Level 4 — ApplicationElectricity & Charge Basics

Electricity & Charge Basics

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time limit: 60 minutes Total marks: 50

Question 1 — LED Dropper Resistor & Energy Budget (12 marks)

A designer wants to drive a single LED from a $5.0\ \text{V}$ regulated supply. The LED datasheet specifies a forward voltage of $2.1\ \text{V}$ at a forward current of $18\ \text{mA}$ .

(a) Determine the resistance of the series "dropper" resistor required so that the LED operates at its rated current. (3)

(b) Calculate the power dissipated in the resistor. (2)

(d) The designer only has resistors rated for $0.125\ \text{W}$ maximum. State whether the resistor in (a) is safe, and justify numerically. (2)

(e) State the direction of conventional current through the resistor relative to the direction of electron flow. (2)

Question 2 — Charge, Current, and Electron Counting (10 marks)

A conductor carries a steady current of $250\ \text{mA}$ .

(a) Calculate the total charge that passes a point in the conductor over $4.0$ minutes. (3)

(b) Using the elementary charge $e = 1.602\times10^{-19}\ \text{C}$ , determine the number of electrons that pass that point in the $4.0$ minutes. (3)

(c) A different wire must deliver $9.0\ \text{C}$ of charge in $30\ \text{s}$ . State the current, in amperes, and name its unit's defining relationship (charge–time). (2)

(d) Explain why a copper wire and a rubber sheath behave differently when the same voltage is applied, referring to charge carriers. (2)

Question 3 — Power Delivery and Material Choice (10 marks)

Two resistors are available: $R_A = 12\ \Omega$ and $R_B = 47\ \Omega$ . Each is connected, one at a time, directly across a $9.0\ \text{V}$ battery.

(a) For each resistor, calculate the current drawn and the power dissipated. (4)

(b) Which resistor dissipates more power, and by what factor (to 2 significant figures)? (2)

(c) A third component is described as a semiconductor. Explain in what way its conduction behaviour differs from both a conductor and an insulator. (2)

(d) The $9.0\ \text{V}$ source is now replaced by an AC source whose voltage varies sinusoidally between $+9.0\ \text{V}$ and $-9.0\ \text{V}$ . State one measurable difference in the current through $R_A$ compared with the DC case, and one property that stays the same on average being zero. (2)

Question 4 — Capacitor Charge Storage (9 marks)

A capacitor of capacitance $470\ \mu\text{F}$ is charged to a potential difference of $12\ \text{V}$ .

(a) Calculate the charge stored on the capacitor. (3)

(b) Calculate the energy stored, using $E = \tfrac{1}{2}CV^2$ . (3)

(d) The same $12\ \text{V}$ is now applied across an inductor of $2.0\ \text{H}$ . State what physical quantity an inductor primarily opposes and give the unit of inductance. (2)

Question 5 — Schematic Interpretation & Field Reasoning (9 marks)

A schematic shows a battery, a switch, one resistor, and one capacitor connected in a single series loop.

(a) State the standard schematic symbol description (in words) for: the battery, the resistor, and the capacitor. (3)

(b) Two parallel metal plates are separated by $2.0\ \text{mm}$ and have a uniform electric field of magnitude $1.5\times10^{4}\ \text{V/m}$ between them. Calculate the potential difference across the plates. (Use $V = E d$ .) (3)

(c) A proton is placed in this field. State the direction of the force on it relative to the field direction, and explain briefly using the definition of electric field. (3)

Answer keyMark scheme & solutions

Question 1

(a) Voltage across resistor $= 5.0 - 2.1 = 2.9\ \text{V}$ (1 — KVL: supply splits between LED and resistor). $R = V/I = 2.9 / 0.018 = 161.1\ \Omega \approx 161\ \Omega$ (2 — Ohm's law applied).

(b) $P = VI = 2.9 \times 0.018 = 0.0522\ \text{W} \approx 52\ \text{mW}$ (2). (Or $I^2R = 0.018^2\times161.1$ .)

(c) $E = P t = 0.0522 \times 3600 = 187.9\ \text{J} \approx 188\ \text{J}$ (3 — energy = power × time; $1\ \text{h}=3600\ \text{s}$ ).

(d) $52\ \text{mW} = 0.052\ \text{W} < 0.125\ \text{W}$ , so the resistor is safe (2 — comparison + conclusion).

(e) Conventional current flows from + terminal through the resistor toward the LED; electron flow is in the opposite direction (2).

Question 2

(a) $t = 4.0\times60 = 240\ \text{s}$ ; $Q = It = 0.250 \times 240 = 60\ \text{C}$ (3).

(b) $N = Q/e = 60 / 1.602\times10^{-19} = 3.75\times10^{20}$ electrons (3).

(c) $I = Q/t = 9.0/30 = 0.30\ \text{A}$ (1). One ampere = one coulomb per second (1).

(d) Copper is a conductor with many free (delocalised) electrons that move under the applied field → current flows; rubber is an insulator whose electrons are tightly bound → negligible free carriers → almost no current (2).

Question 3

(a) $R_A$ : $I = 9.0/12 = 0.75\ \text{A}$ ; $P = VI = 9.0\times0.75 = 6.75\ \text{W}$ (2). $R_B$ : $I = 9.0/47 = 0.1915\ \text{A}$ ; $P = 9.0\times0.1915 = 1.723\ \text{W}$ (2).

(b) $R_A$ dissipates more; factor $= 6.75/1.723 = 3.9$ (2). (Equivalently $47/12=3.9$ since $P=V^2/R$ .)

(c) A semiconductor conducts more than an insulator but less than a conductor; its conductivity is intermediate and can be strongly altered by temperature, doping, or applied fields (2).

(d) Difference: the current alternates direction (reverses sign) periodically rather than being steady/unidirectional (1). Same on average: the mean current over a full cycle is zero (1).

Question 4

(a) $Q = CV = 470\times10^{-6} \times 12 = 5.64\times10^{-3}\ \text{C} = 5.64\ \text{mC}$ (3).

(b) $E = \tfrac12 CV^2 = 0.5 \times 470\times10^{-6} \times 12^2 = 0.03384\ \text{J} \approx 33.8\ \text{mJ}$ (3).

(c) One farad is the capacitance storing one coulomb of charge per volt of potential difference ( $C = Q/V$ ) (1).

(d) An inductor opposes a change in current (1). Unit of inductance: the henry (H) (1).

Question 5

(a) Battery: long thin line (+) and short thick line (−), often multiple pairs (1). Resistor: rectangular box (or zig-zag) in the line (1). Capacitor: two parallel lines (one may be curved for polarised) (1).

(b) $V = E d = 1.5\times10^{4} \times 2.0\times10^{-3} = 30\ \text{V}$ (3).

(c) The force is in the same direction as the field (1). Electric field is defined as the force per unit positive charge; a proton is positive, so force $= qE$ points along $\vec E$ (2).

[
  {"claim":"Q1a dropper resistor ~161 ohm","code":"R=(5.0-2.1)/0.018; result = abs(R-161.11) < 0.5"},
  {"claim":"Q1c energy in one hour ~188 J","code":"P=2.9*0.018; E=P*3600; result = abs(E-187.92) < 1"},
  {"claim":"Q2b electron count 3.75e20","code":"Q=0.250*240; N=Q/1.602e-19; result = abs(N-3.7453e20)/3.7453e20 < 0.01"},
  {"claim":"Q3b power ratio ~3.9","code":"PA=9.0**2/12; PB=9.0**2/47; result = abs(PA/PB-3.9167) < 0.05"},
  {"claim":"Q4b capacitor energy ~33.8 mJ","code":"E=0.5*470e-6*12**2; result = abs(E-0.03384) < 1e-4"},
  {"claim":"Q5b plate voltage 30 V","code":"V=1.5e4*2.0e-3; result = abs(V-30) < 1e-6"}
]