الأسبوع 14 · الفصل الرابع (٣): دائرتا (R–C) و(R–L–C) على التوالي وزاوية الطور والقدرة، والدائرة المهتزة وترددها، وظاهرة الرنين ودائرة التوليف في المذياع · 3 دروس
3
Resonance and the tuned radio circuit
Physics Sec 3 Term 1, Chapter 4: pp. 107–113
GoalExplain resonance in a series LCR circuit, find the resonant frequency, and describe how a radio is tuned by changing C.
New words
resonance · the state in which XL = XC, the impedance is smallest (Z = R) and the current is largestresonant frequency f₀ · the frequency at which XL = XC: f₀ = 1 ÷ (2π√(LC))tuning · changing C (or L) so that the circuit’s f₀ matches the frequency of the wanted station
Explanation
Connect a resistor, a coil and a capacitor in series to a source whose frequency can be changed, and keep its voltage fixed. When f is small, XC is large and XL is small. As f rises, XL = 2πfL grows and XC = 1 ÷ (2πfC) falls. At one frequency they are equal, XL = XC. Then the net reactance is zero, Z = R is as small as it can be, and the current I = V ÷ R is at its maximum: this is resonance. Setting 2πf₀L = 1 ÷ (2πf₀C) gives f₀ = 1 ÷ (2π√(LC)), the same value as the natural frequency of the oscillating circuit: the source pushes the circuit exactly at the frequency at which it likes to swing. A radio uses this. Its aerial picks up many stations at once, each with its own frequency, and drives currents of all these frequencies in a coil–capacitor circuit. The circuit answers strongly only to the station whose frequency equals its f₀. Turning the tuning knob changes the capacitance C, so f₀ moves to another station. The chosen signal is then amplified. We only calculate and draw here; we do not take a radio apart.
L = 0.1 H, C = 10 µF, R = 20 Ω, V = 10 V
f (Hz)
XL (Ω)
XC (Ω)
Z (Ω)
I (A)
100
62.8
159.2
98.5
0.102
159 (f₀)
100
100
20
0.500
250
157
63.7
95.4
0.105
Worked examples
The table shows a series circuit with L = 0.1 H, C = 10 µF, R = 20 Ω and V = 10 V. √(LC) = √(10⁻⁶) = 10⁻³ s, so f₀ = 1 ÷ (2 × 3.14 × 10⁻³) = 159 Hz. There XL = XC = 100 Ω, Z = 20 Ω and I = 10 ÷ 20 = 0.5 A. At 100 Hz and at 250 Hz the current is only about 0.10 A. The graph has a sharp peak at f₀.
A series circuit with L = 0.1 H and C = 1 µF. LC = 10⁻⁷ s², √(LC) = 3.16 × 10⁻⁴ s, f₀ = 1 ÷ (2 × 3.14 × 3.16 × 10⁻⁴) = 504 Hz.
A radio coil has L = 200 µH = 2 × 10⁻⁴ H. To tune to a station of 1 MHz = 10⁶ Hz, C = 1 ÷ (4π²f²L) = 1 ÷ (4 × 3.14² × 10¹² × 2 × 10⁻⁴) = 1.27 × 10⁻¹⁰ F = 127 pF. For a station of 2 MHz, f is twice as large, so C must be a quarter: 127 ÷ 4 = 31.7 pF.
Class activity
Paper only. For the circuit L = 0.1 H, C = 10 µF, R = 20 Ω make your own table for f = 50, 159 and 400 Hz (π = 3.14) with XL, XC, Z and I at V = 10 V, and check that the current is largest at 159 Hz. Then draw a block picture of a radio: aerial, tuned circuit with a knob, amplifier, loudspeaker.
Practice
1
A series circuit has L = 0.9 H and C = 10 µF. Find the resonant frequency (π = 3.14).
When you turn the tuning knob of a radio from one station to another, which quantity of the circuit changes, and why does the first station fade?
The capacitance C changes, so f₀ = 1 ÷ (2π√(LC)) moves. The first station’s frequency no longer equals f₀, so the circuit no longer answers strongly to it.
