What a Capacitor Does in an AC Circuit
Consider an AC source $v = v_m\sin\omega t$ connected to a capacitor alone — a purely capacitive circuit. The behaviour is best understood by first contrasting it with the DC case. When a capacitor is connected to a DC source, current flows only for the short time required to charge the plates. As charge accumulates, the voltage across the plates rises and opposes the source, until the capacitor is fully charged and the current falls to zero. A capacitor therefore blocks steady DC.
With an AC source the situation is different. The source voltage reverses every half cycle, so the capacitor is alternately charged and discharged. Charge keeps moving back and forth through the connecting wires, which means a continuous alternating current flows in the circuit even though no charge ever crosses the gap between the plates. The capacitor does not completely prevent the flow of charge; it merely limits or regulates it.
The instantaneous voltage across the capacitor relates to its charge by $v = q/C$. Applying Kirchhoff's loop rule, the source voltage equals the capacitor voltage at every instant, so the charge on the plates follows the source sinusoid directly.
An AC source $\varepsilon = v_m\sin\omega t$ driving a single capacitor $C$. The capacitor charges and discharges each half cycle, sustaining an alternating current.
Deriving the Current and Its Phase
From the loop rule, the charge on the capacitor is
$$ \frac{q}{C} = v_m\sin\omega t \quad\Rightarrow\quad q = C\,v_m\sin\omega t. $$
The current is the rate of change of charge, $i = \dfrac{dq}{dt}$. Differentiating,
$$ i = \frac{d}{dt}\big(C\,v_m\sin\omega t\big) = \omega C\,v_m\cos\omega t. $$
Using $\cos\omega t = \sin\!\left(\omega t + \dfrac{\pi}{2}\right)$, this becomes
$$ i = i_m\sin\!\left(\omega t + \frac{\pi}{2}\right), \qquad i_m = \omega C\,v_m. $$
Comparing the current $i = i_m\sin(\omega t + \pi/2)$ with the source $v = v_m\sin\omega t$ shows that the current is $\pi/2$ ahead of the voltage. The current reaches its peak one-quarter of a period before the voltage does. This is the exact opposite of an inductor, where the current lags by $\pi/2$.
Current (coral) peaks a quarter-cycle before the voltage (teal). The current leads the voltage by $\pi/2$.
Capacitive Reactance $X_C$
Rewrite the current amplitude $i_m = \omega C\,v_m$ in the form
$$ i_m = \frac{v_m}{(1/\omega C)}. $$
Comparing this with $i_m = v_m/R$ for a pure resistor, the quantity $(1/\omega C)$ plays exactly the role that resistance plays in a resistive circuit. It is called the capacitive reactance and denoted $X_C$:
$$ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}, \qquad i_m = \frac{v_m}{X_C}. $$
Capacitive reactance has the same dimensions as resistance and its SI unit is the ohm ($\Omega$). It measures the extent to which the capacitor limits the AC current. The crucial difference from resistance is that $X_C$ depends on frequency — it is inversely proportional to both the frequency and the capacitance.
| Quantity | Resistor R | Capacitor C |
|---|---|---|
| Opposition | R (resistance) | X_C = 1/(ωC) (reactance) |
| SI unit | ohm (Ω) | ohm (Ω) |
| Frequency dependence | independent of f | ∝ 1/f |
| Current amplitude | i_m = v_m / R | i_m = v_m / X_C |
| Phase of i w.r.t. v | in phase (0) | leads by π/2 |
| Average power | I²R (dissipated) | zero |
"ICE" — and the four traps that follow from it
Use the mnemonic ICE: in a Capacitor, I (current) comes before E (EMF/voltage) — current leads. Students routinely flip this with the inductor case. Three companion errors cluster around it: treating $X_C$ as increasing with frequency (it decreases), assuming a capacitor passes DC (it blocks DC, since $X_C\to\infty$ as $f\to 0$), and thinking a pure capacitor consumes power (its average power is exactly zero).
ICE: Capacitor → I leads E by 90°. $X_C \downarrow$ as $f \uparrow$. Blocks DC, passes high $f$. Pure C → zero average power.
The mirror-image case is the inductor, where the current lags by $\pi/2$ and $X_L=\omega L$ grows with frequency. See AC Voltage Applied to an Inductor to lock in the symmetry.
The Phasor Diagram
A phasor is a rotating vector whose vertical projection gives the instantaneous value of a sinusoidal quantity. For a capacitor, the current phasor $\mathbf{I}$ sits $\pi/2$ ahead of the voltage phasor $\mathbf{V}$ as both rotate counter-clockwise with angular speed $\omega$. As they rotate, their vertical projections trace out the voltage and current waveforms of Figure 2.
The current phasor $\mathbf{I}$ leads the voltage phasor $\mathbf{V}$ by $90^\circ$ for a pure capacitor.
$X_C$ Versus Frequency
Because $X_C = 1/(2\pi f C)$, the reactance falls as frequency rises. At very low frequency the reactance is large, choking the current; in the DC limit $f\to 0$ it becomes infinite, which is the formal statement of "a capacitor blocks DC." At high frequency the reactance shrinks toward zero and the capacitor behaves almost like a plain wire. NIOS §19.3.2 summarises this as: capacitive reactance decreases with an increase in frequency and capacitance.
$X_C \propto 1/f$: high at low frequency (blocks DC), falling toward zero at high frequency.
Average Power Is Zero
The instantaneous power delivered to the capacitor is the product of instantaneous current and voltage:
$$ p_C = i\,v = i_m\cos(\omega t)\,v_m\sin(\omega t) = \frac{i_m v_m}{2}\sin(2\omega t). $$
Averaging over a complete cycle, the average of $\sin(2\omega t)$ is zero, so the average power is
$$ \overline{P_C} = \left\langle \frac{i_m v_m}{2}\sin(2\omega t)\right\rangle = 0. $$
Physically, energy is stored in the capacitor's electric field while it charges and is completely returned to the source while it discharges; over a full cycle there is no net energy transfer. The current that flows without dissipating energy is called a wattless current. This is identical in spirit to the inductor result — both pure reactive elements have a power factor of zero.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms and peak current. If the frequency is doubled, what happens to the reactance and the current?
Capacitive reactance: $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi(50)(15.0\times10^{-6})} \approx 212\,\Omega$.
rms current: $I = \dfrac{V}{X_C} = \dfrac{220}{212} \approx 1.04\ \text{A}$.
Peak current: $i_m = \sqrt{2}\,I = (1.41)(1.04) \approx 1.47\ \text{A}$, oscillating between $+1.47$ A and $-1.47$ A, and ahead of the voltage by $\pi/2$.
If the frequency is doubled, $X_C$ is halved and the current is doubled.
Worked Examples
A second NCERT illustration (Example 7.3) makes the DC-versus-AC distinction concrete using a lamp in series with a capacitor.
A lamp is connected in series with a capacitor. Predict the observation for DC and AC connections. What happens in each case if the capacitance is reduced?
With a DC source the capacitor charges and then no current flows, so the lamp does not glow; reducing $C$ changes nothing. With an AC source the capacitor offers reactance $1/(\omega C)$ and current flows, so the lamp shines. Reducing $C$ increases $X_C$, so the current falls and the lamp shines less brightly than before.
AC across a capacitor in six lines
- Charge $q = C v_m\sin\omega t$ gives current $i = i_m\sin(\omega t + \pi/2)$ — current leads voltage by $\pi/2$.
- Current amplitude $i_m = \omega C v_m = v_m/X_C$.
- Capacitive reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$, in ohms.
- $X_C \propto 1/f$: large at low $f$, infinite for DC (blocks DC), small at high $f$.
- Phasor: $\mathbf{I}$ is $90^\circ$ ahead of $\mathbf{V}$.
- Average power over a cycle is zero — a wattless current. Mnemonic: ICE (I leads E in a capacitor).