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Question 1. Which law states that the force between two static electric charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them? A) Coulomb’s Law B) Ohm’s Law C) Faraday’s Law D) Ampère’s Law Answer: A Explanation: Coulomb’s Law describes the electrostatic force between point charges, following the inverse‑square relationship. Question 2. In a conductor at electrostatic equilibrium, the electric field inside the material is: A) Zero B) Uniform and non‑zero C) Varying linearly with distance D) Equal to the external field Answer: A Explanation: Charges redistribute on the surface of a conductor until the internal electric field cancels, leaving zero field inside. Question 3. The unit “ampere” measures: A) Electric potential difference B) Electric current C) Electrical resistance D) Electric power Answer: B
Explanation: An ampere is the SI unit for electric current, defined as one coulomb of charge passing a point per second. Question 4. Ohm’s Law is expressed as V = I R. If a resistor has a voltage of 12 V and a current of 3 A, what is its resistance? A) 4 Ω B) 36 Ω C) 0.25 Ω D) 15 Ω Answer: A Explanation: R = V / I = 12 V / 3 A = 4 Ω. Question 5. In a series circuit containing three resistors of 2 Ω, 4 Ω, and 6 Ω, the total resistance is: A) 12 Ω B) 8 Ω C) 10 Ω D) 6 Ω Answer: A Explanation: Series resistances add directly: 2 + 4 + 6 = 12 Ω. Question 6. In a parallel circuit with two branches, one containing a 10 Ω resistor and the other a 20 Ω resistor, the equivalent resistance is: A) 30 Ω B) 6.67 Ω C) 15 Ω
D) Zero unless the wire is moving. Answer: C Explanation: According to the right‑hand rule, a steady current generates concentric circular magnetic field lines around the conductor. Question 10. A solenoid with 500 turns per meter carrying 2 A produces a magnetic field of approximately: A) 0.001 T B) 0.001 T × π C) 0.001 T × 2π D) 0.001 T × 500 Answer: C Explanation: B = μ₀ n I = (4π × 10⁻⁷ T·m/A) × 500 × 2 ≈ 1.26 × 10⁻³ T (≈0.001 T × 2π). Question 11. In an electromagnet, increasing the number of turns on the coil while keeping current constant will: A) Decrease the magnetic field strength. B) Increase the magnetic field strength. C) Not affect the magnetic field. D) Reverse the polarity. Answer: B Explanation: Magnetic field B ∝ n I; more turns (higher n) increase B. Question 12. The principle of electromagnetic induction states that a changing magnetic flux through a circuit induces: A) A constant voltage.
B) An electric current proportional to the flux. C) An electromotive force proportional to the rate of change of flux. D) No effect unless the circuit is moving. Answer: C Explanation: Faraday’s law: ε = – dΦ/dt; the induced emf is proportional to the time rate of change of magnetic flux. Question 13. Self‑induction in a coil is primarily caused by: A) Mutual coupling with another coil. B) The coil’s own changing magnetic field. C) External electric fields. D) Static charge accumulation. Answer: B Explanation: A coil’s changing current creates a changing magnetic field that links with the same coil, inducing an emf opposing the change (self‑induction). Question 14. In a step‑up transformer, the ratio of secondary to primary voltage equals the ratio of: A) Primary turns to secondary turns. B) Secondary turns to primary turns. C) Primary resistance to secondary resistance. D) Primary current to secondary current. Answer: B Explanation: Vₛ/Vₚ = Nₛ/Nₚ; a higher turn count on the secondary yields a higher voltage. Question 15. A transformer with a turns ratio of 1:5 (primary:secondary) will:
Question 18. Eddy currents are minimized in transformer cores by: A) Using solid iron plates. B) Laminating the core material. C) Increasing the core thickness. D) Decreasing the number of turns. Answer: B Explanation: Laminations break up circulating paths for induced currents, reducing eddy‑current losses. Question 19. In the main X‑ray circuit, the line compensator serves to: A) Increase the kVp. B) Adjust the incoming line voltage to a constant 220 V. C) Convert AC to DC. D) Provide filament heating. Answer: B Explanation: The compensator stabilizes the mains voltage, ensuring the transformer receives a steady 220 V input. Question 20. The autotransformer in an X‑ray unit is primarily used as: A) The high‑voltage step‑up transformer. B) The kVp selector, providing a tapped winding for voltage selection. C) The filament power supply. D) The AEC control module. Answer: B Explanation: The autotransformer provides a variable tap that sets the kilovoltage peak (kVp) before the main step‑up transformer.
Question 21. The kVp meter measures voltage before it is stepped up because: A) It needs to display the final output voltage. B) It measures the voltage across the autotransformer tap, which determines the final kVp. C) It monitors the filament current. D) It calibrates the exposure timer. Answer: B Explanation: The kVp meter reads the voltage at the autotransformer output (pre‑step‑up), which directly correlates to the selected kVp. Question 22. Which type of exposure timer is most accurate for modern X‑ray units? A) Mechanical timer B) Synchronous timer C) Electronic timer D) Falling‑load timer Answer: C Explanation: Electronic timers use digital circuitry to control exposure time with high precision and repeatability. Question 23. An mAs timer is typically used with: A) Constant‑potential generators. B) Falling‑load generators. C) High‑frequency generators. D) Portable X‑ray units. Answer: B
C) 30 pulses/s D) 240 pulses/s Answer: B Explanation: Full‑wave rectification produces two pulses per AC cycle; 60 Hz × 2 = 120 pulses/s. Question 27. The MA (milliamperes) meter on the secondary side of the X‑ray generator is grounded at the center of the transformer because: A) It measures filament current. B) It provides a reference point for accurate current measurement and safety. C) It controls the kVp. D) It supplies power to the exposure timer. Answer: B Explanation: Grounding the meter at the transformer’s center tap creates a stable reference, improving measurement accuracy and protecting the operator. Question 28. In the filament circuit, the rheostat (or variable resistor) is used to: A) Select the focal spot size. B) Adjust filament temperature, thereby controlling filament current (mA). C) Rectify the high‑voltage output. D) Modulate the kVp. Answer: B Explanation: Changing the resistance alters the filament heating current, which sets the filament temperature and electron emission rate. Question 29. The step‑down transformer in the filament circuit serves to:
A) Increase filament voltage. B) Decrease filament voltage while increasing current to provide sufficient heating power. C) Convert AC to DC. D) Provide high‑frequency excitation. Answer: B Explanation: Lower voltage with higher current is ideal for heating the filament (thermionic emission). Question 30. Switching between small and large focal spots is achieved by: A) Changing the kVp. B) Changing the filament current only. C) Selecting different filament‑anode connections that alter the effective focal spot size. D) Adjusting the exposure time. Answer: C Explanation: The circuit routes current to either the small or large filament, changing the size of the electron source and thus the focal spot. Question 31. Thermionic emission from the tungsten filament occurs because: A) Photons strike the filament. B) The filament is heated to a temperature where electrons have enough kinetic energy to overcome the work function. C) Magnetic fields pull electrons out. D) The filament is under high voltage. Answer: B Explanation: Heating raises electron energy; when it exceeds the material’s work function, electrons are emitted.
Explanation: (100 kVp × 200 mA) / 1000 = 20 kW. Question 35. In X‑ray tube construction, the anode’s rotating design primarily serves to: A) Increase tube voltage. B) Distribute heat over a larger area, reducing focal spot damage. C) Decrease electron velocity. D) Produce higher kVp. Answer: B Explanation: Rotation spreads the heat generated by electron impact, allowing higher power without overheating the focal spot. Question 36. Which material is most commonly used for the X‑ray tube target (anode) due to its high atomic number and melting point? A) Aluminum B) Tungsten C) Copper D) Lead Answer: B Explanation: Tungsten (Z = 74) provides efficient bremsstrahlung production and can withstand high temperatures. Question 37. The focusing cup in the cathode assembly: A) Generates the high‑voltage potential. B) Shapes the electron beam toward the focal spot. C) Acts as a filament.
D) Provides cooling for the anode. Answer: B Explanation: The cup’s negative potential draws and narrows the electron stream, improving focal spot definition. Question 38. Bremsstrahlung radiation is produced when: A) Electrons are captured by inner‑shell electrons. B) Electrons are decelerated in the Coulomb field of the target nucleus. C) Photons are emitted from the filament. D) The anode rotates. Answer: B Explanation: “Braking radiation” occurs as high‑energy electrons lose kinetic energy interacting with the nucleus’s electric field, yielding a continuous spectrum. Question 39. Characteristic X‑ray peaks arise from: A) Electron‑nucleus collisions. B) Transitions of electrons filling K‑shell vacancies. C) Thermal radiation from the filament. D) Scattering of external photons. Answer: B Explanation: When a K‑shell electron is ejected, electrons from higher shells drop down, emitting photons with discrete energies characteristic of the target element. Question 40. Increasing the kVp of an X‑ray tube will: A) Decrease the maximum photon energy.
Question 43. In a constant‑potential (CP) X‑ray generator, the voltage across the tube remains essentially: A) Constant throughout the exposure. B) Variable, following the mains waveform. C) Decreasing linearly. D) Zero. Answer: A Explanation: CP generators use a high‑frequency inverter and feedback control to maintain a steady voltage, minimizing voltage ripple. Question 44. The primary function of the induction motor in a rotating‑anode X‑ray tube is to: A) Generate high voltage. B) Rotate the anode at high speed for heat distribution. C) Control filament temperature. D) Provide the exposure timer. Answer: B Explanation: The motor drives the anode’s rotation, spreading the heat over a larger surface area. Question 45. Hysteresis loss in a transformer core is proportional to: A) The square of the frequency. B) The frequency and the area of the hysteresis loop. C) Only the core material’s resistivity. D) The number of turns only. Answer: B Explanation: Hysteresis loss = k f B_max^n, where f is frequency and the loop area reflects magnetic properties.
Question 46. The term “self‑inductance” (L) of a coil is measured in: A) Ohms B) Farads C) Henrys D) Webers Answer: C Explanation: Inductance is expressed in henries (H). Question 47. The back‑EMF induced in a coil opposing the change in current is described by: A) ε = L dI/dt B) ε = R I C) ε = V I D) ε = C dV/dt Answer: A Explanation: Faraday’s law for a self‑inducing coil gives ε = – L dI/dt; the negative sign indicates opposition to change. Question 48. In a high‑frequency X‑ray generator, the voltage ripple is typically less than: A) 100 % B) 13 % C) 3 % D) 1 % Answer: D
Answer: C Explanation: kV determines photon energy, mA·s (mAs) determines quantity; together they define the total exposure. Question 52. The term “half‑wave rectifier” refers to a circuit that: A) Converts both halves of an AC cycle to DC. B) Uses two diodes to produce full‑wave output. C) Allows only one half of the AC waveform to pass, producing pulsating DC. D) Inverts the polarity of the AC input. Answer: C Explanation: A single diode conducts during one polarity, blocking the opposite half, resulting in half‑wave rectification. Question 53. In a constant‑potential generator, the high‑frequency inverter operates at approximately: A) 50 Hz B) 60 Hz C) 10 kHz to 30 kHz D) 1 MHz Answer: C Explanation: CP generators typically use a high‑frequency oscillator in the 10– 30 kHz range to achieve low ripple. Question 54. The “falling‑load” characteristic of an X‑ray generator means that: A) Voltage rises as the load decreases. B) Current decreases linearly with time during exposure.
C) The output voltage drops as the filament heats up. D) The generator supplies a constant mAs regardless of kVp. Answer: B Explanation: In a falling‑load system, the current (or mA) declines during the exposure due to the decreasing load on the high‑voltage circuit. Question 55. Which component in the X‑ray circuit directly protects against over‑current conditions? A) Line compensator B) Main circuit breaker C) Autotransformer D) AEC chamber Answer: B Explanation: The circuit breaker trips when current exceeds a preset limit, preventing overload. Question 56. The purpose of the “grounded center tap” in the secondary of the high‑voltage transformer is to: A) Provide a reference for the MA meter and improve safety. B) Increase the output voltage. C) Reduce filament heating. D) Control the exposure timer. Answer: A Explanation: Grounding the center tap creates a stable reference point, allowing accurate current measurement and safety isolation. Question 57. In an X‑ray tube, the “focal spot size” influences:
