Week 14 · Chapter 4 (C): the series RC and LCR circuits (Z = √(R² + (XL − XC)²), the phase angle and the three cases, power only in the resistor), the oscillating LC circuit and f = 1/2π√(LC), resonance and the tuned circuit of a radio; the summary and exercises (pp. 114–116) are not used · 3 lessons
2
The oscillating circuit
Physics Sec 3 Term 1, Chapter 4: pp. 107–113
GoalDescribe how energy swings in an LC circuit, explain why the oscillation dies away, and calculate its frequency f = 1 ÷ (2π√(LC)).
New words
oscillating circuit · a coil and a capacitor joined in a loop, in which charge and current swing to and fronatural frequency · the frequency at which such a circuit oscillates by itself, f = 1 ÷ (2π√(LC))damping · the dying away of the oscillation, because the resistance turns energy into heat
Explanation
Charge a capacitor from a cell, then join it to a coil. The capacitor starts to discharge, but the coil opposes any quick growth of the current, so the current rises gradually. When the capacitor is empty, the current is at its largest and all the energy is in the magnetic field of the coil. The current cannot stop at once: the self-induced emf of the coil, which opposes the fall of the current, keeps it going and charges the capacitor again, now the other way round. When the capacitor is full again, the current is zero and the energy is back in its electric field. Then everything repeats in the opposite direction. The energy swings between the electric field and the magnetic field. The wires have some resistance, which turns a little energy into heat in every cycle, so the swings get smaller: the oscillation is damped. The circuit oscillates at its natural frequency f = 1 ÷ (2π√(LC)), with L in henry and C in farad. A larger L or C gives a lower frequency. Use π = 3.14 and convert mH and µF first.
Stage
Capacitor
Current in the coil
Energy stored mainly in
1
fully charged
zero, about to grow
the electric field of the capacitor
2
empty
largest
the magnetic field of the coil
3
fully charged the other way
zero
the electric field of the capacitor
4
empty
largest, in the opposite direction
the magnetic field of the coil
Worked examples
L = 25 mH and C = 40 µF. L = 0.025 H, C = 40 × 10⁻⁶ F, so LC = 1.0 × 10⁻⁶ s² and √(LC) = 1.0 × 10⁻³ s. f = 1 ÷ (2 × 3.14 × 1.0 × 10⁻³) = 159 Hz. The period is T = 1 ÷ 159 = 6.28 ms.
C
f
40 µF
159 Hz
10 µF
318 Hz
To double this frequency without changing L, f must be twice as large, so √C must be half as large and C a quarter: 40 ÷ 4 = 10 µF. Check: LC = 0.025 × 10 × 10⁻⁶ = 2.5 × 10⁻⁷, √(LC) = 5.0 × 10⁻⁴ s, and f = 1 ÷ (2 × 3.14 × 5.0 × 10⁻⁴) = 318 Hz, which is twice 159 Hz.
Keep C = 40 µF and raise L from 25 mH to 100 mH. LC = 0.1 × 40 × 10⁻⁶ = 4 × 10⁻⁶, √(LC) = 2 × 10⁻³ s, and f = 1 ÷ (2 × 3.14 × 2 × 10⁻³) = 79.6 Hz: L is four times larger and f is half as large. The graph shows the damping of such an oscillation.
Class activity
Paper only; no coil or capacitor is needed. Draw four small pictures of the four stages of the table: a capacitor with plus and minus signs and a coil with an arrow for the current. Under each write where the energy is. Then explain with the help of the graph what would change if the resistance were larger.
Practice
1
L = 1 H and C = 25 µF. Find the natural frequency (π = 3.14).
