Specification: AQA A-level Physics (7408) 3.13.3.1 Electronics option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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Resonant frequency f₀: the frequency at which an LC circuit naturally oscillates. For a parallel LC circuit used as a filter, the impedance is a maximum at f₀, so the voltage across it is greatest at this frequency.
Energy exchange: in an LC circuit energy is transferred back and forth between the electric field of the capacitor and the magnetic field of the inductor. Resistance dissipates energy, so real oscillations are damped.
Tuning: a radio selects one station by adjusting C (a variable capacitor) or L so that f₀ equals the carrier frequency of that station.
The equation
resonant frequency = 1 ÷ (2π × √(inductance × capacitance)) f₀ = 1 ÷ (2π√(LC))
| Quantity | Symbol | Unit |
|---|---|---|
| resonant frequency | f₀ | hertz (Hz) |
| inductance | L | henries (H) |
| capacitance | C | farads (F) |
Rearranged: LC = 1 ÷ (4π²f₀²) | C = 1 ÷ (4π²f₀²L) | L = 1 ÷ (4π²f₀²C) | f₀ ∝ 1 ÷ √C, so halving the frequency needs four times the capacitance. Convert mH, μH, μF, nF and pF to H and F before you substitute.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An LC circuit has an inductance of 0.25 H and a capacitance of 4.0 × 10−6 F. Calculate its resonant frequency, to 3 significant figures.
Show solution
F: f₀ = 1 ÷ (2π√(LC))
I: f₀ = 1 ÷ (2π√(0.25 × 4.0 × 10−6)) = 1 ÷ (2π × 1.0 × 10−3)
F: f₀ is already the subject
A: 159.15… Hz = 159 Hz (3 s.f.)
Q2 (Core) A filter circuit uses an inductor of 2.2 × 10−3 H and a capacitor of 4.7 × 10−9 F. Calculate the resonant frequency in kHz, to 2 significant figures.
Show solution
F: f₀ = 1 ÷ (2π√(LC))
I: LC = 2.2 × 10−3 × 4.7 × 10−9 = 1.034 × 10−11, so √(LC) = 3.215… × 10−6
F: f₀ = 1 ÷ (2π × 3.215… × 10−6)
A: 49 494… Hz = 49 kHz (2 s.f.)
Q3 (Core) A long-wave radio receiver has a 2.5 mH inductor in its tuning circuit. Calculate the capacitance needed to tune it to a station broadcasting at 198 kHz. Give your answer in pF to 3 significant figures.
Show solution
Convert first: 2.5 mH = 2.5 × 10−3 H; 198 kHz = 1.98 × 105 Hz
F: f₀ = 1 ÷ (2π√(LC))
I: 1.98 × 105 = 1 ÷ (2π√(2.5 × 10−3 × C))
F: C = 1 ÷ (4π²f₀²L) = 1 ÷ (4π² × (1.98 × 105)² × 2.5 × 10−3)
A: 2.584… × 10−10 F = 258 pF (3 s.f.)
Watch out: forgetting to square f₀ (and the 2π) when rearranging. Square both sides of the equation first, then make C the subject.
Q4 (Core) A 330 pF capacitor is used in a circuit that must resonate at 1.0 MHz. Calculate the inductance needed, in μH, to 3 significant figures.
Show solution
Convert first: 330 pF = 330 × 10−12 F; 1.0 MHz = 1.0 × 106 Hz
F: f₀ = 1 ÷ (2π√(LC))
I: 1.0 × 106 = 1 ÷ (2π√(L × 330 × 10−12))
F: L = 1 ÷ (4π²f₀²C) = 1 ÷ (4π² × (1.0 × 106)² × 330 × 10−12)
A: 7.675… × 10−5 H = 76.8 μH (3 s.f.)
Q5 (Exam) The tuning circuit of a medium-wave radio has a fixed 220 μH inductor in parallel with a variable capacitor whose capacitance can be set anywhere from 50 pF to 500 pF. Calculate the range of frequencies the radio can be tuned to, to 2 significant figures.
Show solution
Convert first: 220 μH = 2.2 × 10−4 H; 50 pF = 5.0 × 10−11 F; 500 pF = 5.0 × 10−10 F
Step 1 – lowest frequency (largest C)
F: f₀ = 1 ÷ (2π√(LC))
I: f₀ = 1 ÷ (2π√(2.2 × 10−4 × 5.0 × 10−10)) = 1 ÷ (2π × 3.316… × 10−7)
A: 479 870… Hz = 480 kHz
Step 2 – highest frequency (smallest C)
C is 10 times smaller, so f₀ is √10 times larger: 479 870… × √10 = 1 517 482… Hz = 1.5 MHz
A: from 480 kHz to 1.5 MHz (2 s.f.)
Watch out: the largest capacitance gives the lowest frequency, not the highest.
Q6 (Exam) A radio is tuned to a station at 909 kHz when its variable capacitor is set to 200 pF. The inductance does not change. Calculate the capacitance needed to tune to a station at 1215 kHz, to 3 significant figures.
Show solution
F: f₀ = 1 ÷ (2π√(LC)), so with L fixed, f₀ ∝ 1 ÷ √C and C ∝ 1 ÷ f₀²
I: C₂ ÷ 200 = (909 ÷ 1215)² = 0.5597…
F: C₂ = 200 × 0.5597…
A: 111.94… pF = 112 pF (3 s.f.)
Watch out: scaling C in simple inverse proportion to f gives 150 pF. Capacitance depends on 1 ÷ f₀², so square the frequency ratio.
Q7 (Exam) A 10 μF capacitor is charged and then connected across a 40 mH inductor. Assume the circuit has negligible resistance. Calculate (a) the frequency of the oscillations and (b) the time taken for the capacitor to discharge fully for the first time, both to 2 significant figures.
Show solution
Convert first: 10 μF = 1.0 × 10−5 F; 40 mH = 4.0 × 10−2 H
Step 1 – (a) frequency
F: f₀ = 1 ÷ (2π√(LC))
I: f₀ = 1 ÷ (2π√(4.0 × 10−2 × 1.0 × 10−5)) = 1 ÷ (2π × 6.324… × 10−4)
A: 251.6… Hz = 250 Hz (2 s.f.)
Step 2 – (b) time to discharge
period T = 1 ÷ f₀ = 3.973… × 10−3 s. The energy moves from the capacitor’s electric field to the inductor’s magnetic field in one quarter of a cycle, so t = T ÷ 4 = 9.934… × 10−4 s
A: 0.99 ms (2 s.f.)
Watch out: the capacitor is fully discharged after T ÷ 4, not T ÷ 2. At T ÷ 2 it is fully charged again with the opposite polarity.
Q8 (Exam) A long-wave station broadcasts with a wavelength of 1500 m. A receiver’s tuning circuit has a 2.0 mH inductor. Calculate the capacitance needed to tune to this station, to 2 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – carrier frequency
f = c ÷ λ = 3.00 × 108 ÷ 1500 = 2.0 × 105 Hz
Step 2 – capacitance
F: f₀ = 1 ÷ (2π√(LC))
I: 2.0 × 105 = 1 ÷ (2π√(2.0 × 10−3 × C))
F: C = 1 ÷ (4π²f₀²L) = 1 ÷ (4π² × (2.0 × 105)² × 2.0 × 10−3) = 1 ÷ (3.158… × 109)
A: 3.166… × 10−10 F = 320 pF (2 s.f.)
Q9 (Stretch) An FM receiver must tune across the band 88 MHz to 108 MHz using a fixed 0.10 μH inductor. The circuit wiring adds a stray capacitance of 5.0 pF in parallel with the variable tuning capacitor. Calculate the range of values the variable capacitor must cover, to 3 significant figures.
Show solution
Convert first: 0.10 μH = 1.0 × 10−7 H; 88 MHz = 8.8 × 107 Hz; 108 MHz = 1.08 × 108 Hz
Step 1 – total capacitance at 88 MHz
F: C = 1 ÷ (4π²f₀²L)
I: C = 1 ÷ (4π² × (8.8 × 107)² × 1.0 × 10−7)
A: C = 3.270… × 10−11 F = 32.70… pF
Step 2 – total capacitance at 108 MHz
C = 1 ÷ (4π² × (1.08 × 108)² × 1.0 × 10−7) = 2.171… × 10−11 F = 21.71… pF
Step 3 – subtract the stray capacitance
capacitances in parallel add, so Cvariable = Ctotal − 5.0 pF
A: from 16.7 pF to 27.7 pF (3 s.f.)
Watch out: ignoring the stray capacitance. It is in parallel, so it adds to the variable capacitor and must be subtracted from the total you calculate.
Q10 (Stretch) A 100 nF capacitor is charged to 6.0 V and then connected across a 25 mH inductor. Assume there is no resistance in the circuit. Calculate (a) the frequency of the oscillations, (b) the maximum current in the inductor, and (c) the time after connection at which the current is first a maximum. Give each answer to 2 significant figures.
Show solution
Convert first: 100 nF = 1.0 × 10−7 F; 25 mH = 2.5 × 10−2 H
Step 1 – (a) frequency
F: f₀ = 1 ÷ (2π√(LC))
I: f₀ = 1 ÷ (2π√(2.5 × 10−2 × 1.0 × 10−7)) = 1 ÷ (2π × 5.0 × 10−5)
A: 3183.0… Hz = 3.2 kHz (2 s.f.)
Step 2 – (b) maximum current from energy conservation
energy stored in capacitor = ½CV² = ½ × 1.0 × 10−7 × 6.0² = 1.8 × 10−6 J
all of this is in the inductor’s magnetic field when the current is a maximum: ½LI² = 1.8 × 10−6
I² = 2 × 1.8 × 10−6 ÷ 2.5 × 10−2 = 1.44 × 10−4, so I = 0.012 A
A: 12 mA (2 s.f.)
Step 3 – (c) time to maximum current
the current is a maximum when the capacitor has fully discharged, a quarter of a cycle: t = 1 ÷ (4f₀) = 7.853… × 10−5 s
A: 79 μs (2 s.f.)
Watch out: the current is a maximum when the capacitor voltage is zero, not when it is a maximum. The current is a quarter of a cycle out of step with the charge on the capacitor.
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