Turn the tuning knob of an old transistor radio slowly. Most of the dial gives only a hiss. At certain places a voice comes in clearly, a news bulletin at one point and film songs at another. Dozens of stations fill the air of every Indian city at the same time. Yet the radio plays one of them and ignores the rest.
A capacitor emptying into a coil
Charge a capacitor of capacitance \(C\) to a charge \(q_0\). Then join it, through a switch, across a coil of inductance \(L\) whose resistance is negligible. The circuit is an LC circuit. Close the switch at \(t = 0\). The capacitor starts to discharge through the coil. The coil opposes any change in its current, so the current does not jump; it grows gradually.
As the charge on the plates falls, the current grows. When the capacitor is empty, the current is at its largest. The coil now opposes the fall of its current and keeps it flowing. So the charge piles up on the plates again, the other way round. The current falls to zero just as the capacitor is full again, with its signs reversed.
Then everything runs backwards, and the charge returns to where it started. With no resistance, the swing repeats for ever. Charge and current rise and fall like the displacement and the velocity of a block on a spring.
The capacitor has lost all its charge for the first time. What is the current in the coil at that instant?
In the figure above, press play. The plates start with \(100\,\mu\text{C}\), and the bars show where the energy is. At a quarter period the plates are empty and the field lines fill the coil. At half a period the plates are full again, with the lower plate positive. On the graph the two curves are a quarter period apart. The current is zero whenever the charge is largest, and largest whenever the charge is zero.
The equations of the swing
Let \(q\) be the charge on the upper plate and \(i = dq/dt\) the rate at which it grows. The charge swings like a block on a spring, so it is a cosine, starting at \(q_0\). The current is its derivative. The next topic proves that the angular frequency is \(\omega = 1/\sqrt{LC}\).
\[ q = q_0\cos\omega t \qquad i = \dfrac{dq}{dt} = -q_0\omega\sin\omega t \]
The minus sign says that the current first flows so as to empty the upper plate. The largest current is \(I_0 = q_0\omega\). The current reaches it a quarter period ahead of the charge: \(-\sin\omega t = \cos(\omega t + \pi/2)\), so \(i\) leads \(q\) by \(90°\).
A \(10\,\mu\text{F}\) capacitor holding \(100\,\mu\text{C}\) is joined across a \(0.1\,\text{H}\) coil at \(t = 0\). Find (a) \(\omega\), \(f\) and \(T\), (b) the largest current, and (c) the instants when the current is largest, with the charge then.
(a) \(\omega = 1/\sqrt{LC}\) \(= 1/\sqrt{0.1\times10^{-5}} = 1000\,\text{rad/s}\). So \(f = \omega/2\pi = 159\,\text{Hz}\) and \(T = 2\pi/\omega = 6.28\,\text{ms}\).
(b) \(I_0 = q_0\omega = 10^{-4}\times1000 = 0.1\,\text{A}\).
(c) \(i = -q_0\omega\sin\omega t\) is largest in size when \(\sin\omega t = \pm1\). That is at \(t = T/4 = 1.57\,\text{ms}\), then \(3T/4 = 4.71\,\text{ms}\), and so on. The charge is zero at those instants.
Values used: the sheet's circuit, \(q_0 = 100\,\mu\text{C}\), \(C = 10\,\mu\text{F}\), \(L = 0.1\,\text{H}\).
A larger capacitor or a larger coil makes the swing slower. With \(40\,\mu\text{F}\) in place of \(10\,\mu\text{F}\), \(LC\) is four times as large. So \(\omega\) halves to \(500\,\text{rad/s}\), and the period doubles to \(12.6\,\text{ms}\).
Now the radio dial. Its knob turns a variable capacitor in an LC circuit, and so sets the circuit's natural frequency. A coil of \(1\,\text{mH}\) oscillates at \(1\,\text{MHz}\) with \(C = 1/(4\pi^2f^2L) = 25\,\text{pF}\). Medium-wave stations lie between about \(530\) and \(1600\,\text{kHz}\). With the same coil, the capacitor must then range from about \(87\) down to \(10\,\text{pF}\).
Two curves are drawn for the circuit of the example. Curve A starts at its largest value at \(t = 0\); curve B starts at zero. Which is which?
The \(10\,\mu\text{F}\) capacitor of the example is replaced by a \(40\,\mu\text{F}\) capacitor, with the same \(0.1\,\text{H}\) coil. What is the new period, in milliseconds?
A tuning circuit has \(L = 1\,\text{mH}\) and must oscillate at \(1\,\text{MHz}\). What capacitance does it need, in picofarads?
The equation of the LC circuit
Go once round the circuit in the direction of \(i\). Through the coil the potential falls by \(L\,di/dt\), the voltage across an inductor. Across the capacitor, from the upper plate to the lower, it falls by \(q/C\). Kirchhoff's loop rule, from the chapter Circuits: Kirchhoff's Laws, Networks and RC Transients, says the two falls add to zero.
\[ L\dfrac{di}{dt} + \dfrac{q}{C} = 0 \]
Why. Kirchhoff's loop rule, going round in the direction of \(i\); there is no resistance and no cell.
\[ \dfrac{di}{dt} = \dfrac{d^2q}{dt^2} \]
Why. \(i = dq/dt\) is the rate at which the upper plate gains charge.
\[ L\dfrac{d^2q}{dt^2} + \dfrac{q}{C} = 0 \]
Why. put step 2 into step 1.
\[ \dfrac{d^2q}{dt^2} = -\dfrac{1}{LC}\,q \]
Why. divide by \(L\). This is the form \(d^2x/dt^2 = -\omega^2x\) of simple harmonic motion.
\[ q = q_0\cos\omega t, \quad \omega = \dfrac{1}{\sqrt{LC}} \]
Read it. the solution of SHM with \(q = q_0\) and \(i = 0\) at \(t = 0\), and \(\omega^2 = 1/LC\).
Put this equation beside the block on a spring of the chapter Dynamics of SHM, \(m\,d^2x/dt^2 + kx = 0\). The two have exactly the same form. The charge takes the place of the displacement, and so the current takes the place of the velocity. The inductance takes the place of the mass, and \(1/C\) takes the place of the spring constant. Every result for the block then holds for the circuit.
A coil is sluggish in the way a heavy block is. The block resists a change of velocity; the coil resists a change of current. A small capacitor is stiff in the way a strong spring is. A little charge on it already gives a large voltage, which pushes the charge back hard. A heavy block on a soft spring swings slowly, and so does a large coil with a large capacitor.
In the analogy between an LC circuit and a block on a spring, what plays the part of the spring constant \(k\)?
Energy in two fields
The energy of the circuit sits in two places. The capacitor holds \(q^2/2C\) in the electric field between its plates. The coil holds \(\tfrac12Li^2\) in its magnetic field. The chapter Capacitance and Dielectrics found the first, and the chapter Eddy Currents, Generators and Inductance the second. As the charge and the current swing, the energy passes from one field to the other and back.
\[ \dfrac{q^2}{2C} = \dfrac{q_0^2}{2C}\cos^2\omega t \]
Why. put \(q = q_0\cos\omega t\) into the energy of the capacitor.
\[ \tfrac12Li^2 = \tfrac12Lq_0^2\omega^2\sin^2\omega t \]
Why. put \(i = -q_0\omega\sin\omega t\) into the energy of the coil.
\[ \tfrac12Lq_0^2\omega^2\sin^2\omega t = \dfrac{q_0^2}{2C}\sin^2\omega t \]
Why. \(\omega^2 = 1/LC\), so \(L\omega^2 = 1/C\).
\[ \dfrac{q^2}{2C} + \dfrac12Li^2 = \dfrac{q_0^2}{2C}\left(\cos^2\omega t + \sin^2\omega t\right) = \dfrac{q_0^2}{2C} \]
Read it. add steps 1 and 3; \(\cos^2 + \sin^2 = 1\) at every instant.
Each energy goes as a square, \(\cos^2\omega t\) or \(\sin^2\omega t\), and a square repeats every half period. So the energy passes back and forth at twice the frequency of the charge, \(2f\). For the circuit of the example that is \(2\times159 = 318\,\text{Hz}\). The energy is first all magnetic at \(T/4 = 1.57\,\text{ms}\).
For the circuit of the example (\(100\,\mu\text{C}\) on \(10\,\mu\text{F}\), \(0.1\,\text{H}\)), find the total energy and check it against \(\tfrac12LI_0^2\). Then find the charge when the energy is shared equally between the capacitor and the coil.
At \(t = 0\) all the energy is in the capacitor. At the moment of equal sharing, each field holds half the total.
\(U_0 = q_0^2/2C\). For equal sharing, \(q^2/2C = \tfrac12\,q_0^2/2C\).
\(U_0 = (10^{-4})^2/(2\times10^{-5}) = 5\times10^{-4}\,\text{J}\). Check: \(\tfrac12\times0.1\times0.1^2 = 5\times10^{-4}\,\text{J}\). Then \(q = q_0/\sqrt2 = 100/1.414 = 71\,\mu\text{C}\).
The circuit holds \(0.5\,\text{mJ}\). It is shared equally when the charge is \(71\,\mu\text{C}\), first at \(t = T/8 = 0.79\,\text{ms}\).
A capacitor charged to \(60\,\mu\text{C}\) oscillates with a coil. What is the charge on it when the energy is shared equally between the capacitor and the coil, in microcoulombs?
The charge on the capacitor of an LC circuit oscillates at \(159\,\text{Hz}\). At what frequency does the energy pass from the capacitor to the coil and back?
Why real oscillations die away
A real coil is made of wire, and wire has resistance. The current heats it in every cycle, so a little energy leaves the circuit as heat each time. The amplitude of the charge falls cycle by cycle, like the swing of a pendulum in air. This is damping, as in the chapter Superposition, Damped and Forced Oscillations. A circuit that oscillates also sends out a little energy as electromagnetic waves, the subject of the next two lessons.
To keep an oscillation going, energy must be fed in at the right moments. A radio transmitter does this with an electronic circuit, as a child's swing is kept going by well-timed pushes. An alternating source joined to the circuit drives it hardest when its frequency matches the natural frequency \(1/2\pi\sqrt{LC}\). That is resonance, which the chapter Alternating Current and the LCR Circuit studies in detail.
Now the radio dial is explained. Every station's wave reaches the aerial and drives the tuning circuit a little. Only the station whose frequency matches the circuit's natural frequency drives it strongly. Turning the knob changes \(C\), and with it the one station that the radio hears.
The oscillations of a real LC circuit die away after a while. Where does the energy go?
Real coils and wires have resistance, so the current heats them and some energy leaves as heat in every cycle. The oscillating charges also radiate a little energy as electromagnetic waves. The amplitude therefore falls, unless a source feeds energy in at the right moments.
- Expecting the current to be largest when the capacitor is full.The current is zero when the charge is largest and largest when the capacitor is empty; they are a quarter period apart.
- Taking \(C\) as the analogue of the spring constant.The restoring voltage is \(q/C\), so \(1/C\) plays \(k\); \(L\) plays the mass.
- Giving the energy the same frequency as the charge.Energy goes as \(q^2\) and \(i^2\), so it repeats every half period, at \(2f\).
- Taking equal energies at \(q = q_0/2\).Energy goes as the square of the charge; equal sharing is at \(q = q_0/\sqrt2\).