NCEA Level 3 Physics 91526 Electrical Systems: past papers, practice and marking
- NCEA Level 3
- Physics
- AS91526
- 6 credits
- External exam
About 91526
AS91526, Demonstrate understanding of electrical systems. DC circuits, capacitors, inductors and induction, AC circuits and resonance: the other 6-credit external, and the one where naming the right effect earns the marks.
What 91526 covers
- DC circuitsEmf, internal resistance, terminal voltage, Kirchhoff’s laws
- CapacitorsCapacitance, parallel plates, dielectrics, series and parallel, energy stored
- Charging and dischargingTime constant τ = RC, charging and discharging graphs, lamps with capacitors
- InductorsFlux, Faraday and Lenz, back emf, τ = L/R, energy stored, transformers
- AC circuitsRMS and peak, reactance, impedance, effect of frequency and iron cores
- Phase relationshipsPhasor diagrams, phase angle, which leads
- ResonanceX_L = X_C, Z = R, f₀, finding L or C, internal resistance
91526 past papers
Open a 91526 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Physics past papers
Practice questions
Question 1: Capacitors
A camera flash uses a 2200 μF capacitor charged to 9.0 V. When triggered, it discharges through the flash tube in about 2.0 ms.
- Calculate the charge and the energy stored.(up to Achieved)
- Calculate the average power delivered to the flash tube during the discharge.(up to Achieved)
Check your answer
Part a: what gets the marks
- Charge: Q = CV = 2200×10⁻⁶ × 9.0 = 0.0198 C
- Energy: E = ½CV² = ½ × 2200×10⁻⁶ × 81 = 0.0891 J
A full answer: Q = 1.98 × 10⁻² C and E = 8.91 × 10⁻² J
Part b: what gets the marks
- Power is energy per unit time: P = E/t = 0.0891 / 2.0×10⁻³ = 44.6 W
A full answer: P = 44.6 W
What gets each grade
Question 2: Resonance
A series circuit contains a 0.350 H inductor and a 22.0 μF capacitor.
- Derive the condition for resonance and calculate the resonant frequency.(up to Merit)
- State the impedance of the circuit at resonance if the resistance is 100 Ω, and explain why the current is at a maximum there.(up to Merit)
Check your answer
Part a: what gets the marks
- State the condition: At resonance the reactances are equal: X_L = X_C
- Write both out and equate: 2π f L = 1/(2π f C)
- Rearrange for f: 4π² f² L C = 1 f² = 1/(4π² L C) f = 1/(2π √(LC))
- Substitute: f = 1/(2π √(0.350 × 22.0×10⁻⁶)) = 1/(2π √(7.70×10⁻⁶)) = 1/(2π × 2.775×10⁻³) = 57.4 Hz
A full answer: f₀ = 57.4 Hz
Part b: what gets the marks
- Find the net reactance at resonance: X_L − X_C = 0 → Z = √(R² + 0²) = R = 100 Ω
- Impedance is at its lowest possible value: nothing but the resistance opposes the current, since the inductive and capacitive effects cancel exactly. Since I = V/Z, the smallest Z gives the largest current.
A full answer: At resonance X_L = X_C, so the net reactance is zero:
Z = √(R² + 0²) = R = 100 Ω
The inductive and capacitive effects cancel exactly, so only the resistance opposes the current. This is the smallest possible impedance, and since I = V/Z, the current is a maximum.
What gets each grade
Question 3: Phase relationships
A series RL circuit has a resistance of 80.0 Ω and an inductive reactance of 60.0 Ω at the operating frequency.
- Calculate the impedance and the phase angle.(up to Achieved)
- Explain why R and X_L combine as the sides of a right-angled triangle rather than simply adding.(up to Excellence)
Check your answer
Part a: what gets the marks
- Impedance combines R and X at right angles: Z = √(R² + X_L²) = √(80.0² + 60.0²) = √(6400 + 3600) = √10000 = 100 Ω
- Phase angle from the same triangle: tan φ = X_L / R = 60.0 / 80.0 = 0.750 φ = 36.9°
A full answer: Z = 100 Ω and φ = 36.9°, with the current lagging the voltage.
Part b: what gets the marks
- The voltage across the resistor is in phase with the current. The voltage across the inductor leads the current by 90°.
- The two voltages peak at different times, so they never reach their maxima together. Adding their peak values arithmetically would overstate the total.
- Set out the correct addition: The two voltages are 90° apart, so they add like perpendicular vectors: V_total = √(V_R² + V_L²) Dividing through by the common current gives Z = √(R² + X_L²)
A full answer: The resistor voltage is in phase with the current, while the inductor voltage leads the current by 90°.
So the two voltages never peak at the same instant, and adding their peak values would overstate the total.
Quantities 90° apart add like perpendicular vectors:
V_total = √(V_R² + V_L²)
The same current flows through both in series, so dividing by it gives Z = √(R² + X_L²).