Level 2 — RecallElectricity & Charge Basics

Electricity & Charge Basics

30 minutes40 marksprintable — key stays hidden on paper

Chapter: 1.1 Electricity & Charge Basics Level: 2 — Recall (definitions, standard problems, short derivations) Time Limit: 30 minutes Total Marks: 40


Instructions: Answer all questions. Show working for calculation questions. Use gg-free SI units throughout.


Q1. Define each of the following, including the SI unit where applicable: (4 marks) (a) Electric charge (b) The coulomb (c) Current (d) Resistance

Q2. Explain the difference between a conductor, an insulator, and a semiconductor, giving one example of each. (3 marks)

Q3. A charge of 12 C12\ \text{C} passes through a wire in 4 s4\ \text{s}. (a) Calculate the current. (2 marks) (b) State how many electrons this corresponds to, given e=1.6×1019 Ce = 1.6\times10^{-19}\ \text{C}. (2 marks)

Q4. State Ohm's Law and use it to answer the following: (4 marks) A resistor carries a current of 0.5 A0.5\ \text{A} when connected across a 9 V9\ \text{V} supply. Find its resistance.

Q5. A 240 V240\ \text{V} appliance draws a current of 5 A5\ \text{A}. (4 marks) (a) Calculate the power dissipated using P=VIP = VI. (b) Using P=I2RP = I^2R, verify the power by first finding RR.

Q6. Distinguish between DC and AC signals. Give one everyday source of each. (3 marks)

Q7. Explain the difference between conventional current and electron flow, stating the direction of each relative to the positive terminal of a battery. (3 marks)

Q8. Define the following quantities and give the SI unit of each: (6 marks) (a) Voltage (potential difference) (b) Capacitance (c) Inductance

Q9. A 100 Ω100\ \Omega resistor dissipates energy for 30 s30\ \text{s} while carrying a current of 0.2 A0.2\ \text{A}. (4 marks) (a) Calculate the power dissipated. (b) Calculate the total energy consumed, and state the difference between energy and power.

Q10. Identify the component represented by each schematic symbol description: (5 marks) (a) Two parallel lines of equal length (b) A zig-zag line (or rectangle, IEC style) (c) A long line and a short parallel line (d) A coil / series of loops (e) A circle with a wavy line inside

Answer keyMark scheme & solutions

Q1. (4 marks — 1 each)

  • (a) Electric charge: a fundamental property of matter that experiences a force in an electric field; can be positive or negative. (Concept of the source of electrical force.)
  • (b) The coulomb (C): the SI unit of charge; equal to the charge transported by a current of 1 A in 1 s. (Links charge to current × time.)
  • (c) Current: the rate of flow of electric charge; I=Q/tI = Q/t, unit ampere (A). (Defines flow.)
  • (d) Resistance: opposition to the flow of current; R=V/IR = V/I, unit ohm (Ω\Omega). (Ratio of voltage to current.)

Q2. (3 marks — 1 each)

  • Conductor: has many free electrons, allows current to flow easily (e.g. copper).
  • Insulator: very few free charge carriers, resists current flow (e.g. rubber/glass).
  • Semiconductor: conductivity between the two, controllable by doping/temperature (e.g. silicon).

Q3. (4 marks)

  • (a) I=Q/t=12/4=3 AI = Q/t = 12/4 = \mathbf{3\ \text{A}}. (Definition of current.) (2)
  • (b) n=Q/e=12/(1.6×1019)=7.5×1019 electronsn = Q/e = 12 / (1.6\times10^{-19}) = \mathbf{7.5\times10^{19}\ \text{electrons}}. (Total charge ÷ charge per electron.) (2)

Q4. (4 marks)

  • Statement: The current through a conductor is directly proportional to the voltage across it, provided temperature is constant: V=IRV = IR. (2)
  • R=V/I=9/0.5=18 ΩR = V/I = 9/0.5 = \mathbf{18\ \Omega}. (2)

Q5. (4 marks)

  • (a) P=VI=240×5=1200 WP = VI = 240 \times 5 = \mathbf{1200\ \text{W}}. (2)
  • (b) R=V/I=240/5=48 ΩR = V/I = 240/5 = 48\ \Omega; then P=I2R=52×48=25×48=1200 WP = I^2R = 5^2 \times 48 = 25 \times 48 = \mathbf{1200\ \text{W}} — consistent. (2)

Q6. (3 marks)

  • DC (direct current): charge flows in one constant direction; magnitude typically steady (1). Source: battery (0.5).
  • AC (alternating current): direction and magnitude reverse periodically (1). Source: mains/wall socket (0.5).

Q7. (3 marks)

  • Conventional current: assumed direction of positive charge flow, from + terminal to − terminal externally (1.5).
  • Electron flow: actual movement of electrons, from − terminal to + terminal — opposite to conventional current (1.5).

Q8. (6 marks — 2 each: 1 definition + 1 unit)

  • (a) Voltage: energy per unit charge between two points (V=W/QV = W/Q); unit volt (V).
  • (b) Capacitance: ability to store charge per unit voltage (C=Q/VC = Q/V); unit farad (F).
  • (c) Inductance: property opposing change in current, producing an emf (V=LdI/dtV = L\,dI/dt); unit henry (H).

Q9. (4 marks)

  • (a) P=I2R=0.22×100=0.04×100=4 WP = I^2R = 0.2^2 \times 100 = 0.04 \times 100 = \mathbf{4\ \text{W}}. (1.5)
  • (b) E=P×t=4×30=120 JE = P \times t = 4 \times 30 = \mathbf{120\ \text{J}}. (1.5)
  • Difference: Power is the rate of energy transfer (J/s = W); energy is the total transferred (J). (1)

Q10. (5 marks — 1 each)

  • (a) Capacitor
  • (b) Resistor
  • (c) Battery / cell (DC source)
  • (d) Inductor
  • (e) AC source
[
  {"claim":"Q3a current = 3 A", "code":"I=12/4; result = (I==3)"},
  {"claim":"Q3b electrons = 7.5e19", "code":"n=12/(1.6e-19); result = abs(n-7.5e19) < 1e15"},
  {"claim":"Q4 resistance = 18 ohm", "code":"R=9/0.5; result = (R==18)"},
  {"claim":"Q5 power both methods = 1200 W", "code":"P1=240*5; R=240/5; P2=5**2*R; result = (P1==1200 and P2==1200)"},
  {"claim":"Q9 power=4W and energy=120J", "code":"P=0.2**2*100; E=P*30; result = (P==4 and E==120)"}
]