Electricity & Charge Basics

Question 1 (10 marks)

(a) Power = rate of energy transfer: $P = \dfrac{dW}{dt}$. (1) Using $W = QV$ so $dW = V\,dQ$ (V constant): $P = V\dfrac{dQ}{dt}$. (2) Since $I = \dfrac{dQ}{dt}$: $\boxed{P = VI}$. (1)

(b) Substitute $V = IR$ into $P = VI$: $P = (IR)I = I^2R$. (1.5) Substitute $I = V/R$: $P = V\left(\dfrac{V}{R}\right) = \dfrac{V^2}{R}$. (1.5)

(c) $R = \dfrac{P}{I^2} = \dfrac{12}{0.5^2} = \dfrac{12}{0.25} = 48\ \Omega$. (2) $V = IR = 0.5 \times 48 = 24\text{ V}$ (or $V = P/I = 24$ V). (1)

Question 2 (10 marks)

(a) $e = 1.602\times10^{-19}\text{ C}$. (1) One coulomb is the charge transported by a current of one ampere in one second: $1\text{ C} = 1\text{ A}\cdot\text{s}$. (2)

(b) $Q = It = 2 \times (3\times 60) = 2 \times 180 = 360\text{ C}$. (2) $N = \dfrac{Q}{e} = \dfrac{360}{1.602\times10^{-19}} \approx 2.25\times10^{21}$ electrons. (2)

(c) One coulomb corresponds to ~$6.25\times10^{18}$ electron charges; since each electron carries a minuscule charge ($10^{-19}$ C), an enormous number are needed to make up 1 C — hence the coulomb is large relative to a single electron. (3)

Question 3 (8 marks)

(a) Conventional current: direction of flow of positive charge, from + to − terminal in the external circuit. (1.5) Electron flow: actual movement of electrons from − to + terminal. (1.5) They are opposite because Franklin defined current direction (positive) before the electron was discovered; electrons turned out to be negative. (1)

(b) Circuit laws (Ohm's, Kirchhoff's) are consistent as long as one convention is applied throughout; a negative charge moving one way is electrically equivalent to a positive charge moving the opposite way. Voltage/current relationships and power calculations give identical results, so the convention choice never changes the answer. (4)

Question 4 (12 marks)

(a) Capacitance = charge stored per unit voltage: $C = \dfrac{Q}{V}$. (1) Rearranged: $Q = CV$. (1) Farad in base units: $1\text{ F} = 1\text{ C/V} = \text{A}^2\text{s}^4\text{kg}^{-1}\text{m}^{-2}$ (accept $\text{C/V}$ or $\text{A·s/V}$). (2)

(b) Inductance = property opposing change in current, defined by $V = L\dfrac{dI}{dt}$. (2) Henry: $1\text{ H} = 1\text{ V·s/A}$ (one henry induces 1 V for a current change of 1 A/s). (2)

(c) $Q = CV = 100\times10^{-6} \times 9 = 9\times10^{-4}\text{ C} = 0.9\text{ mC}$. (2) $E = \tfrac{1}{2}CV^2 = 0.5 \times 100\times10^{-6} \times 81 = 4.05\times10^{-3}\text{ J} = 4.05\text{ mJ}$. (2)

Question 5 (10 marks)

(a)

• Conductor: many free charge carriers (delocalised electrons) → easy current flow; e.g. copper. (2)
• Insulator: negligible free carriers, electrons tightly bound → blocks current; e.g. rubber/glass. (2)
• Semiconductor: few free carriers, intermediate conductivity, controllable by doping/temperature; e.g. silicon. (2)

(b)

• Resistor: rectangle (IEC) or zig-zag (ANSI). (1.5)
• Battery/cell: long thin line (+) and short thick line (−), pairs for multiple cells. (1.5)
• Capacitor: two parallel lines (plates) with gap. (1)

Question 6 (10 marks)

(a) DC: current/voltage constant in magnitude and one fixed direction — flat horizontal waveform. (2) AC: magnitude and direction vary periodically (typically sinusoidal), reversing polarity — sine wave crossing zero. (2)

(b) Power (W) is the rate of energy use per second; energy (J) is the total accumulated over time ($E = P\times t$). Analogy: power is the speed of a car (rate); energy is total distance travelled. (3)

(c) $E = Pt = 60 \times (5\times60) = 60 \times 300 = 18000\text{ J} = 18\text{ kJ}$. (3)

[
  {"claim":"Q1c resistance R=48 ohm and V=24 V","code":"P=12; I=0.5; R=P/I**2; V=I*R; result=(R==48) and (V==24)"},
  {"claim":"Q2b charge=360 C and electrons approx 2.25e21","code":"Q=2*180; N=Q/1.602e-19; result=(Q==360) and (abs(N-2.247e21)<1e19)"},
  {"claim":"Q4c capacitor charge 0.9mC and energy 4.05mJ","code":"C=100e-6; V=9; Q=C*V; E=0.5*C*V**2; result=(abs(Q-9e-4)<1e-9) and (abs(E-4.05e-3)<1e-9)"},
  {"claim":"Q6c energy=18000 J","code":"P=60; t=5*60; E=P*t; result=(E==18000)"}
]