Corrected exercises: RC, RL and RLC circuits (203-SN2-RE)

At $t=0^{+}$, the capacitor is discharged: $u_{C}=0$, it behaves like a WIRE, and the whole voltage falls across $R$: $i(0)=\frac{E}{R}=\frac{12}{1500}=8.0$ mA, the current is at its MAXIMUM from the moment of closing. In steady state, the capacitor is full: no more charge flows, $i=0$, it behaves like an OPEN SWITCH, and $u_{C}=E=12$ V (no more drop across $R$).

Loop rule: $E=Ri+u_{C}$, with $i=\frac{dq}{dt}=C\frac{du_{C}}{dt}$, hence $RC\frac{du_{C}}{dt}+u_{C}=E$. Substituting $u_{C}=E(1-e^{-t/\tau})$: $\frac{du_{C}}{dt}=\frac{E}{\tau}e^{-t/\tau}$, so $\frac{RC}{\tau}Ee^{-t/\tau}+E-Ee^{-t/\tau}=E$, which requires $\frac{RC}{\tau}=1$: the solution works if and only if $\tau=RC$. The initial condition $u_{C}(0)=0$ is indeed satisfied.

$\tau=RC=(1500)(20\times 10^{-6})=0.030$ s $=30$ ms. Reading 1: $u_{C}(\tau)=12(1-e^{-1})\approx 7.6$ V, that is 63% of 12 V, the horizontal line on the figure. Reading 2: the tangent at the origin, with slope $\frac{E}{\tau}$, reaches the asymptote $E$ exactly at $t=\tau=30$ ms. Reading 3: $e^{-5}\approx 0.7$% remains at $t=5\tau=150$ ms, the right edge of the figure: charging is over to 99% and a bit more.

The battery delivers the total charge $Q=CE$ under the constant voltage $E$: it supplies $W=QE=CE^{2}=(20\times 10^{-6})(144)=2.88$ mJ. The capacitor stores only $\frac{1}{2}CE^{2}=1.44$ mJ: exactly HALF. The other half went to heat in the resistor, and this split does not depend on $R$: a small resistance dissipates through large, brief currents, a large one through small, long ones, but the bill is the same. Charging a capacitor at constant voltage always costs twice what is stored in it: an efficiency of 50%, incompressible.

$W=\frac{1}{2}CU_{0}^{2}=\frac{1}{2}(1200\times 10^{-6})(300)^{2}=54$ J: the energy of a litre of water lifted five and a half metres, stored in a component that fits in a pocket.

With no battery, the loop rule gives $Ri+u=0$ with $i=-C\frac{du}{dt}$ (the capacitor is emptying): $RC\frac{du}{dt}+u=0$. With $u=U_{0}e^{-t/\tau}$: $-\frac{RC}{\tau}U_{0}e^{-t/\tau}+U_{0}e^{-t/\tau}=0$, satisfied if $\tau=RC=(10)(1200\times 10^{-6})=12$ ms. Initial current: $I_{0}=\frac{U_{0}}{R}=\frac{300}{10}=30$ A: thirty amperes through a lamp.

$u=\frac{U_{0}}{2}$ when $e^{-t/\tau}=\frac{1}{2}$, that is $t_{1/2}=\tau\ln 2\approx 8.3$ ms. From any voltage $u_{1}$ reached at time $t_{1}$: $u(t_{1}+t_{1/2})=u_{1}e^{-t_{1/2}/\tau}=\frac{u_{1}}{2}$: the exponential halves every $8.3$ ms, wherever you start, exactly like radioactive decay. The 110 V line is $\frac{U_{0}}{e}\approx 0.37\times 300$ V: the voltage at $t=\tau=12$ ms, the 37% reading.

$P_{0}=U_{0}I_{0}=\frac{U_{0}^{2}}{R}=\frac{90000}{10}=9000$ W: nine kilowatts, the power of three kettles. On average over $5\tau=60$ ms: $P_{\text{avg}}=\frac{54}{0.060}=900$ W. Yet recharging the same capacitor, through a large resistance and a small current, takes several seconds at a few watts: the capacitor takes in energy slowly and gives it back abruptly. It has not created a single joule, it has only CONCENTRATED them in time: a power amplifier, never an energy amplifier.

The total CHARGE is conserved: the two capacitors form an isolated system, and no charge can escape from the circuit. The ENERGY has no reason to be conserved, since the circuit contains a resistor that dissipates. Initial charge: $Q=C_{1}U_{0}=(100\times 10^{-6})(100)=1.0\times 10^{-2}$ C. In steady state there is no more current, so the two capacitors are at the same voltage: $U_{f}=\frac{Q}{C_{1}+C_{2}}=\frac{1.0\times 10^{-2}}{200\times 10^{-6}}=50$ V, half the starting voltage.

Before: $W_{i}=\frac{1}{2}C_{1}U_{0}^{2}=\frac{1}{2}(100\times 10^{-6})(10^{4})=0.50$ J. After: $W_{f}=\frac{1}{2}(C_{1}+C_{2})U_{f}^{2}=\frac{1}{2}(200\times 10^{-6})(2500)=0.25$ J. Exactly HALF of the energy has disappeared, although not one charge has been lost. The reason lies in the square: the charge has spread over twice the capacitance, so the voltage has been halved, and the energy, which goes as $U^{2}$ for a given charge, has been halved again.

The calculation in b) used only conservation of charge and the final state: $R$ appears nowhere. A small resistance gives a violent, brief transfer, a large one a slow, weak transfer, but in both cases $\int Ri^{2}dt$ has the same value: it is the gap between the two states, fixed by the physics of the states, not by the path. In general, $W_{diss}=W_{i}-W_{f}=\frac{1}{2}C_{1}U_{0}^{2}-\frac{1}{2}\frac{(C_{1}U_{0})^{2}}{C_{1}+C_{2}}=\frac{1}{2}\frac{C_{1}C_{2}}{C_{1}+C_{2}}U_{0}^{2}$. We recognize the SERIES capacitance of the two capacitors. Check: $\frac{1}{2}(50\times 10^{-6})(10^{4})=0.25$ J.

$U_{f}=\frac{1.0\times 10^{-2}}{400\times 10^{-6}}=25$ V, $W_{f}=\frac{1}{2}(400\times 10^{-6})(625)=0.125$ J: only a quarter remains, the loss reaches 75%. The relative loss is $\frac{C_{2}}{C_{1}+C_{2}}$: it grows with $C_{2}$ and tends to 100% as $C_{2}\to\infty$ (pouring into an infinite reservoir loses everything), and to 0 as $C_{2}\to 0$ (transferring nothing loses nothing). There is therefore no interior maximum: the only way to limit the damage is to transfer LITTLE at a time, which is exactly what a charge pump does by multiplying small transfers.

No, a perfect wire saves nothing: the missing energy does not depend on $R$, it must go somewhere whatever happens. With $R\to 0$, every real circuit keeps a stray inductance: the setup becomes an LC circuit, the charge oscillates instead of settling, and the energy leaves as electromagnetic radiation (the circuit becomes a small antenna) until the oscillation dies out on the final state. To transfer WITHOUT loss, you must instead add a DELIBERATE inductor and open the switch at the right moment, after half a cycle: the coil then acts as an inertial buffer that carries the energy instead of burning it, exactly as in switching power supplies. The same mechanism is already at work in exercise 1: charging a capacitor from a FIXED voltage source always wastes half of the energy supplied, for the same deep reason, whatever the resistance.

Any increase in $i$ increases the flux of the coil's own field through itself: by Lenz's law, the induced emf $e=-L\frac{di}{dt}$ OPPOSES this increase. A sudden jump in current would require an infinite emf: the current therefore builds up gradually. At $t=0^{+}$: $i=0$, the inductor behaves like an OPEN SWITCH (the whole voltage is across it). In steady state: $\frac{di}{dt}=0$, no emf any more, and the ideal inductor is a plain WIRE: $I_{\infty}=\frac{E}{R}=\frac{12}{6.0}=2.0$ A. This is the exact mirror of the capacitor, which acts as a wire at the start and an open circuit at the end.

$[L]=$ V·s/A (an emf per rate of change of current) and $[R]=$ V/A: the quotient $\frac{L}{R}$ is in seconds. $\tau=\frac{L}{R}=\frac{0.30}{6.0}=50$ ms, and $i(t)=I_{\infty}(1-e^{-t/\tau})=2.0(1-e^{-t/0.050})$: the same exponential as the charging of the capacitor, with the same readings of $\tau$ on the curve (63%, tangent at the origin).

$i(\tau)=2.0(1-e^{-1})\approx 1.26$ A. Energy in steady state: $W=\frac{1}{2}LI_{\infty}^{2}=\frac{1}{2}(0.30)(2.0)^{2}=0.60$ J, stored in the magnetic field of the inductor.

Opening forces $i$ from 2.0 A to 0 in $\Delta t\approx 1$ ms: $|e|=L\frac{\Delta i}{\Delta t}\approx(0.30)\frac{2.0}{0.001}=600$ V, fifty times the battery voltage! This spike breaks down the air in the half-open switch: that is the spark, the inductor's revenge for having its current cut. Application: a car's ignition coil pushes the process to tens of thousands of volts to make the spark jump at the spark plug. Protection: a flyback (freewheeling) diode in parallel gives the current a path to fall back on and absorbs the spike (relays, motors, electronics).

Turn density: $n=\frac{500}{0.25}=2000$ turns/m. Field: $B=\mu_{0}nI=(4\pi\times 10^{-7})(2000)(3.0)\approx 7.54\times 10^{-3}$ T. Flux per turn: $\Phi=BA=(7.54\times 10^{-3})(4.0\times 10^{-4})\approx 3.02\times 10^{-6}$ Wb. Inductance: $L=\frac{N\Phi}{I}=\frac{(500)(3.02\times 10^{-6})}{3.0}\approx 5.03\times 10^{-4}$ H, about 0.50 mH. The current cancels: $\Phi$ is proportional to it, so the quotient keeps no trace of it. Inductance is a property of GEOMETRY, not of the operating conditions.

$B=\mu_{0}nI$, $\Phi=\mu_{0}nIA$, and $N=n\ell$: $L=\frac{N\Phi}{I}=\frac{(n\ell)(\mu_{0}nIA)}{I}=\mu_{0}n^{2}A\ell$. Check: $(4\pi\times 10^{-7})(2000)^{2}(4.0\times 10^{-4})(0.25)\approx 5.03\times 10^{-4}$ H, identical. Doubling $N$ at fixed length doubles $n$, so it QUADRUPLES $L$: we go to about 2.0 mH.

$W=\frac{1}{2}LI^{2}=\frac{1}{2}(5.03\times 10^{-4})(9.0)\approx 2.26\times 10^{-3}$ J. By the field: $u=\frac{B^{2}}{2\mu_{0}}=\frac{(7.54\times 10^{-3})^{2}}{2(4\pi\times 10^{-7})}\approx 22.6$ J/m³, and the inner volume is $A\ell=(4.0\times 10^{-4})(0.25)=1.0\times 10^{-4}$ m³, hence $W=uV\approx 2.26\times 10^{-3}$ J: the books balance exactly. The energy of an inductor is therefore not stored “in the wire”: it is stored in the FIELD, between the turns, exactly as that of a capacitor is stored in the electric field between its plates.

The flux created by the solenoid also passes through the second winding: $M=\frac{N_{2}\Phi}{I}=\frac{(100)(3.02\times 10^{-6})}{3.0}\approx 1.01\times 10^{-4}$ H, which is $\frac{N_{2}}{N_{1}}L$, one fifth of $L$. Voltage across the secondary: $|e_{2}|=M\frac{dI}{dt}=(1.01\times 10^{-4})(200)\approx 2.0\times 10^{-2}$ V. In the primary, self-induction gives $|e_{1}|=L\frac{dI}{dt}=(5.03\times 10^{-4})(200)\approx 0.10$ V, five times more: the ratio of the voltages is the ratio of the numbers of turns, which is precisely the principle of the transformer.

Two effects multiply. Adding turns first increases the field, hence the flux through EACH turn, in proportion to $N$; but this flux is then counted once per turn, which multiplies by $N$ again. Hence the $N^{2}$: the coil induces itself, and each new turn acts as both source and receiver. In a bifilar winding, the wire goes out and comes back right alongside itself: the two strands carry the same current in OPPOSITE directions, their fields cancel almost everywhere, the self flux is nearly zero and so is $L$. This is how to make a resistor that remains a pure resistor up to high frequencies, without reacting to changes in current.

Loop rule on the closed capacitor-inductor circuit: $u_{C}=L\frac{di}{dt}$ is balanced by nothing else; with $u_{C}=\frac{q}{C}$ and $i=-\frac{dq}{dt}$ (the capacitor is emptying), we get $\frac{q}{C}=-L\frac{d^{2}q}{dt^{2}}$, that is $\frac{d^{2}q}{dt^{2}}=-\frac{1}{LC}\,q$: the acceleration of the charge is proportional to the charge and of opposite sign, the exact signature of the harmonic oscillator, with $\omega_{0}^{2}=\frac{1}{LC}$.

$\omega_{0}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(0.050)(20\times 10^{-6})}}=\frac{1}{\sqrt{10^{-6}}}=1000$ rad/s. Frequency: $f_{0}=\frac{\omega_{0}}{2\pi}\approx 159$ Hz. Period: $T_{0}=2\pi\sqrt{LC}\approx 6.28$ ms: this is indeed the position of the vertical line on the figure, where the cosine closes its first cycle.

The total energy is that of the start: $W=\frac{1}{2}CU_{0}^{2}=\frac{1}{2}(20\times 10^{-6})(144)=1.44$ mJ. When the current is at its maximum, the capacitor is empty and ALL the energy is magnetic: $\frac{1}{2}LI_{\max}^{2}=W$, hence $I_{\max}=U_{0}\sqrt{\frac{C}{L}}=12\sqrt{\frac{20\times 10^{-6}}{0.050}}=0.24$ A. Consistency: $Q_{0}=CU_{0}=2.4\times 10^{-4}$ C and $\omega_{0}Q_{0}=(1000)(2.4\times 10^{-4})=0.24$ A, identical. The current peaks when $u_{C}=0$: at the quarter and three-quarter points of the period, each time the voltage crosses zero on the figure.

The correspondences: $q\leftrightarrow x$ (the displacement), $i\leftrightarrow v$ (the velocity), $L\leftrightarrow m$ (the INERTIA: the inductor resists changes in current as the mass resists changes in velocity), $\frac{1}{C}\leftrightarrow k$ (the STIFFNESS: a small capacitor is quickly “stretched”, like a stiff spring). The energies answer each other: $\frac{1}{2}Li^{2}\leftrightarrow\frac{1}{2}mv^{2}$ and $\frac{q^{2}}{2C}\leftrightarrow\frac{1}{2}kx^{2}$, and $\omega_{0}=\frac{1}{\sqrt{LC}}$ is the counterpart of $\omega=\sqrt{\frac{k}{m}}$. The LC circuit is a mass-spring system in which the energy swings between electric field and magnetic field twice per period.

The resistor dissipates part of the energy as heat, by Joule heating, on every back-and-forth of the charge: the total energy of the circuit decreases, and so does the amplitude. Time constant of the envelope: $\frac{2L}{R}=\frac{2(0.050)}{20}=5.0$ ms. Time to halve the amplitude: $t=\frac{2L}{R}\ln 2\approx 3.5$ ms: on the figure, the grey envelope goes from 12 V to 6 V in three and a half milliseconds.

$R<R_{c}$: UNDERDAMPED regime, the circuit oscillates under a decaying envelope (the figure). $R=R_{c}$: CRITICAL regime, the quickest possible return to zero without oscillating. $R>R_{c}$: OVERDAMPED regime, a slow return without oscillation. Here $R_{c}=2\sqrt{\frac{0.050}{20\times 10^{-6}}}=2\sqrt{2500}=100$ Ω.

$\frac{R}{2L}=\frac{20}{0.10}=200$ s$^{-1}$, so $\omega'=\sqrt{1000^{2}-200^{2}}=\sqrt{9.6\times 10^{5}}\approx 980$ rad/s and $T'=\frac{2\pi}{\omega'}\approx 6.41$ ms, against $T_{0}=6.28$ ms: a lengthening of barely 2%. Meanwhile, the amplitude has been halved EVERY 3.5 ms: damping kills the amplitude long before it touches the rhythm. This is why a pendulum running out of energy beats nearly the same second right to the end.

Critical damping is the return to rest that is fastest without a rebound: car shock absorbers (the body comes back without pumping), door closers (the door closes quickly but does not slam), the needles of measuring instruments (the galvanometer settles on the value without swinging around it). Less damping: you oscillate, the car bounces, the needle overshoots and comes back. More damping: no more rebound but a sluggish return, the door takes ten seconds to close. Critical is the exact optimum between the two faults.

$\omega=2\pi f=2\pi(60)\approx 377$ rad/s. $X_{L}=\omega L=(377)(0.10)\approx 37.7$ Ω. $X_{C}=\frac{1}{\omega C}=\frac{1}{(377)(40\times 10^{-6})}\approx 66.3$ Ω. The inductor opposes CHANGES in current: the higher the frequency, the faster the current changes, the more it reacts, hence $X_{L}$ increasing (at very high frequency, an inductor is an open circuit). The capacitor, for its part, must fill and empty: the higher the frequency, the less time it has to charge, the less voltage it opposes, hence $X_{C}$ decreasing (at very high frequency, a capacitor is a wire). In direct current it is exactly the reverse, as the steady state of the Part A exercises reminds us.