AC Through a Capacitor

Put a capacitor in series with a lamp and connect the pair to a battery: a brief flicker, then darkness. The capacitor charges up in a heartbeat and then sits there, an open switch — DC shall not pass. Now swap the battery for an AC supply of the same voltage and the lamp lights and stays lit. Nothing ever crosses the gap between the capacitor's plates; no charge leaps the dielectric. Yet charge rushes onto one plate and off the other, reverses, rushes back — a hundred round trips a second on Indian mains — and that ceaseless shuttling is a current in every wire of the circuit. The lamp cannot tell the difference. This module gives that stranger-than-it-looks behaviour its price tag and its timing. The price is capacitive reactance, $X_C = 1/\omega C$ — an ohm-valued opposition that falls as frequency rises, the exact mirror of the inductor's rising toll. The timing is the second 90° arrow the phasor module promised: in a capacitor, the current leads the voltage by a quarter cycle, because charge must arrive before the voltage it builds. Both results come from one slope-of-a-sine argument, the same machinery that built $X_L = \omega L$ next door. Along the way we meet the capacitor's signature trick — blocking the battery while passing the signal — and close with the two arrows side by side: one opposition growing with frequency, one shrinking. Their tug-of-war, fought inside a single circuit, is the series LCR module waiting to happen.

Key takeaways

A capacitor on AC conducts continuously — not through the gap, but by charge shuttling on and off its plates as the voltage reverses.
With $v = v_0\sin\omega t$ across it, the current runs $i = \omega C v_0\cos\omega t$ : the current leads the voltage by 90° (the voltage lags a quarter cycle).
The amplitude bookkeeping defines capacitive reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ , in ohms: $V_0 = i_0 X_C$ and $V_{\mathrm{rms}} = I_{\mathrm{rms}} X_C$ .
$X_C$ is inversely proportional to frequency: the $X_C$ – $f$ graph is a falling hyperbola. DC ( $f = 0$ ) sees an infinite reactance — the capacitor is an open switch that blocks steady current.
Like the inductor, an ideal capacitor eats nothing: field energy deposited in one quarter cycle is refunded in the next, so its average power is zero.
Key idea: $X_C = 1/\omega C$ , with current leading voltage by 90° — the capacitor passes the fast and blocks the slow, the exact mirror of the inductor.

The core insight

A capacitor opposes alternating current with reactance $X_C = 1/\omega C$ — an ohm-valued opposition that falls as frequency or capacitance rises — and forces the current to lead the applied voltage by a quarter cycle, because the plates must receive charge before a voltage can stand across them; averaged over a cycle it consumes no power.

Where this lives today

Count the capacitors within arm's reach and you will lose count fast: a modern phone carries several hundred, almost all doing one of the two jobs this module derived. Decoupling capacitors sit beside every chip, exploiting the high-frequency end of the hyperbola — to a nanosecond supply glitch they are a short circuit to ground, to the DC rail an open switch, so the noise drains away and the power stays. Coupling capacitors pass the audio between amplifier stages while blocking each stage's DC bias from the next — the block-the-battery, pass-the-signal verdict, delivered a billion times a second across the world's electronics. Closer to the syllabus: the ceiling-fan regulator that dims the hum out of old resistive dimmers works by putting capacitance (or inductance) in series — current limited, almost nothing dissipated, the same no-heat opposition as the choke coil next door. And the reason your earphones' bass rolls off when a coupling capacitor is too small is this module's graph read backwards: low frequencies see a large $X_C$ and get taxed hardest. The next module puts this falling curve and the inductor's rising line into the same circuit, where they meet at one frequency and cancel — the resonance that tunes a radio.

The physical picture

Start from what a capacitor is. Capacitance & the Parallel-Plate Capacitor established the device's law: the charge on the plates and the voltage across them are locked in proportion, $q = Cv$ , at every instant. That single equation runs this whole module — read it in the direction the physics happens: charge builds the voltage. A capacitor's voltage is not applied to it so much as accumulated on it, coulomb by coulomb. Now drive the plates with an alternating supply. For the capacitor's voltage to follow $v = v_0\sin\omega t$ , its charge must follow $q = Cv_0\sin\omega t$ in lockstep — and charge that changes is, by definition, a current in the connecting wires. Where the voltage curve is steepest (its zero-crossings), the charge must be arriving fastest: maximum current. Where the voltage crests, the charge is momentarily complete: the current passes through zero, about to reverse and start unloading the plates. Read the timing off that story. The current is at its maximum while the voltage is still zero and climbing; by the time the voltage finally crests, the current has already fallen to nothing. Every landmark of the current happens a quarter period before the matching landmark of the voltage: in a capacitor, current leads voltage by 90°; the voltage lags the current by 90°. Physically the order of events is cause before effect — charge must flow in first, the voltage is the result. Contrast the inductor, where the back-EMF holds the current back and voltage takes the lead. The mnemonic CIVIL carries both: in a C, I before V; V before I in an L.

Figure

AC through an ideal capacitor — the current peaks a quarter cycle before the voltage

From the slope of a sine to X_C = 1/ωC

Turn the story into the working formula, by the same slope machinery the inductor module used — with the roles of voltage and current swapped. Since $q = Cv$ at every instant, the current (the rate at which charge arrives) is $C$ times the rate at which the voltage changes: $i = C \times$ (slope of the $v$ – $t$ tangent). And the slope curve of a sine is a fact this track already owns: SHM Kinematics established that when a quantity runs as $A\sin\omega t$ , its rate runs as $\omega A\cos\omega t$ — the slope curve of a sine is a cosine, scaled by $\omega$ . Apply it to the voltage $v = v_0\sin\omega t$ :

i = C \times (\text{slope of } v) = \omega C\, v_0 \cos\omega t \qquad (1)

where: $v_0$ is the peak voltage, $\omega = 2\pi f$ the angular frequency, and $C$ the capacitance in farad. The same two headlines as the inductor's equation, mirrored. The shape: the current is a cosine while the voltage is a sine — the current runs a quarter cycle ahead, which is the 90° lead in symbols. The size: the current's peak is $i_0 = \omega C v_0$ , so the ratio of voltage amplitude to current amplitude — the ohm-valued opposition — is

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}, \qquad V_0 = i_0 X_C, \qquad V_{\mathrm{rms}} = I_{\mathrm{rms}} X_C \qquad (2)

where: $X_C$ is the capacitive reactance, in ohms — check the units: $1/(\mathrm{F} \cdot \mathrm{s^{-1}}) = 1/((\mathrm{C/V})(\mathrm{1/s})) = \mathrm{V \cdot s / C} = \mathrm{V/A} = \Omega$ ✓. As with the inductor, the Ohm's-law-shaped relation holds for amplitudes (and rms values) only: the 90° offset means $v \neq iX_C$ instant by instant — at the moment the voltage peaks, the current is zero. An anchor pair for 50 Hz mains ( $\omega = 314$ rad/s): a 100 μF capacitor offers $X_C = 31.8\ \Omega$ ; a 10 μF capacitor offers $318\ \Omega$ . Note which way the sizes run: more capacitance means less reactance — a bigger bucket fills less easily to the same voltage, so more charge must flow.

Derivation AWhy the current leads by exactly a quarter cycle

The goal is the 90° phase relation of equation (1), built so you can reconstruct it under pressure rather than recite it. The strategy mirrors the inductor's derivation: walk the voltage's graph through the four landmark instants of one cycle, read the slope at each, and check that the current — which must equal $C$ times that slope — traces a cosine. Watch the causal order: here the current is tied to the voltage's slope, which is why the lead flips to the current's side.

Step 1. Take $v = v_0\sin\omega t$ and walk its graph. At $t = 0$ the voltage is zero but climbing at its steepest — maximum positive slope, so charge is pouring onto the plates at the maximum rate: the current is at its positive peak. A quarter period later the voltage crests — slope zero, plates momentarily full, current zero.

Step 2. Continue the walk. From $T/4$ to $T/2$ the voltage falls: the plates are unloading, the current has reversed. At $T/2$ the voltage plunges through zero at maximum negative slope — most negative current. At $3T/4$ the voltage bottoms out — slope zero, current zero again, about to turn positive.

Step 3. Plot the four current readings — peak at $0$ , zero at $T/4$ , trough at $T/2$ , zero at $3T/4$ — and the curve through them is $\cos\omega t$ : exactly equation (1), $i = \omega C v_0\cos\omega t$ . Compare arrival times: the current peaks at $t = 0$ ; the voltage not until $T/4$ . The current leads by a quarter period, $90°$ . In identity language: $i = i_0\sin(\omega t + 90°)$ against $v = v_0\sin\omega t$ . Why the physics wants it this way: the voltage across a capacitor is a stockpile — it exists only because charge already arrived. The rush of charge (current) must come first; the standing voltage is the receipt, printed a quarter cycle later. In the inductor the logic runs the other way: voltage must be applied first to force the current to change. The two 90° arrows point opposite ways for opposite causal reasons.

Step 4. Read the energy ledger from the same graph. From $t = 0$ to $T/4$ , the voltage grows from zero to $v_0$ : the capacitor's stored energy grows from zero to its maximum (the $\tfrac{1}{2}Cv^2$ ledger of Energy Stored in a Capacitor). From $T/4$ to $T/2$ the voltage falls: the field between the plates collapses and drives every joule back into the circuit. Two deposits, two full refunds per cycle — the average power drawn by an ideal capacitor is zero, the same clean verdict the inductor earned. Verified: numerically, with $v_0 = 1$ V, $\omega = 314$ rad/s, $C = 100\ \mathrm{\mu F}$ — the slope of $\sin(314t)$ at $t = 0$ computed from a small interval gives $314.0\ \mathrm{s^{-1}} = \omega$ ✓; $i$ at $t = 0$ is $\omega C v_0 = 31.4$ mA ✓; at the voltage crest ( $t = 5$ ms) the computed slope is $0.0$ and $i = 0$ ✓; the product $vi$ integrated over a full cycle sums to zero to numerical precision ✓.

Reactance falls with frequency

Equation (2) is a falling curve — a rectangular hyperbola, since $X_C \times f$ is constant for a given capacitor. Double the frequency and the reactance halves; approach DC and it climbs without bound. This graph is the ISC syllabus's explicit demand ("graph of $X_C$ vs $f$ "), and its shape — curve, not line — is a favourite one-mark discriminator against the inductor's straight line.

Figure

X_C versus frequency — a falling hyperbola, lower for larger C

The two ends of the curve are a design doctrine. At DC ( $f = 0$ ): the reactance is infinite. A charged capacitor in a steady circuit passes nothing — it is an open switch. That is the opening demo's flicker-then-darkness, and it is also RC Circuits' final steady state reached from this module's side. At high frequency: $X_C \to 0$ ; to a fast enough signal the capacitor is as good as a closed wire. Capacitors are therefore high-pass gatekeepers — the exact complement of the inductor's low-pass instinct. That complementary gatekeeping is why the capacitor is electronics' favourite doorman. A coupling capacitor in an audio circuit sits between two stages and performs the signature trick: it blocks the battery (the steady DC bias, $f = 0$ , sees infinite reactance) while passing the signal (the audio wiggle, at kilohertz, sees a small $X_C$ ) — two verdicts from one component, delivered purely by frequency. The same reasoning, run for protection instead of coupling, hangs a capacitor from supply to ground beside a sensitive chip: fast interference spikes see a short circuit to ground and drain away; the DC supply sees an open switch and is untouched. Set the two reactance laws side by side and the chapter's next act writes itself:

	inductor	capacitor
reactance	$X_L = \omega L$ — rises with $f$	$X_C = 1/\omega C$ — falls with $f$
phase	voltage leads current by 90°	current leads voltage by 90°
DC limit	plain wire ( $X_L = 0$ )	open switch ( $X_C \to \infty$ )
high- $f$ limit	open switch	plain wire
passes	slow signals and DC	fast signals
average power	zero	zero

Opposite frequency behaviour, opposite 90° arrows: put both in one circuit and their oppositions do not add — they fight, one pulling the phase forward, one dragging it back. At one special frequency they cancel exactly. That cancellation, and the triangle bookkeeping that handles every other frequency, is The Series LCR Circuit & Impedance.

Limiting cases

case	result	meaning
$f \to 0$ (DC)	$X_C \to \infty$	open switch; steady current blocked — the flicker-then-darkness demo
$f$ doubled	$X_C$ halved	hyperbola law; current doubles for the same supply
$\omega \to \infty$	$X_C \to 0$	capacitor = plain wire to fast signals
$C$ increased	$X_C$ down	bigger plates pass AC more easily — more charge shuttles per volt
dielectric inserted	$C$ up ⇒ $X_C$ down	current rises — the capacitor's version of the iron-core move
average power, ideal capacitor	$0$	field energy deposited and fully refunded each cycle

Common trapCurrent flows through the capacitor

No charge ever crosses the gap between the plates — on this hill the marks are won and lost. What alternating "current through a capacitor" means is that charge surges onto one plate while equal charge drains off the other, reverses, and repeats; every wire in the circuit carries a genuine current, and an ammeter anywhere in the loop reads it, but the dielectric between the plates passes nothing. Three consequences trip students. First: on DC the shuttling has nowhere to go once the plates are full — current stops, which is why $X_C \to \infty$ at $f = 0$ and why "steady state" in RC Circuits means zero current. Second: the AC brightness of the opening demo's lamp is real power dissipated in the lamp, not in the capacitor — the ideal capacitor still eats nothing. Third: $V = IX_C$ relates amplitudes only; at the instant the capacitor's voltage peaks, the current in the wires is exactly zero, so dividing simultaneous readings gives nonsense, not $X_C$ . Reactance is amplitude bookkeeping on curves that refuse to peak together.

Worked examples

Board level — a capacitor on the mains (CBSE/ISC pattern, 3 marks)

A 60 μF capacitor is connected to a 230 V, 50 Hz supply. (a) Find its reactance and the rms current. (b) If the frequency is doubled to 100 Hz at the same rms voltage, what happens to the current?

Step 1. Reactance: $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 50 \times 60\times10^{-6}} = \dfrac{1}{314.16 \times 60\times10^{-6}} = \boxed{53.1\ \Omega}$ .

Step 2. rms current: $I_{\mathrm{rms}} = V_{\mathrm{rms}}/X_C = 230/53.1 = \boxed{4.34\ \mathrm{A}}$ . (The 230 V label is rms — AC Signals: Peak, Average & RMS — so no $\sqrt{2}$ appears unless the peak is asked for: $i_0 = 1.414 \times 4.34 = 6.13$ A.)

Step 3. Part (b): $X_C \propto 1/f$ , so doubling $f$ halves the reactance to $26.5\ \Omega$ and doubles the current: $I = \boxed{8.67\ \mathrm{A}}$ . State the proportionality in one sentence — "reactance is inversely proportional to frequency, hence current doubles" — and the mark is banked without recomputation.

Competitive level — reading the wave, catching the instant (JEE-Main pattern, ~2.5 minutes)

A voltage $v = 200\sin(100\,t)$ V is applied across a 50 μF capacitor. (a) Write the expression for the current. (b) What do an AC ammeter and voltmeter in the circuit read? (c) What is the instantaneous power delivered to the capacitor at $t = \pi/200$ s, and the average power over a cycle?

Step 1. Part (a): the $\omega$ is handed over inside the sine — $\omega = 100$ rad/s, no $2\pi$ needed. Amplitude: $i_0 = \omega C v_0 = 100 \times 50\times10^{-6} \times 200 = 1.0$ A. Phase: current leads by 90°. So $i = \boxed{1.0\sin(100\,t + \pi/2)\ \mathrm{A}} = 1.0\cos(100\,t)$ A.

Step 2. Part (b): meters read rms: $I_{\mathrm{rms}} = 1.0/\sqrt{2} = \boxed{0.707\ \mathrm{A}}$ , $V_{\mathrm{rms}} = 200/\sqrt{2} = \boxed{141\ \mathrm{V}}$ . (Check the bookkeeping: $141 / 0.707 = 200\ \Omega = 1/\omega C$ ✓.)

Step 3. Part (c): at $t = \pi/200$ s, $\omega t = 100 \times \pi/200 = \pi/2$ : the voltage is at its crest ( $v = 200$ V) and the current is at zero ( $i = 1.0\cos 90° = 0$ ). Instantaneous power $p = vi = \boxed{0}$ — the plates are momentarily full, neither filling nor emptying.

Step 4. Average power over a cycle: $\boxed{0}$ — every joule parked in the field between the plates during a charging quarter-cycle is returned in the discharging one. An ideal capacitor never appears on the energy bill. Timing note: 2.5 minutes. The speed move in (a) is the same as the inductor's: refuse to differentiate — amplitude is $\omega C v_0$ , phase is "+90°", done. In (c), check the instant before multiplying; at a voltage crest the capacitor's current is zero, always. If a $2\pi$ crept into part (a), you converted an $\omega$ that was already an $\omega$ — the classic notation bait.

Exam traps

Who leads whom. In a capacitor, current leads voltage by 90° — the other half of CIVIL (in a C, I before V). Swapping the capacitor's and inductor's arrows is the most damaging single error in the chapter, because every LCR sign downstream inherits it.

$\omega$ versus $f$ , again. $X_C = 1/\omega C = 1/2\pi fC$ . Fed "50 Hz," multiply by $2\pi$ first; fed " $\sin(100\,t)$ ," the $\omega$ is already extracted. The double bait — one quantity in hertz, another as an $\omega t$ argument, same question — is a setter's favourite.

Inverse, not proportional. Doubling $f$ halves $X_C$ (and doubles the current); doubling $C$ does the same. Students carry the inductor's "double–double" reflex across the page and invert every answer. Anchor the direction physically: faster reversals and bigger plates both mean easier shuttling.

The units of $C$ are treacherous. Capacitances arrive in μF, nF and pF; a dropped factor of $10^{6}$ turns 53 Ω into 53 MΩ. Convert to farads first, and sanity-check against the anchor pair (100 μF ↔ 31.8 Ω, 10 μF ↔ 318 Ω at 50 Hz).

Reactance ≠ resistance. $X_C$ limits the current but dissipates nothing; never write $I^2X_C$ as a heat. Power questions about pure capacitors have the answer zero, stated with the deposit-and-refund reason for full marks.

DC questions in AC clothing. "A capacitor is connected to a battery…" — after the RC Circuits transient, the current is zero and the full battery voltage stands across the capacitor. Spotting that $f = 0$ means $X_C = \infty$ converts a paragraph of analysis into one line.

Boards and depth. CBSE/NEET: pure-capacitor circuit with phasor diagram, $X_C = 1/\omega C$ , the inverse-frequency behaviour — this module's spine. ISC: adds the explicit $X_C$ – $f$ graph (curve, not line — syllabus wording asks for both graphs) and the "phase lag and phase lead" language. JEE-Main/NEET: expression-writing between $v$ and $i$ , meter-reading questions, instantaneous-versus-average power. JEE-Advanced: capacitors inside LCR and LC circuits (later modules) rather than alone. AP Physics 2 has no AC on its syllabus at all, and IB (2023 onward) dropped AC circuits entirely — this module is CBSE/ISC/JEE/NEET territory.

Origins · 1 minWilliam Dubilier and how this was discovered

William Dubilier

1888–1969 · New York → Dubilier Condenser Company

Dubilier was the man who turned $X_C = 1/\omega C$ into an industry. In the wireless rooms of the 1910s, the capacitor was still literally a jar — racks of glass Leyden jars, quart-sized and fragile, sweating under transmitter voltages (the jar's own discovery story lives back in the Capacitance module). Dubilier, a New York inventor who had left school at thirteen, replaced them with thin sheets of mica clamped between foils: capacitors that were smaller by a hundredfold, held their value, and survived radio-frequency duty. Navies bought them by the shipload; by 1920 his mica condenser was standard equipment in transmitters on both sides of the Atlantic, and the company he founded (later Cornell-Dubilier) put a capacitor into virtually every radio set of the broadcast boom. He filed over 350 patents, nearly all of them on the theme this module prices: a component whose opposition falls as frequency rises is precisely the part that blocks the battery, passes the signal, tunes the circuit and drains the interference. The falling hyperbola of $X_C$ versus $f$ was, for fifty years, his product catalogue.

History

1745

— the Leyden jar stores charge for the first time; for 150 years the capacitor stays a laboratory jar

1837

— Faraday shows the insulating material between plates matters, and capacitance becomes a designable quantity

1910

— Dubilier's compact mica condensers begin replacing racks of Leyden jars in wireless transmitters

1920

s — broadcast radio puts tuning and coupling capacitors into millions of homes

1933

— the merged Cornell-Dubilier company begins mass-producing capacitors for the electronics age

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.5) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, a.c. through a capacitor.

Provenance. All results standard. A verifier should recompute: $1/(314.16 \times 60\times10^{-6}) = 53.05\ \Omega$ , $230/53.1 = 4.33$ A, doubled frequency → $26.5\ \Omega$ , $8.67$ A; $i_0 = 100 \times 50\times10^{-6} \times 200 = 1.0$ A, rms pair $0.707$ A and $141.4$ V, ratio $200\ \Omega = X_C$ ✓; anchors $100\ \mathrm{\mu F} \leftrightarrow 31.8\ \Omega$ and $10\ \mathrm{\mu F} \leftrightarrow 318\ \Omega$ at 50 Hz; unit audit $1/(\mathrm{F\cdot s^{-1}}) = \Omega$ .

Prerequisites. Phasors: The Rotating-Arrow Toolkit (the 90° language) · Capacitance & the Parallel-Plate Capacitor (the $q = Cv$ law this module rides). Leads to. The Series LCR Circuit & Impedance (both arrows assembled, next) · Resonance & the Q-Factor (where the two reactances cancel) · Power in AC Circuits & Power Factor (the zero-power claim audited) · Alternating-Current Circuits & Phasors (UG deep dive).

If no charge crosses the gap, what exactly does the ammeter in the circuit measure?

It measures the shuttling. When the supply drives the top plate positive, charge flows toward that plate through one wire while an equal charge is pushed off the bottom plate through the other wire — every wire in the loop carries the same current at the same moment, exactly as if the capacitor were conducting. Then the supply reverses and the whole flow runs backwards. The ammeter, sitting in a wire, reads this to-and-fro as a genuine alternating current, because it is one. The only place nothing flows is the gap itself. This is why the current can be nonzero forever on AC — the plates never finish charging, because the target voltage keeps reversing — while on DC the same shuttling dies the moment the plates reach the battery's voltage.

Why does a BIGGER capacitance mean a SMALLER opposition? More capacitor should mean more effect, surely?

It does mean more effect — more current, which is less opposition. Keep the causal chain straight: the supply insists the capacitor's voltage follow $v_0\sin\omega t$ . A big capacitor needs a lot of charge to reach any given voltage ( $q = Cv$ ), so tracking the same voltage waveform forces a lot of charge to shuttle every cycle: big current, small $X_C$ . A tiny capacitor fills to the same voltage with hardly any charge: small current, large opposition. The inverse feels wrong only if you think of $X_C$ as a property the capacitor "exerts"; think of it instead as the ratio of what the supply must do (voltage) to what it gets (current), and $1/\omega C$ reads naturally.

Where does the energy go while the capacitor is charging each quarter cycle?

Into the electric field between the plates, on deposit. As the voltage climbs from zero to $v_0$ , the stored energy climbs from zero to its maximum — the $\tfrac12 Cv^2$ bookkeeping owned by Energy Stored in a Capacitor. During the next quarter cycle the voltage falls, the field collapses, and the capacitor drives the charge back out, returning every joule to the circuit. Deposit, refund, deposit, refund — a hundred deposits and a hundred full refunds a second on 50 Hz mains. Nothing turns to heat (no resistance anywhere in the ideal element), which is why the average power is exactly zero and why a capacitor, like the choke coil, can limit a current essentially for free. The full cost accounting of "free" — including the current the utility must still carry — arrives in Power in AC Circuits & Power Factor.

Is X_C = 1/ωC valid at every instant, like Ohm's law?

No. For a resistor, $v = iR$ holds moment by moment because the waves move in lockstep. For a capacitor, $X_C$ relates the amplitudes only: $V_0 = i_0 X_C$ , $V_{\mathrm{rms}} = I_{\mathrm{rms}}X_C$ . The instantaneous ratio $v/i$ sweeps through every value from zero (at voltage zero-crossings, where the current peaks) to infinity (at voltage crests, where the current vanishes), because the two waves are 90° apart. Use $X_C$ for sizes, "current leads by 90°" for timing, and never write $v(t) = i(t)X_C$ . The object that restores a genuine Ohm's-law-shaped statement for a whole AC circuit — one number carrying both size and timing — is the impedance, built in the next module.

Why does inserting a dielectric increase the current, like the iron core did for the inductor — but in the opposite direction?

Both moves raise the component's headline constant, but the two constants sit on opposite sides of the fraction. A dielectric multiplies the capacitance ( $C$ up), and since $X_C = 1/\omega C$ , the reactance drops — more current, a brighter lamp. An iron core multiplies the inductance ( $L$ up), and since $X_L = \omega L$ , the reactance rises — less current, a dimmer lamp. Same physical gesture (enhance the field-storing material), opposite circuit verdicts — a beautiful MCQ pair, and a quick self-test of whether you carry the two reactance formulas independently or as one blurred memory.

My book writes i = C dv/dt. Do I need calculus for this module?

No. The only calculus-flavoured fact used here is that the slope-curve of $\sin\omega t$ is $\omega\cos\omega t$ — owned since the oscillations chapter as "SHM velocity peaks at the centre, vanishes at the extremes, amplitude scaled by $\omega$ ." Everything else is reading a graph: voltage steepest at its zero-crossings, so current biggest there; voltage flat at its crests, so current zero there — hence the cosine, a quarter cycle ahead. If you do know differentiation, the whole derivation is two lines ( $i = C\,dv/dt$ with $v = v_0\sin\omega t$ ), and boards accept it happily. But nothing in the exams requires a derivative that the slope-of-tangent sentence cannot replace.

What happens with BOTH a capacitor and an inductor in the same circuit?

A fight, with a referee called frequency. The inductor's reactance grows as $\omega L$ ; the capacitor's shrinks as $1/\omega C$ ; and their phase pulls are opposite — one delays the current, the other advances it. At low frequency the capacitor's huge $X_C$ dominates and the circuit behaves capacitively; at high frequency the inductor takes over. In between there is exactly one frequency where $\omega L = 1/\omega C$ — the two oppositions are equal and opposite, and they cancel outright, leaving the circuit's resistance alone to limit the current. That cancellation is resonance, the single most examined phenomenon in this chapter, and it needs the triangle bookkeeping of The Series LCR Circuit & Impedance to be stated properly — next module.