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Question 1. What is the SI unit of electric charge? A) Volt B) Ampere C) Coulomb D) Ohm Answer: C Explanation: The coulomb (C) is defined as the amount of charge transferred by a current of one ampere in one second. Question 2. Which particle carries a negative electric charge? A) Proton B) Neutron C) Electron D) Positron Answer: C Explanation: Electrons have a charge of – 1 e, whereas protons are positively charged and neutrons are neutral. Question 3. If a glass rod is rubbed with silk, the glass becomes: A) Positively charged B) Negatively charged C) Neutral D) Magnetised Answer: A Explanation: Electrons are transferred from the glass to the silk, leaving the glass with a deficit of electrons (positive charge). Question 4. In an electric field, the direction of the field lines is:
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A) From negative to positive B) From positive to negative C) From high potential to low potential D) Random Answer: B Explanation: Electric field lines are defined to point away from positive charges and toward negative charges. Question 5. A gold‑leaf electroscope shows the leaves diverging. This indicates that the electroscope is: A) Uncharged B) Negatively charged C) Positively charged D) Either B or C Answer: D Explanation: Diverging leaves mean the electroscope carries a net charge, which could be either positive or negative. Question 6. Current is defined as: A) Charge per unit time B) Voltage per unit resistance C) Energy per unit charge D) Power per unit time Answer: A Explanation: Electric current I = Q / t, the rate at which charge flows. Question 7. The conventional direction of current flows: A) From negative to positive
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D) R = V / I²
Answer: A Explanation: The relationship between voltage, current and resistance for an ohmic conductor is V = I R. Question 11. If the resistance of a wire is doubled while the voltage across it remains constant, the current will: A) Double B) Halve C) Remain unchanged D) Quadruple Answer: B Explanation: I = V / R, so doubling R halves I. Question 12. Which factor does NOT affect the resistance of a uniform metallic wire? A) Length B) Cross‑sectional area C) Temperature D) Color of the insulation Answer: D Explanation: Resistance depends on length, area, material and temperature, not the colour of the covering. Question 13. A filament lamp shows a non‑linear V‑I characteristic because: A) Its resistance is constant B) Its temperature changes with current C) It is a perfect conductor D) It is powered by AC only Answer: B
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Explanation: As current increases, the filament heats up, raising its resistance, giving a curved V‑I graph. Question 14. Which symbol represents a variable resistor (rheostat) in a circuit diagram? A) A zig‑zag line with an arrow across it B) A straight line with a coil C) A circle with a diagonal line D) Two parallel lines with a gap Answer: A Explanation: The symbol for a variable resistor is a resistor symbol with an arrow indicating adjustability. Question 15. A light‑dependent resistor (LDR) has its resistance decrease when: A) Temperature rises B) Light intensity increases C) Voltage increases D) It is cooled Answer: B Explanation: LDRs are made of photoconductive material; more light creates more charge carriers, lowering resistance. Question 16. A thermistor with a negative temperature coefficient will: A) Increase resistance as temperature rises B) Decrease resistance as temperature rises C) Remain constant regardless of temperature D) Become a perfect conductor at high temperature Answer: B Explanation: NTC thermistors have resistance that falls with increasing temperature. Question 17. In a series circuit, the total resistance is:
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D) To make the circuit easier to switch off Answer: B Explanation: In parallel each lamp gets the full mains voltage, so they operate at normal brightness even if other lamps are switched off. Question 21. The electrical power dissipated in a resistor can be expressed as: A) P = V / I B) P = I² R C) P = V + I R D) P = V × I² Answer: B Explanation: Using P = VI and Ohm’s law (V = IR) gives P = I²R. Question 22. A 60 W light bulb is connected to a 240 V supply. The current drawn is: A) 0.25 A B) 2.5 A C) 4 A D) 14.4 A Answer: B Explanation: I = P / V = 60 W / 240 V = 0.25 A (Oops correct calculation: 60/240 = 0.25 A). Wait answer B is 2.5 A, not correct. The correct answer should be A. Let's correct. Answer: A Explanation: I = P / V = 60 W / 240 V = 0.25 A. Question 23. If a device uses 1500 J of energy in 30 s, its power is: A) 5 W B) 30 W C) 45 W
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D) 1500 W
Answer: B Explanation: P = E / t = 1500 J / 30 s = 50 W (Oops). Actually 1500/30 = 50 W, not 30 W. Need correct answer. Answer: C (45 W) is also wrong. Let's recalc: 1500/30 = 50 W, not among options. Need a correct option. Let's modify options: A) 5 W B) 30 W C) 50 W D) 1500 W. Then answer C. Answer: C Explanation: Power is energy per unit time, so 1500 J / 30 s = 50 W. Question 24. A household appliance rated at 2 kW runs for 3 h. The energy consumed is: A) 6 kWh B) 0.667 kWh C) 600 Wh D) 3 kWh Answer: A Explanation: Energy (kWh) = Power (kW) × time (h) = 2 kW × 3 h = 6 kWh. Question 25. If electricity costs $0.20 per kWh, how much will it cost to run the 2 kW appliance for 3 h? A) $0. B) $1. C) $3. D) $6. Answer: D Explanation: 6 kWh × $0.20/kWh = $1.20 (Oops). Actually 6 kWh × 0.20 = $1.20, not $6.00. Let's correct options: A) $0.60 B) $1.20 C) $2.40 D) $12.00. Answer B. Answer: B Explanation: 6 kWh × $0.20 = $1.20.
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C) Direction of force on a current‑carrying conductor in a magnetic field D) Direction of electron flow in a circuit Answer: C Explanation: The left‑hand rule gives the relation between field (first finger), current (second finger), and force (thumb). Question 30. In a DC motor, the split‑ring commutator is required to: A) Increase the magnetic field strength B) Reverse the direction of current in the armature each half‑turn C) Provide a constant voltage source D) Reduce friction in the bearings Answer: B Explanation: The commutator reverses the current direction in the coil so that the torque remains in the same rotational direction. Question 31. When a straight conductor moves perpendicular to a magnetic field, the induced emf is given by: A) ε = B l v B) ε = B l / v C) ε = B v / l D) ε = B l² v Answer: A Explanation: Faraday’s law for a moving conductor gives ε = B l v (magnetic flux change per unit time). Question 32. If the magnetic flux through a coil is doubled in 0.5 s, the average induced emf is: A) 0.5 B t B) 2 B / 0. C) ΔΦ / Δt = (2Φ – Φ)/0.5 s = Φ / 0.5 s
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D) Not enough information Answer: C Explanation: ε = – ΔΦ / Δt; doubling the flux means ΔΦ = Φ, so ε = Φ / 0.5 s. Question 33. An AC generator uses slip rings instead of a commutator because: A) Slip rings allow the output voltage to change direction each half‑turn B) Slip rings reduce the resistance of the coil C) Slip rings increase the magnetic field strength D) Slip rings are cheaper to manufacture Answer: A Explanation: Slip rings provide a continuous electrical connection while the coil rotates, producing alternating polarity. Question 34. In a step‑up transformer, if the primary has 100 turns and the secondary 400 turns, the secondary voltage is: A) ¼ of the primary voltage B) Equal to the primary voltage C) Four times the primary voltage D) Zero Answer: C Explanation: V_s / V_p = N_s / N_p = 400 / 100 = 4, so the secondary voltage is four times higher. Question 35. The efficiency of a transformer is 95 %. If the input power is 200 W, the output power is: A) 190 W B) 210 W C) 195 W D) 180 W
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Question 39. Earthing a metal appliance protects a user by: A) Raising the appliance’s temperature B) Providing a low‑resistance path for fault current to the ground C) Increasing the appliance’s resistance D) Storing excess electrical energy Answer: B Explanation: If the live wire contacts the metal case, the earth wire carries the fault current safely to ground, reducing shock risk. Question 40. In a standard UK three‑pin plug, the brown wire is: A) Neutral B) Live (phase) C) Earth D) Protective conductor Answer: B Explanation: Colour coding: brown = live, blue = neutral, green/yellow = earth. Question 41. The frequency of the UK mains supply is: A) 50 Hz B) 60 Hz C) 100 Hz D) 120 Hz Answer: A Explanation: The United Kingdom uses 50 Hz alternating current for household supply. Question 42. The peak voltage of a 230 V RMS mains supply is: A) 230 V B) 325 V
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C) 460 V
D) 115 V
Answer: B Explanation: V_peak = √2 × V_RMS ≈ 1.414 × 230 V ≈ 325 V. Question 43. If a resistor of 10 Ω has a potential difference of 5 V across it, the current through it is: A) 0.2 A B) 0.5 A C) 2 A D) 5 A Answer: B Explanation: I = V / R = 5 V / 10 Ω = 0.5 A. Question 44. A capacitor in a DC circuit initially draws a large current because: A) It behaves like an open circuit at first B) It behaves like a short circuit initially, allowing charge to accumulate C) It stores magnetic energy D) It changes the polarity of the source Answer: B Explanation: When uncharged, a capacitor initially offers little resistance, acting like a short, then the current falls as voltage builds. Question 45. In a parallel circuit with three branches carrying 2 A, 3 A and 5 A respectively, the total current supplied by the source is: A) 10 A B) 5 A C) 2 A D) 3 A
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A) Transformer B) Electric heater C) DC motor D) Photovoltaic cell Answer: C Explanation: A motor uses electromagnetic forces to produce rotation, turning electrical energy into mechanical work. Question 49. When a current‑carrying coil is placed inside a solenoid, the magnetic field inside the coil is: A) Unaffected by the solenoid B) Strengthened by the solenoid’s field C) Opposite to the solenoid’s field D) Zero Answer: B Explanation: The solenoid’s uniform field adds to the field produced by the coil, increasing the total magnetic field. Question 50. The magnetic flux through a single loop of area 0.02 m² in a uniform field of 0.5 T is: A) 0.01 Wb B) 0.02 Wb C) 0.04 Wb D) 0.1 Wb Answer: A Explanation: Φ = B A = 0.5 T × 0.02 m² = 0.01 Wb. Question 51. If the number of turns in a coil is increased from 50 to 200 while the magnetic field and speed remain constant, the induced emf will: A) Remain the same
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B) Double C) Quadruple D) Increase by a factor of 8 Answer: C Explanation: ε ∝ N, so increasing turns by factor 4 multiplies emf by 4. Question 52. A diode allows current to flow: A) In both directions equally B) Only when reverse‑biased C) Only when forward‑biased D) Only at high frequencies Answer: C Explanation: A diode conducts when forward‑biased (positive voltage on the anode relative to cathode) and blocks reverse bias. Question 53. In a circuit, an ammeter must be connected: A) In parallel with the component whose current is to be measured B) In series with the component C) Across the power source only D) Between two resistors in a parallel branch Answer: B Explanation: An ammeter measures the current through a component, so it must be placed in series so the same current passes through it. Question 54. A voltmeter must have: A) Very low internal resistance B) Very high internal resistance C) The same resistance as the source
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Answer: C Explanation: For parallel identical resistors, 1/R_total = 1/R + 1/R = 2/R → R_total = R/2. Question 58. The term “electromotive force” (emf) is best described as: A) An actual force that pushes charges B) The potential difference produced by a source when no current flows C) The resistance of a battery D) The magnetic field around a wire Answer: B Explanation: emf is the open‑circuit voltage of a source; it is not a mechanical force. Question 59. A battery of emf 9 V and internal resistance 1 Ω supplies a lamp that draws 0.5 A. The terminal voltage across the lamp is: A) 9 V B) 8.5 V C) 8 V D) 7 V Answer: B Explanation: V_terminal = emf – I r = 9 V – (0.5 A × 1 Ω) = 8.5 V. Question 60. If a conductor is moved parallel to magnetic field lines, the induced emf is: A) Maximum B) Zero C) Depends on speed only D) Negative Answer: B Explanation: No magnetic flux change occurs when motion is parallel to the field, so ε = 0.
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Question 61. In a DC circuit, the power dissipated by a resistor can also be written as: A) P = V / R B) P = I R² C) P = V² / R D) P = I + V Answer: C Explanation: Using P = IV and Ohm’s law (V = IR) gives P = V² / R. Question 62. A 12 V battery is connected to a series circuit of three resistors: 2 Ω, 3 Ω and 5 Ω. The current in the circuit is: A) 0.6 A B) 1 A C) 1.2 A D) 2 A Answer: A Explanation: Total R = 2 + 3 + 5 = 10 Ω. I = V / R = 12 V / 10 Ω = 1.2 A (Oops, that's 1.2 A). Option C matches. Answer: C Explanation: Total resistance is 10 Ω, so I = 12 / 10 = 1.2 A. Question 63. If the same three resistors are connected in parallel, the total resistance is: A) 0.5 Ω B) 1 Ω C) 2 Ω D) 10 Ω Answer: B Explanation: 1/R_total = 1/2 + 1/3 + 1/5 = (15+10+6)/30 = 31/30 → R_total ≈ 30/31 ≈ 0.97 Ω ≈ 1 Ω (closest).
