Capacitors

The charge follows directly from the defining relation:

The energy:

Doubling the voltage to 40 V:

Charge doubles; energy quadruples. This is the same distinction that made power non-superposable in Set 11 — linear quantities and quadratic ones behave differently under scaling.

The three relations to keep straight, and the alternative forms of the energy:

All equivalent via \(q = Cv\). The last is worth noting: the energy is half the product of charge and final voltage, because the voltage rose from zero as the charge accumulated. Problem 17 shows where the other half goes.

A note on scale. A picofarad is tiny — 3 pF at 20 V stores 600 pJ, about the energy of a single visible photon multiplied by a few billion. Practical energy-storage capacitors are farads, twelve orders of magnitude larger.

The definition. A capacitor stores charge in proportion to the voltage applied:

with \(+q\) on one plate and \(-q\) on the other, so the net charge is always zero.

Current is the rate of flow of charge, so differentiating gives the element law directly:

assuming \(C\) constant — the linearity assumption again. A varactor diode has \(C\) depending on voltage, and none of Sets 9 to 13 applies to it.

The energy. Integrate the absorbed power from an uncharged state:

The change of variable from \(t\) to \(v\) is again the key step, and it is why the result depends only on the final voltage, not the path.

Does current pass through the dielectric? No — no charge crosses the gap. What happens is that charge accumulates on one plate and an equal amount leaves the other, so the currents in the two connecting wires are equal and the capacitor appears, from outside, to conduct.

Maxwell's account. The changing electric field between the plates constitutes a displacement current

exactly equal to the conduction current in the wires. This is what makes KCL hold at a capacitor plate: without the displacement term, current would appear to vanish at the dielectric, and Kirchhoff's current law — which Set 3 derived from charge conservation — would fail.

Differentiate, remembering that the chain rule brings down a factor of \(\omega = 6000\):

The phase. Writing the result as a cosine makes the relationship visible:

The current leads the voltage by a quarter cycle — the opposite of an inductor, where the voltage leads. A useful mnemonic: in a capacitor, current comes first; in an inductor, voltage does.

The amplitude ratio defines the capacitive reactance:

Note the inverse dependence: reactance falls with frequency, again the opposite of an inductor. Problem 15 develops this.

Sanity check by inspection. The current is largest where the voltage is changing fastest — at its zero crossings — and zero where the voltage is stationary, at its peaks. Setting \(t = 0\): the voltage is at its maximum of 10 V and the current is indeed zero.

Integrate:

The final voltage is 1 V, approached asymptotically as the current dies away. The time constant is \(1/3000 = 333\ \mu\text{s}\), so the capacitor is within 1% of its final value after about five of those — 1.7 ms.

The charge delivered is the total area under the current curve:

Consistent, and obtained without solving the integral as a function of time — often the quicker route when only the final value is wanted.

Reading the waveform. The current starts at 6 mA and decays; the voltage starts at zero and rises, most steeply at the beginning where the current is largest. Integration smooths: a discontinuous current would still give a continuous voltage, which is Problem 13's result.

First check the waveform is continuous — it must be, by Problem 13:

Consistent. The voltage rises to \(+50\ \text{V}\), falls to \(-50\ \text{V}\), then returns to zero.

The current is the slope times \(C\). The slopes are \(+50\), \(-50\) and \(+50\ \text{V/s}\):

A triangular voltage gives a rectangular current — differentiation turns straight-line segments into constants and corners into steps:

Check by charge. The net charge delivered over the whole waveform must be zero, since the voltage starts and ends at zero:

Note the duality with Set 16, Problem 7. There a trapezoidal current in an inductor gave a rectangular voltage; here a triangular voltage across a capacitor gives a rectangular current. The same picture with the axis labels exchanged.

Series capacitors carry the same current, so the 3 µF sees the full source current and can be treated on its own:

aFirst interval, \(0 < t < 1\):

bSecond interval, starting from that value:

Check both ends:

Confirm the final value by charge. The current waveform is a triangle peaking at 90 mA over 2 s, so the total charge is its area:

Note the magnitude. Thirty kilovolts across a 3 µF capacitor stores \(\tfrac12(3\times10^{-6})(30{,}000)^2 = 1.35\ \text{kJ}\) — enough to be lethal. Pushing 90 mC into a small capacitor produces voltages far beyond anything the source's own rating would suggest, which is why capacitor banks are treated with respect and always discharged before handling.

Parallel capacitors share a voltage and add:

So the pair may be replaced by a single 10 µF capacitor to find \(v\), and the branch currents recovered afterwards.

Integrate interval by interval, carrying each final value forward:

Parallel rule. Capacitors in parallel share a voltage, and KCL adds their currents:

Equivalently by charge: the same voltage puts \(C_1v\) on one and \(C_2v\) on the other, so the total charge is \((C_1+C_2)v\).

Series rule. Capacitors in series carry the same current, hence the same charge, and KVL adds their voltages:

Note this is the reverse of resistors and inductors. Capacitance behaves like conductance, so anywhere you would add resistances you add reciprocal capacitances.

Now apply them. The 20 µF and 5 µF in series:

That 4 µF is in parallel with the 6 µF and 20 µF, so they add:

Finally, in series with the 60 µF:

Two sanity checks that catch most errors in these problems: a series combination is always smaller than either capacitor, and a parallel combination is always larger than both. Here \(4 < 5\) and \(20 < 30\) for the series steps, and \(30 > 20\) for the parallel one.

Reduce first. The parallel pair combines by addition:

leaving three capacitors — 20, 30 and 60 µF — in series across the source.

The series equivalent:

The key step: the charge is common. Series elements carry the same current, so they accumulate the same charge:

Every one of the three series elements holds exactly 300 µC, whatever its capacitance.

The voltages then follow from \(v = q/C\):

Voltage divides inversely with capacitance — the smallest capacitor takes the largest voltage. Compare the resistive divider, where the largest resistance takes the largest share:

The charge on the parallel pair divides between its members in proportion to capacitance: \(40\ \mu\text{F} \times 5\ \text{V} = 200\ \mu\text{C}\) and \(20 \times 5 = 100\ \mu\text{C}\), totalling the required 300 µC.

What DC steady state means for a capacitor. All voltages are constant, so

Zero current for any voltage — the definition of an open circuit. In DC steady state, remove every capacitor. This is the exact dual of Set 16, Problem 13, where inductors became shorts.

Redraw without the capacitors. The 6 mA now divides between the 3 kΩ branch and the \(2 + 4 = 6\ \text{k}\Omega\) branch:

Current division with the opposite resistance on top, as in Set 10.

Each capacitor holds the voltage of the resistor it shunts:

The stored energies:

Eight times as much in the second, from twice the capacitance and twice the voltage — the quadratic dependence of Problem 1 at work.

Note what the capacitance did not affect. Neither \(C\) entered the current or voltage calculation; they matter only for the energy. In steady state the capacitor values are irrelevant to every voltage and current in the circuit — exactly as inductance was in Set 16.

Convert to SI units first — the commonest source of error here:

Note that \(1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\), not \(10^{-2}\) — areas scale as the square.

The capacitance:

Charge and energy at 100 V:

What the formula says about design. To increase capacitance: more area, less separation, or a higher permittivity. Each has a limit:

The breakdown limit is the binding one. Mica withstands about \(100\ \text{MV/m}\), so at \(d = 10^{-4}\ \text{m}\) the maximum voltage is

Halving \(d\) would double the capacitance but also halve the voltage rating — and since \(w = \tfrac12CV^2\), the stored energy would fall. Thinner is not automatically better.

The energy density is the invariant that reveals this. With \(E = V/d\) and volume \(Ad\):

Energy density depends only on the material's permittivity and the field it can withstand — not on the geometry at all. That is why capacitor energy density has improved so slowly: it is a materials problem, not a design one.

Charge is conserved, because no source is present and the two capacitors form an isolated island. Problem 13 will show that connecting them demands an impulsive current, and it is that impulse — equal and opposite in the two branches — which forces charge conservation:

After connection the two are in parallel and share that charge at a common voltage:

The energy accounting:

Sixty per cent of the stored energy has vanished, in a circuit containing nothing but two ideal capacitors and a wire.

The general result. With \(C_2\) initially uncharged:

Here \(6/10 = 60\%\), matching. The fraction depends only on the capacitance ratio — not on \(V_0\), and not on the resistance of the connecting wire.

Where the energy goes. The ideal model cannot say, but any real mechanism accounts for exactly the same 3 mJ:

The independence from \(R\) is the striking part, and it is the same phenomenon as Problem 17's missing half.

The proof mirrors Set 16, Problem 14 exactly. Across an interval of length \(\varepsilon\) about any instant \(t_0\):

If the current is bounded by \(I_{\max}\):

so \(v_C(0^+) = v_C(0^-)\) — the result used at every switching instant in Sets 18 and 19.

The condition for a jump is an impulsive current:

An impulse of strength \(Q\) coulombs delivers that charge instantaneously and steps the voltage by \(Q/C\). Nothing else will.

Which connections are impossible. Any that would force a voltage discontinuity:

Every "impossible" case is resolved the same way — an impulsive current flows, charge redistributes instantly, and energy is lost.

Contrast with the inductor, where the forbidden connection is the series one. The pattern is:

The general statement. Since the stored energy is a function of the state alone:

If \(v(t_2) = v(t_1)\) the integral is zero, whatever happened in between. The capacitor returns every joule it absorbed.

Applied to Problem 5's triangular waveform: the voltage starts and ends at zero, so the net energy absorbed over the four seconds is exactly zero — despite substantial currents flowing throughout.

The sign of the power tells the story within the interval:

Reconciling with Problem 12. There the energy was not lost in the capacitors — each ended with more energy than the pure-storage bookkeeping might suggest, and the deficit went elsewhere:

The capacitors themselves remain lossless. What dissipated the 3 mJ was the resistance, arc or radiation of the connecting path — elements the ideal model omitted, not the capacitors.

The distinction to hold onto: "the capacitor is lossless" is a statement about the element. "Energy was lost" in Problem 12 is a statement about the circuit, and the loss occurred in a component the idealisation had made invisible.

The current:

The current leads the voltage by a quarter cycle — largest where the voltage is changing fastest, zero where the voltage peaks.

The reactance:

A capacitor blocks DC and passes high frequencies — the exact opposite of the inductor, and the basis of coupling and bypass capacitors.

The average power is zero, as Problem 14 requires:

Again a sinusoid at twice the supply frequency, symmetric about zero.

The two elements side by side:

The opposite phase shifts are what make resonance possible. At the frequency where \(\omega L = 1/\omega C\) the two reactances cancel exactly, leaving a purely resistive circuit — Set 29's subject, and already visible in Set 16, Problem 18's self-resonant coil.

A real capacitor is three elements in series: the capacitance itself, the ESR of the plates, foil and electrolyte, and the ESL of the leads and internal winding. There is also a parallel leakage resistance, usually large enough to ignore except over long times.

The self-resonant frequency, where the capacitive and inductive reactances cancel:

At that frequency the impedance is purely the ESR — its minimum value, 0.1 Ω.

Tabulating the three contributions:

Above 113 kHz the component is an inductor. Its impedance rises with frequency, and it no longer does the job it was fitted for.

The design consequence is the standard practice of paralleling capacitors of different types. A 100 µF electrolytic handles low-frequency bulk energy; a 100 nF ceramic, with perhaps 1 nH of ESL, self-resonates near 16 MHz and covers the range where the electrolytic has become inductive:

Together they hold the impedance low across several decades, which is why every digital board carries both.

ESR also sets the ripple-current rating. A ripple current \(I_{\text{rms}}\) dissipates \(I_{\text{rms}}^2 \times \text{ESR}\) inside the capacitor, heating the electrolyte. Excess ripple is the usual cause of electrolytic failure — and ESR rises as the electrolyte dries, so the failure accelerates itself.

The energy delivered by the source. An ideal source holds \(V\) constant while charge \(Q = CV\) passes through it:

The energy stored in the capacitor at the end:

Exactly half. The capacitor's voltage rose from zero, so on average it was only at \(V/2\) while the charge flowed — which is the physical content of the factor \(\tfrac12\) noted in Problem 1.

The resistor's share. The charging current is \(i = (V/R)e^{-t/RC}\), so

The \(R\) cancels completely. A smaller resistor gives a larger, briefer current and dissipates precisely the same total energy.

The accounting closes:

Charging a capacitor from a constant-voltage source through any resistance is exactly 50% efficient.

Why reducing \(R\) does not help. It is tempting to think a superconducting wire would avoid the loss. It does not — as \(R \to 0\) the dissipation stays at \(\tfrac12CV^2\) while the time taken tends to zero, so the power diverges. With no resistance at all, the energy goes into radiation and the ringing of stray inductance, as in Problem 12.

How the loss is actually avoided: by not using a constant-voltage source. Charging through an inductor, or with a current source whose voltage tracks the capacitor's, can approach 100%. This is exactly what a switched-mode converter does, and why it beats a linear regulator.

The proof. Over one period \(T\), with \(v(T) = v(0)\):

The average current is zero — amp-second balance, the exact dual of Set 16's volt-second balance. Charge in must equal charge out over every cycle, or the voltage would drift.

Applying it to the smoothing capacitor. Between conduction peaks the rectifier delivers nothing and the capacitor alone supplies the load. For a full-wave rectifier at 100 Hz the discharge lasts roughly one half-period:

The charge removed in that interval, and the voltage drop it causes:

The estimate is conservative — the diode conducts for part of the cycle, so the true ripple is somewhat less — but it is the standard design figure.

The design relation in general:

Halving the ripple requires doubling \(C\), or doubling the frequency. This is why switch-mode supplies, running at tens of kilohertz rather than 100 Hz, need capacitors hundreds of times smaller for the same ripple.

The charge must balance over a cycle. The capacitor supplies 20 mC while discharging, so the rectifier must return exactly 20 mC during its brief conduction interval. If that interval is only 2 ms, the peak diode current is around 10 A — five times the load current, and the reason rectifier diodes are rated well above the apparent DC draw.

The capacitive prediction. Equal charge on both, so voltage divides inversely with capacitance:

The smaller capacitor takes the larger voltage, as Problem 9 established.

Why this is only the initial answer. Every real capacitor has a finite leakage resistance in parallel with it. At DC the capacitors are open circuits (Problem 10), so after the transient dies away the voltage division is determined entirely by those leakage resistances:

Leakage resistances are poorly controlled — they vary by a factor of several between nominally identical parts and change strongly with temperature and age.

The danger. If one leakage resistance happens to be four times the other, that capacitor takes 240 V of the 300 V. Should it be rated at 200 V, it fails — and its failure (usually to a short) puts the full 300 V across the survivor, which then fails too. Series strings fail cumulatively.

The remedy: balancing resistors. Fit a resistor across each capacitor, small enough to swamp the leakage:

With equal balancing resistors the voltage divides equally at 150 V each, regardless of leakage. A common rule is to size them to draw about ten times the worst-case leakage current.

The cost. Balancing resistors dissipate continuously. With 150 V across each and \(R_{\text{bal}} = 100\ \text{k}\Omega\):

A permanent standing loss, and the design trade is between tighter balancing and lower dissipation. They also usefully discharge the string when the supply is removed.

The dictionary, extending Set 8's duality to the storage elements:

Verifying it works. Set 16, Problem 16 and Problem 12 here are the same proof: two elements forced into the connection that would demand a discontinuity, resolved by an impulse, conserving the linear quantity and losing energy. Set 16's buck converter and Problem 18's ripple estimate are the two balance conditions. Nothing in either set had to be proved twice.

Where the duality is exact. Everywhere in the mathematics. The two element laws are identical under the substitution, so every algebraic consequence carries across without exception.

Where it breaks: the physics. The elements are not equally realisable.

This asymmetry is why active filters, switched-capacitor circuits and gyrators exist: it is often easier to simulate an inductor with capacitors and amplifiers than to build one.

And one deeper asymmetry. Magnetic energy is stored in the space where \(\mu\) is small — the air gap, not the iron, as Problem 11's figures showed. Electric energy is stored where \(\varepsilon\) is large — in the dielectric itself. The mathematics is symmetric; the materials are not.