Electromagnetic Oscillations and Alternating Current
A capacitor stores energy in an electric field; an inductor stores energy in a magnetic field. Wire the two together and the energy does not simply sit still — it sloshes back and forth between them, exactly as a mass on a spring trades kinetic for potential energy. The result is an LC oscillation, the electrical twin of simple harmonic motion. Add resistance and the oscillation damps; drive it with an external emf and you get the steady alternating current that powers the world. This chapter builds that story end to end: free LC oscillations, the deep mechanical analogy, damped RLC ringing, the driven series RLC circuit with its reactances, impedance, resonance, the delivery of power, and finally the transformer.
- LC oscillations: charge and current oscillate sinusoidally at \(\omega = \dfrac{1}{\sqrt{LC}}\), with \(q = Q\cos(\omega t + \phi)\), while energy sloshes between the capacitor \(U_E = \dfrac{q^{2}}{2C}\) and the inductor \(U_B = \tfrac{1}{2}Li^{2}\), their sum staying constant.
- The electrical–mechanical analogy: \(q \leftrightarrow x\), \(i \leftrightarrow v\), \(1/C \leftrightarrow k\), and \(L \leftrightarrow m\) — the LC circuit is a spring–mass oscillator dressed in electrical clothing.
- Damped RLC oscillations: resistance bleeds energy as heat, so the amplitude decays and the frequency shifts to \(\omega' = \sqrt{\dfrac{1}{LC} - \left(\dfrac{R}{2L}\right)^{2}}\).
- Driven AC circuits: the reactances \(X_C = \dfrac{1}{\omega_d C}\) and \(X_L = \omega_d L\), the impedance \(Z = \sqrt{R^{2} + (X_L - X_C)^{2}}\), current amplitude \(I = \dfrac{\varepsilon_m}{Z}\), and phase \(\tan\phi = \dfrac{X_L - X_C}{R}\).
- Resonance at \(\omega_d = \dfrac{1}{\sqrt{LC}}\), average power \(P_{\text{avg}} = \varepsilon_{\text{rms}} I_{\text{rms}}\cos\phi\) with rms values \(I_{\text{rms}} = I/\sqrt{2}\) and the power factor \(\cos\phi\), and the ideal transformer \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\).
What Is Physics?
We have met three passive circuit elements separately. A resistor dissipates energy; a capacitor stores it in an electric field; an inductor stores it in a magnetic field. In Chapter 30 we watched current rise and fall in an RL circuit, and earlier we watched charge build and drain in an RC circuit. In both, the approach to the final state was a one-way exponential — energy flowed in or out and the story ended.
Put a capacitor and an inductor together with little or no resistance and something far richer happens. Energy no longer settles; it oscillates, pouring out of the capacitor's electric field into the inductor's magnetic field and back again, over and over. This is the seed of every radio tuner, every oscillator, and ultimately of the electromagnetic waves of the next chapter. And once we drive such a circuit with an external sinusoidal source, we arrive at alternating current — the form in which electrical energy is generated, transmitted, and delivered to nearly every wall socket on Earth.
LC Oscillations, Qualitatively
Charge a capacitor and then connect it across an inductor, with no resistance and no battery. At the first instant all the energy sits in the capacitor's electric field, \(U_E = Q^{2}/2C\), and the current is zero. The capacitor begins to discharge through the inductor, but the inductor opposes any sudden change in current, so the current builds gradually. As charge drains, \(U_E\) falls while the magnetic energy \(U_B = \tfrac{1}{2}Li^{2}\) rises.
When the capacitor is fully discharged, the electric field is gone and all the energy has moved into the magnetic field; the current is now maximum. But a current through an inductor cannot stop on a dime — it keeps flowing, now recharging the capacitor with the opposite polarity. The energy slides back into the electric field, the current falls to zero, and the capacitor finds itself charged backwards, ready to discharge the other way. The cycle repeats indefinitely. This back-and-forth is the LC oscillation, and both the charge on the capacitor and the current in the loop vary sinusoidally in time.
Twice each cycle all the energy is electric (capacitor fully charged, current zero) and twice each cycle all of it is magnetic (capacitor empty, current maximum). In between, the two fields share it. With no resistor there is nowhere for the energy to go, so the oscillation never dies — the electrical analogue of a frictionless pendulum.
The Electrical–Mechanical Analogy
The LC oscillation is not merely like a block on a spring — mathematically it is the very same problem. For the spring, energy alternates between kinetic \(\tfrac{1}{2}mv^{2}\) and potential \(\tfrac{1}{2}kx^{2}\). For the circuit, it alternates between magnetic \(\tfrac{1}{2}Li^{2}\) and electric \(q^{2}/2C\). Lining up the two energies term by term reveals the dictionary that translates one system into the other.
The block oscillates at \(\omega = \sqrt{k/m}\). Make the substitutions \(k \to 1/C\) and \(m \to L\) and you instantly read off the angular frequency of the LC circuit — no new derivation needed.
LC Oscillations, Quantitatively
Let us make it rigorous. With no resistor or source, the total energy \(U = U_E + U_B\) is constant, so \(dU/dt = 0\). Writing \(U = q^{2}/2C + \tfrac{1}{2}Li^{2}\) and using \(i = dq/dt\) gives the differential equation of motion.
Differentiating the charge gives the current, which leads the charge by a quarter cycle. The two amplitudes are linked through \(\omega\).
Damped Oscillations in an RLC Circuit
Real circuits always have some resistance. Insert \(R\) in series with \(L\) and \(C\) and the loop dissipates energy at the rate \(i^{2}R\). The total electromagnetic energy now decreases with time, and the oscillation gradually dies away — the exact electrical analogue of a pendulum slowed by air drag. Setting \(dU/dt = -i^{2}R\) yields the damped equation.
Alternating Current
To keep an oscillation going forever we must feed energy in to replace what resistance removes. Connect an external alternating emf \(\varepsilon = \varepsilon_m\sin\omega_d t\) — the kind produced by a rotating coil in a generator. After any initial transient dies out, the circuit settles into a steady forced (driven) oscillation: the current alternates at the driving angular frequency \(\omega_d\), not the circuit's natural frequency \(\omega\).
Three Simple Circuits
Drive each element alone with the source \(\varepsilon_m\sin\omega_d t\) and watch the phase relationship between the voltage across it and the current through it. A handy mnemonic is "ELI the ICE man": in an inductor (L) the emf E leads the current I, while in a capacitor (C) the current I leads the emf E.
Over a full cycle a pure capacitor or inductor returns to the source exactly as much energy as it took in — the average power they consume is zero. They merely shuttle energy back and forth, shifting the current's phase by \(\pm 90^\circ\). Only the resistor actually removes energy from the circuit.
The Series RLC Circuit
Now put all three in series across the source. Because they carry the same current, we add their voltages as phasors — rotating arrows whose lengths are the voltage amplitudes and whose angles encode the phase. The resistor voltage is along the current; the inductor voltage runs \(90^\circ\) ahead; the capacitor voltage \(90^\circ\) behind. Adding them like vectors and matching the source amplitude gives the current amplitude and phase in one stroke.
Setting \(\omega_d L = 1/\omega_d C\) gives the resonance condition \(\omega_d = 1/\sqrt{LC} = \omega\). At resonance the impedance collapses to its minimum \(Z = R\), the current peaks at \(I = \varepsilon_m/R\), and the current is exactly in phase with the emf. This sharp, frequency-selective response is precisely how a radio tunes to one station and rejects the rest.
Power in Alternating-Current Circuits
Energy is dissipated only in the resistor, and the instantaneous rate \(i^{2}R\) rises and falls through each cycle. What matters in practice is the average rate. Because the average of \(\sin^{2}\) over a cycle is \(\tfrac{1}{2}\), it is natural to define root-mean-square values, which let AC quantities be handled with the familiar DC power formulas.
Transformers
Power is delivered most efficiently to a resistive load when its resistance is matched to the source, and transmitted over long lines most efficiently at very high voltage and low current (to minimize \(i^{2}R\) losses in the wires). The device that steps voltage up or down at will — using only the mutual induction of Chapter 30 — is the transformer: two coils wound on a common iron core, a primary of \(N_p\) turns and a secondary of \(N_s\) turns.
Putting It to Work
Problem. A \(1.5\,\mu\mathrm{F}\) capacitor is charged to \(57\,\mathrm{V}\) and then connected across a \(75\,\mathrm{mH}\) inductor (negligible resistance). Find (a) the angular frequency of oscillation and (b) the maximum current in the inductor.
Solution. The frequency follows from \(\omega = 1/\sqrt{LC}\). The charge amplitude is \(Q = CV\), and the current amplitude is \(I = \omega Q\).
Equivalently, energy conservation gives \(\tfrac{1}{2}LI^{2} = \tfrac{1}{2}CV^{2}\), so \(I = V\sqrt{C/L}\) — the same 0.25 A.
Problem. A series RLC circuit has \(L = 12\,\mathrm{mH}\), \(C = 1.6\,\mu\mathrm{F}\), and \(R = 1.5\,\Omega\). By what fraction does the charge amplitude fall after the first 50 complete oscillations?
Solution. The undamped frequency is \(\omega = 1/\sqrt{LC}\); since \(R\) is small, \(\omega' \approx \omega\), and the period is \(T = 2\pi/\omega\). The amplitude envelope is \(e^{-Rt/2L}\).
After 50 cycles only about \(6.6\%\) of the original charge amplitude survives — the ring is nearly gone.
Problem. A series RLC circuit has \(R = 200\,\Omega\), \(L = 230\,\mathrm{mH}\), and \(C = 16.0\,\mu\mathrm{F}\), driven by \(\varepsilon_m = 36.0\,\mathrm{V}\) at \(f_d = 60.0\,\mathrm{Hz}\) (\(\omega_d = 2\pi f_d \approx 377\,\mathrm{rad/s}\)). Find the reactances, the impedance, the current amplitude, and the phase constant.
Solution. Compute the reactances first.
Since \(X_C > X_L\) the circuit is capacitive: \(\phi\) is negative, so the current leads the source emf by about \(22^\circ\).
Problem. For the circuit of Example 3, find the average power dissipated.
Solution. All dissipation occurs in \(R\), so the cleanest route is \(P_{\text{avg}} = I_{\text{rms}}^{2}R\) with \(I_{\text{rms}} = I/\sqrt{2}\).
Check with the power-factor form: \(\varepsilon_{\text{rms}} = 36.0/\sqrt{2} = 25.5\,\mathrm{V}\), \(\cos\phi = \cos(21.6^\circ) = 0.930\), so \(P_{\text{avg}} = (25.5)(0.118)(0.930) \approx 2.8\,\mathrm{W}\) — agreement.
Problem. An ideal transformer steps a 240 V (rms) supply down to 12 V for a halogen lamp drawing 5.0 A on the secondary. The primary has 1000 turns. Find (a) the number of secondary turns and (b) the primary current.
Solution. Voltages scale with turns; power is conserved.
Stepping the voltage down by 20 steps the current up by 20 on the secondary side; the primary draws only 0.25 A, and \(V_p I_p = V_s I_s = 60\,\mathrm{W}\) as required.
Chapter Summary
\(q = Q\cos(\omega t+\phi)\) with \(\omega = \dfrac{1}{\sqrt{LC}}\); current amplitude \(I = \omega Q\).
\(U_E = \dfrac{q^{2}}{2C}\) and \(U_B = \tfrac{1}{2}Li^{2}\); their sum \(Q^{2}/2C\) stays constant.
\(q\!\leftrightarrow\!x\), \(i\!\leftrightarrow\!v\), \(L\!\leftrightarrow\!m\), \(1/C\!\leftrightarrow\!k\) — an electrical spring–mass system.
\(q = Q e^{-Rt/2L}\cos(\omega' t+\phi)\), \(\omega' = \sqrt{\dfrac{1}{LC} - \left(\dfrac{R}{2L}\right)^{2}}\).
\(X_C = \dfrac{1}{\omega_d C}\) (current leads), \(X_L = \omega_d L\) (current lags); resistor in phase.
\(I = \dfrac{\varepsilon_m}{Z}\), \(Z = \sqrt{R^{2}+(X_L-X_C)^{2}}\), \(\tan\phi = \dfrac{X_L-X_C}{R}\).
Peak current at \(\omega_d = \dfrac{1}{\sqrt{LC}}\); \(P_{\text{avg}} = \varepsilon_{\text{rms}} I_{\text{rms}}\cos\phi\), \(I_{\text{rms}} = I/\sqrt{2}\).
\(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\), \(I_s = I_p\dfrac{N_p}{N_s}\); matching \(R_{\text{eq}} = \left(\dfrac{N_p}{N_s}\right)^{2}R\).
Problems
For the free oscillator, lean on the energy bookkeeping: at any instant \(q^{2}/2C + \tfrac{1}{2}Li^{2}\) equals the constant total, and the peaks satisfy \(\tfrac{1}{2}CV^{2} = \tfrac{1}{2}LI^{2}\). For driven circuits, always compute \(X_L\) and \(X_C\) first, then \(Z\), then \(I\) and \(\phi\) — and remember that power lives only in the resistor.
- An LC circuit oscillates at 1.50 kHz with a 12.0 mH inductor. (a) Find the capacitance. (b) If the peak charge is \(2.90\,\mu\mathrm{C}\), find the peak current.
- In an LC circuit with \(L = 25\,\mathrm{mH}\) and \(C = 7.8\,\mu\mathrm{F}\), the peak charge on the capacitor is \(Q\). At the instant the energy is shared equally between the electric and magnetic fields, what is the charge on the capacitor in terms of \(Q\)?
- An LC circuit has a peak current of 1.50 A and stores a maximum of \(1.80\times10^{-3}\,\mathrm{J}\). Find (a) the inductance and (b) the oscillation frequency if \(C = 4.0\,\mu\mathrm{F}\).
- Show that the angular frequency of an LC circuit, \(\omega = 1/\sqrt{LC}\), follows directly from the spring result \(\omega = \sqrt{k/m}\) using the electrical–mechanical correspondence. State the substitutions you make.
- A damped RLC circuit has \(L = 220\,\mathrm{mH}\), \(C = 12.0\,\mu\mathrm{F}\), and \(R = 5.0\,\Omega\). (a) Find \(\omega'\) and compare it with \(\omega\). (b) After how many cycles does the charge amplitude drop to half its initial value?
- What value of resistance makes a circuit with \(L = 8.0\,\mathrm{mH}\) and \(C = 2.0\,\mu\mathrm{F}\) just fail to oscillate (critical damping)?
- A 50.0 Ω resistor, a 0.100 H inductor, and a \(20.0\,\mu\mathrm{F}\) capacitor are in series across a generator of amplitude 30.0 V at 60.0 Hz. Find (a) \(X_L\), (b) \(X_C\), (c) \(Z\), (d) the current amplitude, and (e) the phase constant. Does the current lead or lag?
- For the circuit of Problem 7, at what driving frequency does resonance occur, and what is the current amplitude there?
- A purely capacitive load draws 2.5 A (rms) from a 120 V (rms), 60 Hz line. Find (a) the capacitive reactance and (b) the capacitance.
- An inductor is connected across a 240 V (rms), 50 Hz source and draws 0.80 A (rms). (a) Find the inductive reactance. (b) Find the inductance. (c) What average power does the inductor dissipate?
- A series RLC circuit dissipates 75 W when driven at its resonant frequency by a source of rms emf 60 V. (a) Find the resistance. (b) Find the rms current. (c) What is the power factor?
- In a driven series RLC circuit the rms emf is 75.0 V, the rms current is 1.20 A, and the current lags the emf by \(35.0^\circ\). Find (a) the average power delivered and (b) the resistance of the circuit.
- An ideal transformer has 500 primary turns and 25 secondary turns. The primary is connected to a 120 V (rms) line. (a) Find the secondary voltage. (b) If the secondary delivers 3.0 A to a resistive load, find the primary current and the load resistance.
- A transmission line carries 80 kW to a town. (a) Compare the line loss \(I^{2}R\) when the power is sent at 240 V versus at 24 kV, for a line resistance of \(0.50\,\Omega\). (b) Explain in one sentence why the grid uses transformers to transmit at high voltage.