Part 3 · Chapter 31

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.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 70 min
i What you'll learn
  • 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 \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\).
Section 31-1

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.

Section 31-2

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.

📐
Energy is conserved, and it ping-pongs
In an ideal LC circuit the total electromagnetic energy \(U = U_E + U_B\) is constant; only its form changes.

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.

Section 31-3

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 correspondence between the two oscillators
\[ \tfrac{1}{2}Li^{2} \;\longleftrightarrow\; \tfrac{1}{2}mv^{2}, \qquad \frac{q^{2}}{2C} \;\longleftrightarrow\; \tfrac{1}{2}kx^{2} \]
Charge plays the role of position, \(q \leftrightarrow x\); current the role of velocity, \(i = dq/dt \leftrightarrow v = dx/dt\); inductance the role of mass (inertia), \(L \leftrightarrow m\); and the inverse capacitance the role of the spring constant, \(1/C \leftrightarrow k\). Every result for the spring carries straight over.

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.

Section 31-4

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.

The LC equation and its solution
\[ L\frac{d^{2}q}{dt^{2}} + \frac{1}{C}q = 0, \qquad q = Q\cos(\omega t + \phi), \qquad \omega = \frac{1}{\sqrt{LC}} \]
This is identical in form to the spring equation \(m\,\ddot{x} + kx = 0\). Here \(Q\) is the charge amplitude and \(\phi\) a phase set by the initial conditions. The natural angular frequency depends only on \(L\) and \(C\) — not on how much charge you started with.

Differentiating the charge gives the current, which leads the charge by a quarter cycle. The two amplitudes are linked through \(\omega\).

Current and the two energies
\[ i = \frac{dq}{dt} = -\omega Q\sin(\omega t + \phi), \qquad U_E = \frac{Q^{2}}{2C}\cos^{2}(\omega t+\phi), \qquad U_B = \frac{Q^{2}}{2C}\sin^{2}(\omega t+\phi) \]
The current amplitude is \(I = \omega Q\). Because \(\cos^{2}+\sin^{2}=1\), the two energies always add to the constant \(Q^{2}/2C\); each individually oscillates at twice the circuit frequency, since energy peaks twice per cycle.
Section 31-5

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.

Damped RLC oscillation
\[ L\frac{d^{2}q}{dt^{2}} + R\frac{dq}{dt} + \frac{1}{C}q = 0, \qquad q = Q\,e^{-Rt/2L}\cos(\omega' t + \phi) \]
The charge amplitude decays exponentially with the envelope \(e^{-Rt/2L}\). The same equation governs a block on a spring with a drag force \(-b\,v\), with \(R \leftrightarrow b\). For light damping the oscillation persists for many cycles before fading.
Shifted angular frequency
\[ \omega' = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^{2}} = \sqrt{\omega^{2} - \left(\frac{R}{2L}\right)^{2}} \]
Damping lowers the frequency slightly below the undamped value \(\omega = 1/\sqrt{LC}\). When \(R\) is small the correction is negligible and \(\omega' \approx \omega\). If \(R\) grows large enough that the term under the root turns negative, the circuit no longer oscillates at all — it is overdamped.
Section 31-6

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 \(\omega\).

Driving emf and the steady-state current
\[ \varepsilon = \varepsilon_m\sin\omega_d t, \qquad i = I\sin(\omega_d t - \phi) \]
The current has the same frequency as the source but generally lags or leads it by a phase constant \(\phi\). Our task is to find the current amplitude \(I\) and the phase \(\phi\) in terms of \(R\), \(L\), \(C\), and \(\omega_d\). We build up to it one element at a time.
Section 31-7

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.

Resistive load — voltage and current in phase
\[ v_R = V_R\sin\omega_d t, \qquad i_R = \frac{V_R}{R}\sin\omega_d t \]
For a pure resistor the current is in step with the voltage; their phase difference is zero. The resistance \(R\) relates the two amplitudes through \(V_R = I_R R\), with no frequency dependence.
Capacitive load — current leads by 90°; capacitive reactance
\[ i_C = \omega_d C\,V_C\sin\!\left(\omega_d t + 90^\circ\right), \qquad X_C = \frac{1}{\omega_d C}, \qquad V_C = I_C X_C \]
The current leads the voltage by a quarter cycle (the capacitor must charge before its voltage peaks). The capacitive reactance \(X_C\), in ohms, is large at low frequency (a capacitor blocks DC) and small at high frequency.
Inductive load — current lags by 90°; inductive reactance
\[ i_L = \frac{V_L}{\omega_d L}\sin\!\left(\omega_d t - 90^\circ\right), \qquad X_L = \omega_d L, \qquad V_L = I_L X_L \]
The current lags the voltage by a quarter cycle (the inductor resists the change in current). The inductive reactance \(X_L\), in ohms, grows with frequency — an inductor passes DC freely but chokes high-frequency current.
📐
Reactance is a frequency-dependent "resistance"
Each reactance relates a voltage amplitude to a current amplitude, but unlike resistance it dissipates no energy and depends on \(\omega_d\).

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.

Section 31-8

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.

Current amplitude and impedance
\[ I = \frac{\varepsilon_m}{\sqrt{R^{2} + (X_L - X_C)^{2}}} = \frac{\varepsilon_m}{Z}, \qquad Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \]
The impedance \(Z\) (in ohms) is the AC generalization of resistance: it combines the dissipative \(R\) with the net reactance \(X_L - X_C\) in quadrature. The quantity \(X_L - X_C\) can be positive (mainly inductive) or negative (mainly capacitive).
Phase constant of the current
\[ \tan\phi = \frac{X_L - X_C}{R} \]
If \(X_L > X_C\) the circuit is inductive and the current lags the emf (\(\phi > 0\)); if \(X_L < X_C\) it is capacitive and the current leads (\(\phi < 0\)); if \(X_L = X_C\) the circuit behaves purely resistively (\(\phi = 0\)).
📐
Resonance
The current amplitude is largest when the two reactances cancel, \(X_L = X_C\) — that is, when the driving frequency matches the circuit's natural frequency.

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.

Section 31-9

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.

rms values
\[ I_{\text{rms}} = \frac{I}{\sqrt{2}}, \qquad V_{\text{rms}} = \frac{V}{\sqrt{2}}, \qquad \varepsilon_{\text{rms}} = \frac{\varepsilon_m}{\sqrt{2}} \]
An rms current delivers the same average heating to a resistor as a steady DC current of the same value. This is why household voltages are quoted as rms: a "230 V" supply has an amplitude of about \(230\sqrt{2} \approx 325\,\mathrm{V}\).
Average power and the power factor
\[ P_{\text{avg}} = I_{\text{rms}}^{2}R = \varepsilon_{\text{rms}} I_{\text{rms}}\cos\phi \]
The factor \(\cos\phi\) is the power factor. It is largest (equal to 1) at resonance, where current and voltage are in phase and power transfer is most efficient. When the current is \(90^\circ\) out of phase (pure reactance), \(\cos\phi = 0\) and no net power is delivered. Utilities work hard to keep the power factor near unity.
Section 31-10

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.

Ideal transformer: voltage and current
\[ V_s = V_p\frac{N_s}{N_p}, \qquad I_s = I_p\frac{N_p}{N_s} \]
The same changing flux threads both coils, so each turn carries the same emf — voltage scales with the turns ratio. A step-up transformer (\(N_s > N_p\)) raises voltage but lowers current proportionally, since an ideal transformer conserves power: \(V_p I_p = V_s I_s\).
Impedance matching
\[ R_{\text{eq}} = \left(\frac{N_p}{N_s}\right)^{2} R \]
Seen from the primary, a load resistance \(R\) on the secondary looks like \(R_{\text{eq}}\). By choosing the turns ratio, a transformer can make a load appear to have whatever resistance best matches the source — the principle behind matching a loudspeaker to an amplifier.
Why the grid runs on AC. Transformers work only with a changing flux, so they are useless for steady DC — and that single fact decided the "war of the currents." Because AC voltages can be stepped up for efficient long-distance transmission and stepped back down for safe domestic use, alternating current became the worldwide standard for the power grid.
Worked Examples

Putting It to Work

1 Frequency and peak current of an LC circuit

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 \(I = \omega Q\).

ω = 1/√(LC)
\[ \omega = \frac{1}{\sqrt{(75\times10^{-3})(1.5\times10^{-6})}} \approx 2.98\times10^{3}\,\mathrm{rad/s} \]
I = ωQ = ωCV
\[ I = \omega C V = (2.98\times10^{3})(1.5\times10^{-6})(57) \approx 0.25\,\mathrm{A} \]

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.

2 How many oscillations before damping out

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 \(e^{-Rt/2L}\).

ω and T
\[ \omega = \frac{1}{\sqrt{(12\times10^{-3})(1.6\times10^{-6})}} \approx 7.22\times10^{3}\,\mathrm{rad/s}, \qquad T = \frac{2\pi}{\omega} \approx 8.70\times10^{-4}\,\mathrm{s} \]
Amplitude ratio after t = 50T
\[ \frac{Q(t)}{Q_0} = e^{-Rt/2L} = \exp\!\left[-\frac{(1.5)(50)(8.70\times10^{-4})}{2(12\times10^{-3})}\right] \approx e^{-2.72} \approx 0.066 \]

After 50 cycles only about \(6.6\%\) of the original charge amplitude survives — the ring is nearly gone.

3 Driven series RLC: impedance, current, and phase

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.

X_L = ω_d L; X_C = 1/(ω_d C)
\[ X_L = (377)(0.230) \approx 86.7\,\Omega, \qquad X_C = \frac{1}{(377)(16.0\times10^{-6})} \approx 166\,\Omega \]
Z = √(R² + (X_L − X_C)²); I = ε_m/Z
\[ Z = \sqrt{200^{2} + (86.7 - 166)^{2}} \approx 215\,\Omega, \qquad I = \frac{36.0}{215} \approx 0.167\,\mathrm{A} \]
tan φ = (X_L − X_C)/R
\[ \tan\phi = \frac{86.7 - 166}{200} \approx -0.397 \;\Rightarrow\; \phi \approx -21.6^\circ \]

Since \(X_C > X_L\) the circuit is capacitive: \(\phi\) is negative, so the current leads the source emf by about \(22^\circ\).

4 Average power delivered

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 \(I_{\text{rms}} = I/\sqrt{2}\).

P_avg = (I/√2)² R
\[ I_{\text{rms}} = \frac{0.167}{\sqrt{2}} \approx 0.118\,\mathrm{A}, \qquad P_{\text{avg}} = (0.118)^{2}(200) \approx 2.8\,\mathrm{W} \]

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.

5 A step-down transformer

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.

N_s = N_p (V_s/V_p)
\[ N_s = N_p\frac{V_s}{V_p} = 1000\times\frac{12}{240} = 50\ \text{turns} \]
I_p = I_s (N_s/N_p) = I_s (V_s/V_p)
\[ I_p = I_s\frac{V_s}{V_p} = (5.0)\frac{12}{240} = 0.25\,\mathrm{A} \]

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.

Review

Chapter Summary

LC oscillations

\(q = Q\cos(\omega t+\phi)\) with \(\omega = \dfrac{1}{\sqrt{LC}}\); current amplitude \(I = \omega Q\).

Energy sloshing

\(U_E = \dfrac{q^{2}}{2C}\) and \(U_B = \tfrac{1}{2}Li^{2}\); their sum \(Q^{2}/2C\) stays constant.

Mechanical analogy

\(q\!\leftrightarrow\!x\), \(i\!\leftrightarrow\!v\), \(L\!\leftrightarrow\!m\), \(1/C\!\leftrightarrow\!k\) — an electrical spring–mass system.

Damped RLC

\(q = Q e^{-Rt/2L}\cos(\omega' t+\phi)\), \(\omega' = \sqrt{\dfrac{1}{LC} - \left(\dfrac{R}{2L}\right)^{2}}\).

Reactances

\(X_C = \dfrac{1}{\omega_d C}\) (current leads), \(X_L = \omega_d L\) (current lags); resistor in phase.

Series RLC

\(I = \dfrac{\varepsilon_m}{Z}\), \(Z = \sqrt{R^{2}+(X_L-X_C)^{2}}\), \(\tan\phi = \dfrac{X_L-X_C}{R}\).

Resonance & power

Peak current at \(\omega_d = \dfrac{1}{\sqrt{LC}}\); \(P_{\text{avg}} = \varepsilon_{\text{rms}} I_{\text{rms}}\cos\phi\), \(I_{\text{rms}} = I/\sqrt{2}\).

Transformer

\(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\), \(I_s = I_p\dfrac{N_p}{N_s}\); matching \(R_{\text{eq}} = \left(\dfrac{N_p}{N_s}\right)^{2}R\).

Practice

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.

  1. 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.
  2. 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 \(Q\)?
  3. 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 \(C = 4.0\,\mu\mathrm{F}\).
  4. 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.
  5. 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?
  6. 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)?
  7. 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?
  8. For the circuit of Problem 7, at what driving frequency does resonance occur, and what is the current amplitude there?
  9. 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.
  10. 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?
  11. 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?
  12. 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.
  13. 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.
  14. 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.
Tip: three habits make these problems fall apart cleanly. First, for any LC question reach for energy before reaching for time: the totals \(\tfrac{1}{2}CV_{\max}^{2} = \tfrac{1}{2}LI_{\max}^{2} = Q^{2}/2C\) tie the peaks together without solving any differential equation. Second, in a driven circuit always lay out the trio \(X_L = \omega_d L\), \(X_C = 1/\omega_d C\), \(Z = \sqrt{R^{2}+(X_L-X_C)^{2}}\) in that order, and let the sign of \(X_L - X_C\) tell you immediately whether the current leads (capacitive) or lags (inductive). Third, for power use rms values and remember that only the resistor consumes energy — if a number for average power ever comes out of a capacitor or inductor, you have made a sign or phase error.