AC applied to a capacitor

Plug a fluorescent tube's ballast capacitor into the mains and something strange happens: current flows through it forever, cycle after cycle, even though a capacitor is, at heart, two metal plates with nothing but insulator between them. No wire ever touches the other plate. Yet an AC ammeter in series with that capacitor reads a steady, nonzero current — the same capacitor that would flatly refuse a torch battery's steady current the moment it charged up. The resolution is that alternating current never asks the capacitor to complete a circuit through the gap. It only asks the plates to keep filling and emptying, in step with the source pushing charge one way and then the other. Every quarter-cycle the capacitor is racing to catch up with a voltage that keeps changing beneath it, and that constant chase is what we read as current in the external wire. This article works out that chase quantitatively. From the simple fact that charge on a capacitor is $q = Cv$ , differentiating gives a current that turns out to lead the driving voltage by a full quarter-cycle — a $90°$ phase lead, the mirror image of what an inductor does. It defines capacitive reactance $X_C = 1/(\omega C)$ , the capacitor's frequency-dependent "resistance," and shows why it falls as frequency rises — exactly backwards from an inductor. And because current and voltage are a quarter-cycle out of step, we will see that a pure capacitor, like a pure inductor, dissipates zero average power: it is a wattless element, storing energy in its electric field on the way in and handing all of it back on the way out.

Key takeaways

A capacitor blocks steady DC once fully charged, but passes AC continuously by charging and discharging every half-cycle.
Current in a pure capacitor is $i = \dfrac{dq}{dt}$ with $q = Cv$ ; working this out shows the current leads the voltage by $90°$ ( $\pi/2$ rad) — mnemonic "ICE."
Capacitive reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ plays the role of resistance: $I_0 = V_0/X_C$ , $I_{rms} = V_{rms}/X_C$ .
$X_C$ falls as frequency rises — the opposite of $X_L = \omega L$ ; at $f = 0$ (DC), $X_C \to \infty$ , so a capacitor is an open circuit for steady current.
Because $\phi = 90°$ , $\cos\phi = 0$ , so the average power in a pure capacitor is zero over a full cycle — a wattless current.
Both results (the $90°$ lead and $X_C$ ) are the building blocks used later for the phasor treatment of the series LCR circuit and resonance.
Key idea: differentiating q = Cv for a sinusoidal voltage shows current leads voltage by 90 degrees in a pure capacitor, with peak/RMS current set by capacitive reactance X_C = 1/(omega C), which falls as frequency rises and makes average power zero.

The core insight

In a pure capacitor driven by $v = V_0\sin\omega t$ , the current $i = dq/dt$ works out to $I_0\sin(\omega t + \pi/2)$ — current leads voltage by $90°$ — with $I_0 = V_0/X_C$ where the capacitive reactance $X_C = 1/(\omega C)$ decreases as frequency increases, and the $90°$ phase gap makes the average power dissipated exactly zero.

Where it fits in the syllabus

This is the third of the chapter's basic single-element responses — after "AC applied to a resistor" and "AC applied to an inductor" — and it completes the trio that feeds directly into the phasor analysis of the series LCR circuit. CBSE's 2025-26 prescription treats the LCR circuit purely via phasors, not by solving a differential equation, so the current-leads-voltage result here is meant to be read off the $i = dq/dt$ calculation and then used as a phasor rule, not carried forward as calculus. Everything below — $X_C$ , the $90°$ lead, and zero average power — sits squarely inside the prescribed scope; nothing here reaches into rectifier circuits or three-phase AC, which stay out of bounds.

The physical picture

Take an ideal (resistanceless) capacitor of capacitance $C$ connected straight across an AC source that supplies $v = V_0\sin\omega t$ . At any instant the charge sitting on the capacitor's plates is fixed by the instantaneous voltage across it, because a capacitor's defining relation $q = Cv$ holds at every moment, not just in a steady DC state:

q = Cv = CV_0\sin\omega t

where: $q$ is the instantaneous charge (coulomb), $C$ is the capacitance (farad), and $v$ is the instantaneous source voltage (volt). Physically this says the plates track the source voltage exactly and instantly — whatever voltage the source is pushing right now, the capacitor holds exactly that much charge divided by $C$ . Current is charge in motion, so the current feeding those changing plates is the rate at which $q$ changes:

i = \frac{dq}{dt} = \omega C V_0 \cos\omega t = I_0\sin\left(\omega t + \frac{\pi}{2}\right)

where: $I_0 = \omega C V_0$ is the peak current, and the rewriting of $\cos\omega t$ as $\sin(\omega t + \pi/2)$ makes the phase relationship explicit. The physical reading: current has to run ahead of voltage because charge must arrive on the plates before the potential difference reflects it — the plates cannot show a voltage until charge has already flowed onto them. So the current waveform peaks a quarter-cycle before the voltage waveform does; current leads voltage by $90°$ in a pure capacitor, the mirror image of the inductor, where current lags. Comparing peak values gives the capacitor's counterpart to Ohm's law:

I_0 = \frac{V_0}{X_C}, \qquad X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

where: $X_C$ is the capacitive reactance, measured in ohms just like resistance, but — unlike resistance — it depends on frequency and stores rather than dissipates energy. Because $X_C \propto 1/f$ , raising the frequency shrinks $X_C$ : a capacitor opposes fast-changing voltage less and less. At the DC limit $f \to 0$ , $X_C \to \infty$ , recovering the everyday fact that a capacitor is an open circuit once fully charged by a steady source. At very high frequency, $X_C \to 0$ and the capacitor behaves almost like a short circuit — it barely resists the current at all. The same relation holds between RMS values, $I_{rms} = V_{rms}/X_C$ , since RMS is just the peak divided by $\sqrt2$ on both sides.

Figure

A graph of capacitive reactance against frequency showing a curve that falls steeply from a very large value near zero frequency toward zero at high frequency, alongside a phasor diagram showing the current phasor ninety degrees ahead of the voltage phasor

As with the inductor, the $90°$ phase gap means $\cos\phi = \cos 90° = 0$ , so the average power dissipated in a pure capacitor over a full cycle is zero. The capacitor spends the first quarter-cycle drawing charge in and storing energy in its electric field, then the next quarter-cycle giving that same energy straight back to the source as it discharges — nothing is lost as heat. This, like the inductor's case, is a wattless current: real current flows and the ammeter reads a genuine nonzero value, yet no net energy leaves the source.

Derivation ACurrent leads voltage by 90 degrees in a pure capacitor

The goal is to start from the capacitor's basic charge relation $q = Cv$ and the definition of current as $dq/dt$ , and show — using only differentiation of a sine function, no advanced calculus — that the resulting current waveform is shifted a quarter-cycle ahead of the driving voltage waveform, arriving at both the phase result and the reactance relation $I_0 = V_0/X_C$ that will be quoted throughout the rest of the LCR analysis.

Step 1. Write the source voltage as $v = V_0\sin\omega t$ , and use the capacitor's defining relation $q = Cv$ , valid at every instant for an ideal capacitor. Substituting gives $q(t) = CV_0\sin\omega t$ : the plate charge simply tracks the source voltage, scaled by $C$ .

Step 2. Differentiate to get the current. Since $i = dq/dt$ and the derivative of $\sin\omega t$ is $\omega\cos\omega t$ , $i = CV_0 \cdot \omega\cos\omega t = \omega C V_0\cos\omega t$ . This is the instantaneous current flowing in the connecting wires — remember no charge actually crosses the gap between the plates; this current is charge arriving on one plate and leaving the other as the source pushes and pulls.

Step 3. Rewrite the cosine as a shifted sine so it can be compared directly with $v = V_0\sin\omega t$ . Using the identity $\cos\theta = \sin(\theta + \pi/2)$ , $i = \omega C V_0\sin(\omega t + \pi/2)$ . Defining the peak current $I_0 = \omega C V_0$ , this is $i = I_0\sin(\omega t + \pi/2)$ — the same sinusoidal shape as the voltage, but shifted forward by a phase angle of $\pi/2$ radians, i.e. $90°$ .

Step 4. Read off both results. Comparing the phases of $v = V_0\sin\omega t$ and $i = I_0\sin(\omega t + \pi/2)$ shows current is ahead of voltage by $90°$ — it leads. Comparing the amplitudes, $I_0 = \omega C V_0$ , i.e. $I_0 = V_0/X_C$ with $X_C = 1/(\omega C)$ defined as the capacitive reactance. Both the phase-lead statement and the reactance formula follow from the same two-line differentiation, which is why CBSE is comfortable asking for this derivation even though the full LCR circuit later is treated only by phasors.

Worked examples

Board level - reactance, RMS current and power for a 15 microfarad capacitor

A capacitor of $C = 15\ \mu\text{F}$ is connected to an AC source with RMS voltage $V_{rms} = 200\ \text{V}$ and frequency $f = 50\ \text{Hz}$ . Find (a) the capacitive reactance, (b) the RMS current, and (c) the average power dissipated.

Step 1. Find the angular frequency: $\omega = 2\pi f = 2\pi(50) = 314\ \text{rad/s}$ .

Step 2. Find the capacitive reactance, converting $C$ to farads first ( $15\ \mu\text{F} = 15\times10^{-6}\ \text{F}$ ): $X_C = \dfrac{1}{\omega C} = \dfrac{1}{314 \times 15\times10^{-6}} = \dfrac{1}{4.71\times10^{-3}} \approx 212.3\ \Omega$ .

Step 3. Find the RMS current: $I_{rms} = \dfrac{V_{rms}}{X_C} = \dfrac{200}{212.3} \approx 0.942\ \text{A}$ .

Step 4. Find the average power: $P_{avg} = V_{rms}I_{rms}\cos90° = 200 \times 0.942 \times 0 = 0\ \text{W}$ . Answer: $X_C \approx 212.3\ \Omega$ , $I_{rms} \approx 0.942\ \text{A}$ , and $P_{avg} = 0\ \text{W}$ — nearly a full ampere of current flows, yet no net power is dissipated, because a pure capacitor stores and returns energy rather than consuming it.

Competitive level - working backward from current to capacitance, then to stored charge

A pure capacitor allows an RMS current of $0.5\ \text{A}$ to flow when connected to a $220\ \text{V}$ (RMS), $50\ \text{Hz}$ AC supply. Find (a) its capacitance $C$ , and (b) the peak charge stored on the capacitor at the instant the voltage is at its peak.

Step 1. Find the reactance from the given RMS values: $X_C = \dfrac{V_{rms}}{I_{rms}} = \dfrac{220}{0.5} = 440\ \Omega$ .

Step 2. Relate $X_C$ to $C$ using $X_C = \dfrac{1}{2\pi f C}$ , and solve for $C$ : $C = \dfrac{1}{2\pi f X_C} = \dfrac{1}{2\pi(50)(440)} = \dfrac{1}{138{,}230} \approx 7.23\times10^{-6}\ \text{F} = 7.23\ \mu\text{F}$ .

Step 3. Find the peak voltage and peak charge. $V_0 = \sqrt2\,V_{rms} = \sqrt2(220) \approx 311.1\ \text{V}$ , so the peak charge is $q_0 = CV_0 = (7.23\times10^{-6})(311.1) \approx 2.25\times10^{-3}\ \text{C} = 2.25\ \text{mC}$ . Answer: $C \approx 7.23\ \mu\text{F}$ , and the peak charge on the plates is about $2.25\ \text{mC}$ . Timing note: under two minutes. The chain $X_C = V_{rms}/I_{rms} \to C = 1/(2\pi f X_C)$ is a direct substitution once RMS values are given; the only extra step for the peak-charge part is remembering $V_0 = \sqrt2\,V_{rms}$ before applying $q_0 = CV_0$ — a common shortcut examiners like to tack onto a routine reactance question to test whether RMS and peak are being conflated.

Common traps

Believing a capacitor "blocks" AC the way it blocks DC — a capacitor does pass AC continuously through repeated charging and discharging; it is steady DC alone that it blocks, and only once fully charged.
Reversing the phase direction relative to the inductor — in a capacitor, current leads voltage (mnemonic "ICE": I leads E in a Capacitor), the exact reverse of the inductor's "ELI" (E leads I in an inductor).
Assuming $X_C$ rises with frequency the way $X_L$ does — capacitive reactance falls as frequency rises, an inverse relationship, and it is easy to mix the two formulas up under exam pressure.
Forgetting to convert $\mu\text{F}$ or $\text{pF}$ to farads before substituting into $X_C = 1/(\omega C)$ , which throws the reactance off by a factor of $10^6$ or worse.
Quoting $P_{avg} = V_{rms}I_{rms}$ without the $\cos\phi$ factor — for a pure capacitor $\phi = 90°$ makes this exactly zero, even though both $V_{rms}$ and $I_{rms}$ are individually nonzero.

Common trapA capacitor genuinely carries AC current without any charge crossing the gap between its plates

Students often try to reconcile "current flows through the capacitor" with "the plates never touch" by imagining charge somehow leaking or tunnelling across the gap. Nothing crosses. What flows in the external circuit is ordinary conduction current — electrons moving in the wires — pushed by the source first onto one plate and pulled off the other, then reversed the next half-cycle. As one plate accumulates charge, an equal and opposite charge appears on the facing plate purely by electrostatic induction; the circuit current and the rate of change of that plate charge are numerically equal ( $i = dq/dt$ ) without a single charge carrier ever crossing the insulating gap. This is exactly why a pure capacitor cannot pass true DC: once the plates reach the source's steady voltage, $dq/dt$ becomes zero and the current stops, even though the wires remain perfectly connected.

You should be able to…

Derive, from $q = Cv$ and $i = dq/dt$ , that current leads voltage by $90°$ in a pure capacitor, and state the peak-current relation $I_0 = \omega C V_0$ .
Define capacitive reactance $X_C = 1/(\omega C)$ , explain why it falls as frequency rises, and identify the DC limit $X_C \to \infty$ .
Calculate peak or RMS current given peak or RMS voltage and $X_C$ , and invert the relation to find $C$ or $f$ from measured current and voltage.
Explain why the average power in a pure capacitor is zero over a full cycle, in terms of the $90°$ phase difference and the charge/discharge energy picture.
Contrast a capacitor's frequency dependence and phase behaviour with an inductor's (ICE versus ELI, falling versus rising reactance).
Read a reactance-versus-frequency sketch for a capacitor and identify the low-frequency and high-frequency limiting behaviours from its shape.

Ask the tutor: if you want a side-by-side phasor comparison of how $X_L$ and $X_C$ behave as frequency changes, or a walkthrough combining both into the series LCR circuit, ask the AI tutor to lay it out for you.

If no charge actually crosses the gap between the plates, why does the ammeter in the wire show a current at all?

Because current is defined by what happens in the wire, not inside the gap. Electrons genuinely move through the connecting wires and through the source, piling onto one plate and draining off the other, and this is real conduction current that any ammeter placed in the wire will register. The gap itself never carries current — no charge crosses it — but the rate at which charge builds up on the plate, $dq/dt$ , exactly equals the current flowing in the wire feeding it. So the ammeter is honestly reporting a nonzero current even though the capacitor, taken as a whole, is an open path.

Why does the current lead the voltage instead of the other way round, the way it does for an inductor?

It comes down to which quantity has to "arrive first." In a capacitor, the voltage across the plates is a consequence of charge having already gathered there — you cannot have a voltage until some charge has flowed in, so current necessarily runs ahead. In an inductor it is the opposite: the induced voltage opposes any change in current, so the voltage has to appear first to force the current to build up, meaning voltage leads and current lags. The two elements are mirror images of each other in this sense, which is exactly why the mnemonics ICE (capacitor: I before E) and ELI (inductor: E before I) are opposite.

Is this 90 degree phase lead the same physical size no matter what capacitance or frequency I use?

Yes, the phase angle is always exactly $90°$ for an ideal capacitor, regardless of $C$ or $f$ — that comes purely from differentiating a sine wave, which always shifts it by a quarter-cycle. What changes with $C$ and $f$ is the reactance $X_C = 1/(\omega C)$ , which sets how large the current is for a given voltage, not the phase relationship. So a bigger capacitor or a higher frequency gives a bigger current (smaller $X_C$ ), but the current still peaks exactly a quarter-cycle before the voltage does, every time.

My capacitor got warm during a lab experiment with AC — doesn't that contradict zero average power?

Real capacitors are not perfectly ideal, and this article treats the ideal case CBSE actually examines. In practice a real capacitor has some internal resistance in its leads and dielectric losses in the insulating material, which do dissipate a small amount of real power and show up as mild warming, especially at high current or high frequency. For the ideal, resistance-free capacitor of the syllabus, though, the theoretical average power really is zero — the warming you felt is a real-world imperfection, not a contradiction of the physics being taught here.