because lead-acid batteries and nickel-cadmium batteries have different characteristics and require different servicing equipment. Using the wrong equipment for nickel-cadmium batteries can lead to damage or malfunction. Therefore, it is important to use the appropriate equipment for each type of battery to ensure proper maintenance and avoid any potential issues.

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  • 34. 

    Lead acid batteries are recharged by constant:

    • A.

      Voltage

    • B.

      Current

    • C.

      Ampere-hours

    Correct Answer
    A. Voltage
    Explanation
    Lead acid batteries are recharged by constant voltage. This means that during the recharging process, a constant voltage is applied to the battery terminals. This voltage helps to reverse the chemical reactions that occur during discharge, allowing the battery to regain its stored energy. By maintaining a constant voltage, the charging process can be controlled and optimized to ensure the battery is charged efficiently and safely.

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  • 35. 

    The only accurate and practical way to determine the condition of a nickel-cadmium battery is with a:

    • A.

      Specific gravity check of electrolyte

    • B.

      Measured discharge in the workshop

    • C.

      Check of the terminal voltage

    Correct Answer
    A. Specific gravity check of electrolyte
    Explanation
    The specific gravity check of electrolyte is the only accurate and practical way to determine the condition of a nickel-cadmium battery. This is because the specific gravity of the electrolyte directly correlates with the state of charge and health of the battery. By measuring the specific gravity, one can assess if the battery is fully charged, partially charged, or discharged. This method provides a reliable indication of the battery's condition and is commonly used in the industry.

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  • 36. 

    A cell that can only be charged once is a:

    • A.

      Metal-hydride cell

    • B.

      Secondary cell

    • C.

      Primary cell

    Correct Answer
    C. Primary cell
    Explanation
    A primary cell is a type of cell that can only be charged once. Once its chemical reaction is complete and the reactants are depleted, it cannot be recharged or reused. This is in contrast to secondary cells, which can be recharged multiple times, and metal-hydride cells, which are a type of secondary cell that use a metal-hydride compound as the active material. Therefore, the correct answer is primary cell.

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  • 37. 

    The only time that water is added to the electrolyte of a Ni-Cad battery is:

    • A.

      When fully charged, and the volume of electrolyte is high

    • B.

      When fully discharged, and the volume of electrolyte is high

    • C.

      When fully charged, and the volume of electrolyte is low

    Correct Answer
    C. When fully charged, and the volume of electrolyte is low
    Explanation
    When a Ni-Cad battery is fully charged, the volume of electrolyte is low because the water in the electrolyte has been converted into hydrogen and oxygen gases during the charging process. Adding water to the electrolyte at this point helps to replenish the lost water and maintain the proper electrolyte level. Therefore, the correct answer is when fully charged, and the volume of electrolyte is low.

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  • 38. 

    Referring to the figure, the reason for having an even number of positive plates in a lead-acid battery is because:

    • A.

      Positive plates distort when chemical reactions takes place in both sides

    • B.

      Positive plates distort when chemical reactions takes place on one side

    • C.

      Negative plates distort when chemical reactions takes place in both sides

    Correct Answer
    A. Positive plates distort when chemical reactions takes place in both sides
    Explanation
    In a lead-acid battery, the chemical reactions that occur during the charging and discharging process cause the positive plates to distort. Having an even number of positive plates helps to distribute this distortion evenly throughout the battery, preventing any one plate from becoming excessively distorted. This helps to maintain the structural integrity of the battery and prolong its lifespan.

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  • 39. 

    Referring to the figure, a 20 Ah battery, when discharging at 2 A will be fully discharged at approximately:

    • A.

      2 hours

    • B.

      15 minutes

    • C.

      10 hours

    Correct Answer
    C. 10 hours
    Explanation
    Based on the information given, a 20 Ah battery discharging at a rate of 2 A will take approximately 10 hours to fully discharge. This can be calculated by dividing the battery's capacity (20 Ah) by the discharge rate (2 A), which gives us 10 hours.

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  • 40. 

    Battery capacity is measured in:

    • A.

      Volts

    • B.

      Amperes

    • C.

      Ampere-hours

    Correct Answer
    C. Ampere-hours
    Explanation
    Battery capacity is measured in ampere-hours. Ampere-hours (Ah) is a unit that represents the amount of charge a battery can deliver over a specific period of time. It is a product of the current (in amperes) that a battery can supply and the time (in hours) it can sustain that current. This unit is commonly used to determine the energy storage capacity of batteries, as it takes into account both the current and the duration for which the battery can provide that current.

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  • 41. 

    Two parallel conductors are separated by a distance of 25 mm. Determine the electric field strength if they are fed from a 600V DC supply.

    • A.

      24 kV/m

    • B.

      25 kV/m

    • C.

      48 kV/m

    Correct Answer
    A. 24 kV/m
    Explanation
    The electric field strength between two parallel conductors can be determined using the formula E = V/d, where E is the electric field strength, V is the voltage, and d is the distance between the conductors. In this case, the voltage is given as 600V and the distance is given as 25 mm (which is equivalent to 0.025 m). Plugging these values into the formula, we get E = 600V / 0.025 m = 24,000 V/m = 24 kV/m.

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  • 42. 

    A 68 microFarad is required to store a charge of 160 microCoulumb. What voltage should be applied to the capacitor?

    • A.

      228V

    • B.

      10880V

    • C.

      2.35V

    Correct Answer
    C. 2.35V
    Explanation
    The voltage applied to a capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is given as 160 microCoulomb and the capacitance is given as 68 microFarad. Plugging these values into the formula, we get V = 160 microCoulomb / 68 microFarad = 2.35V. Therefore, the correct answer is 2.35V.

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  • 43. 

    A charge of 1.5 C is transferred to a capacitor in 30 seconds. What current is flowing in the capacitor?

    • A.

      25 mA

    • B.

      50 mA

    • C.

      100 mA

    Correct Answer
    B. 50 mA
    Explanation
    When a charge of 1.5 C is transferred to a capacitor in 30 seconds, we can use the formula I = Q/t to calculate the current flowing in the capacitor. Here, Q represents the charge transferred and t represents the time taken. Plugging in the values, we get I = 1.5 C / 30 s = 0.05 A = 50 mA. Therefore, the correct answer is 50 mA.

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  • 44. 

    A 28V DC aircraft supply delivers a charge of 5 C to a window heater every second. What is the resistance of the heater?

    • A.

      6 ohms

    • B.

      5.6 ohms

    • C.

      6.2 ohms

    Correct Answer
    B. 5.6 ohms
    Explanation
    The resistance of the heater can be calculated using Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is 28V and the charge delivered every second is 5C. Since charge (Q) is equal to current multiplied by time (Q = I * t), we can calculate the current as 5C/1s = 5A. Therefore, the resistance is 28V/5A = 5.6 ohms.

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  • 45. 

    An auxiliary power unit (APU) provides an output of 1.5 kW for 20 minutes. How much energy has it supplied to the aircraft?

    • A.

      1.8 MJ

    • B.

      2 MJ

    • C.

      2.4 MJ

    Correct Answer
    A. 1.8 MJ
    Explanation
    The energy supplied by the auxiliary power unit (APU) can be calculated by multiplying the power output (1.5 kW) by the duration of operation (20 minutes). Converting the power output to megajoules (MJ) by multiplying by the conversion factor of 0.001, the energy supplied to the aircraft is 1.5 kW * 20 minutes * 0.001 = 0.03 MJ. Therefore, the correct answer is 1.8 MJ.

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  • 46. 

    The reservoir capacitor in a power supply is required to store 20 J of energy. How much power is required to store this energy in a time interval of 0.5 sec.?

    • A.

      10 W

    • B.

      20 W

    • C.

      40 W

    Correct Answer
    C. 40 W
    Explanation
    The power required to store energy is given by the equation P = E/t, where P is power, E is energy, and t is time. In this case, the energy to be stored is 20 J and the time interval is 0.5 sec. Plugging these values into the equation, we get P = 20 J / 0.5 sec = 40 W. Therefore, the correct answer is 40 W.

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  • 47. 

    The main aircraft battery is used to start an engine. If the starter demands a current 1000A for 30 sec. and the battery voltage remains at 12V during that period, determine the amount of energy required to start an engine.

    • A.

      100 kJ

    • B.

      360 kJ

    • C.

      420 kJ

    Correct Answer
    B. 360 kJ
    Explanation
    To determine the amount of energy required to start an engine, we can use the formula: Energy = Power x Time. The power can be calculated using the formula: Power = Voltage x Current. Given that the current demanded by the starter is 1000A and the battery voltage remains at 12V for 30 seconds, we can calculate the power as 12V x 1000A = 12000W. Multiplying this power by the time of 30 seconds, we get 12000W x 30s = 360000J, which is equal to 360 kJ. Therefore, the amount of energy required to start the engine is 360 kJ.

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  • 48. 

    An aircraft generator operates at a frequency of 400 Hz. What is the periodic time of the voltage generated?

    • A.

      400 ms

    • B.

      2.5 ms

    • C.

      1 ms

    Correct Answer
    B. 2.5 ms
    Explanation
    The periodic time of a wave is the time it takes for one complete cycle to occur. In this case, the aircraft generator operates at a frequency of 400 Hz, which means it completes 400 cycles per second. To find the periodic time, we can divide 1 second by the frequency. Therefore, the periodic time is 1/400 seconds, which is equal to 2.5 milliseconds (ms).

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  • 49. 

    A sinusoidal current of 40 A peak-to-peak flows in a circuit. What is the RMS value of the current?

    • A.

      14.12 A

    • B.

      13 A

    • C.

      10.5 A

    Correct Answer
    A. 14.12 A
    Explanation
    The RMS value of a sinusoidal current is equal to the peak-to-peak value divided by 2√2. In this case, the peak-to-peak value is 40 A, so the RMS value can be calculated as 40 A / (2√2) = 14.12 A.

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  • 50. 

    A coil is connected to a 50 VAC supply at 400 Hz. If the current supplied to the coil is 200 mA and the coil has a resistance of 6 ohms. What is the value of inductance?

    • A.

      79 mH

    • B.

      95 mH

    • C.

      100 mH

    Correct Answer
    C. 100 mH
    Explanation
    The value of inductance can be calculated using the formula L = (V / I) / (2 * π * f), where V is the voltage, I is the current, and f is the frequency. Plugging in the given values, we get L = (50 V / 0.2 A) / (2 * π * 400 Hz) = 100 mH. Therefore, the value of inductance is 100 mH.

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