AC Voltage Applied to a Capacitor

A Capacitor on DC vs AC

Connect a capacitor to a DC source: current flows only for the brief time the capacitor charges. As charge builds, the plate voltage rises and opposes the current; once fully charged, the current stops. A capacitor blocks steady current.

Connect it to an AC source ($v = v_m\sin\omega t$): the capacitor is alternately charged and discharged as the current reverses every half cycle — it limits the current but never stops it. AC passes; DC doesn't.

With $q$ the charge, $v = q/C$ at every instant (Kirchhoff), so $q = Cv_m\sin\omega t$ and

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

$$\boxed{i_m = \frac{v_m}{X_C}, \qquad X_C = \frac{1}{\omega C}}$$

$X_C$ is the capacitive reactance: dimensions of resistance, unit ohm, and — note the contrast with $X_L$ — inversely proportional to frequency and capacitance.

Current Leads by 90 Degrees

Comparing $i = i_m\sin(\omega t + \pi/2)$ with $v = v_m\sin\omega t$: the current is $\pi/2$ AHEAD of the voltage. The current peaks a quarter period before the voltage does. On the phasor diagram, $\vec{I}$ runs 90 degrees ahead of $\vec{V}$ as both rotate.

Physical picture: current must flow first to deliver charge; only then does the plate voltage build up. Charge flow precedes voltage — the lead is causality, not magic.

Power: again zero on average

$$p_C = iv = \frac{i_m v_m}{2}\sin 2\omega t \quad\Rightarrow\quad \bar{P}_C = 0$$

The capacitor stores energy in its electric field for a quarter cycle, then hands it back — exactly like the inductor's magnetic sloshing, with the phase reversed.

Key Point: 'L lags, C leads' — and neither consumes average power. Only resistance dissipates.

Reading $X_C = 1/\omega C$ Like an Examiner

• DC limit ($\omega \to 0$): $X_C \to \infty$ — infinite opposition; the capacitor is an open switch to steady current. (This is why it 'blocks DC'.)
• High frequency ($\omega \to \infty$): $X_C \to 0$ — the capacitor is nearly a plain wire to fast AC.
• Bigger C, smaller $X_C$: more plate area 'absorbs' charge more easily; the current flows more freely.
• Graphically, $X_C$ vs $\nu$ is a rectangular hyperbola (contrast the straight line of $X_L$).

[NEET Important] NCERT's lamp-in-series-with-capacitor reasoning (Example 7.3): on DC the lamp does not glow at all (capacitor blocks); on AC it glows; reducing C increases $X_C$ and the lamp dims. Asked again and again, in every disguise.

[JEE Tip] $X_L \propto \nu$ rises while $X_C \propto 1/\nu$ falls: at some frequency they must cross — the seed of resonance (Section 5). Keep this picture; it organises the whole chapter.

Solved Examples

Example 1: The 15 microfarad capacitor (NCERT Example 7.4)

A 15.0 $\mu$F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms and peak currents. What happens if the frequency is doubled?

Solution:

1. Reactance: $X_C = \frac{1}{2\pi\nu C} = \frac{1}{2\pi \times 50 \times 15 \times 10^{-6}} = 212\ \Omega$.
2. rms current: $I = V/X_C = 220/212 = 1.04$ A.
3. Peak: $i_m = \sqrt 2 I = 1.47$ A, oscillating between +1.47 A and $-1.47$ A, ahead of the voltage by $\pi/2$.
4. Doubled frequency: $X_C$ halves (106 $\Omega$), so the current doubles (2.08 A).

Example 2: A smaller capacitor [NEET Numerical]

Find the reactance of a 5.0 $\mu$F capacitor on the 50 Hz mains.

Solution:

1. $X_C = \frac{1}{2\pi \times 50 \times 5 \times 10^{-6}} = \frac{1}{1.57 \times 10^{-3}}$.
2. Answer: $X_C \approx 637\ \Omega$ — three times the 15 $\mu$F value: smaller C, larger opposition.

Example 3: Writing i(t) completely [JEE Numerical]

The voltage $v = 311\sin(314\,t)$ V is applied to a 15 $\mu$F capacitor. Write the full current expression.

Solution:

1. $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 15 \times 10^{-6}} = 212\ \Omega$.
2. $i_m = v_m/X_C = 311/212 = 1.47$ A.
3. Current leads by $\pi/2$: $i = 1.47\sin\left(314\,t + \dfrac{\pi}{2}\right)$ A.

Example 4: Peak charge on the plates [JEE Numerical]

For the same circuit, find the maximum charge on the capacitor.

Solution:

1. Charge follows the voltage: $q = Cv$, so $q_m = Cv_m$.
2. $q_m = 15 \times 10^{-6} \times 311 = 4.7 \times 10^{-3}$ C.
3. Answer: about 4.7 mC, reached at each voltage peak — when the current is momentarily zero (the 90-degree story in one line).

Example 5: Frequency halved [NEET Numerical]

The 50 Hz source driving a capacitor is replaced by a 25 Hz source of the same rms voltage. What happens to the reactance and current?

Solution:

1. $X_C \propto 1/\nu$: halving the frequency doubles the reactance.
2. $I = V/X_C$ therefore halves.
3. Lower frequency, lazier charge swapping, weaker current — the exact opposite of an inductor's response.

Example 6: Lead time on the mains [NEET Numerical]

On 50 Hz mains, by how much time does the capacitor current peak before the voltage?

Solution:

1. Lead = quarter period = $T/4$.
2. $T = 20$ ms, so the lead is 5 ms.
3. Same magnitude as the inductor's lag — opposite sign.

Example 7: Lamp and capacitor (NCERT Example 7.3)

A lamp is in series with a capacitor. Predict the observations for DC and AC connections. What changes if C is reduced?

Solution:

1. DC: the capacitor charges briefly, then blocks all current — the lamp never glows. Reducing C changes nothing.
2. AC: the capacitor offers finite reactance $1/\omega C$; current flows and the lamp shines.
3. Reducing C raises $X_C$, lowering the current: the lamp dims.

Example 8: Why does the current lead?

Give the physical reason the capacitor current leads the voltage.

Solution:

1. Plate voltage exists only because charge has already arrived: $v = q/C$.
2. So the charging current must flow first; the voltage follows as charge accumulates.
3. Mathematically, $i = dq/dt$ differentiates $\sin\omega t$ into $\cos\omega t$ — an automatic 90-degree advance.

Example 9: Zero average power, mechanically

Show how the capacitor manages to carry current all cycle yet consume no average power.

Solution:

1. Quarter cycle 1: source charges the capacitor — energy flows into the electric field ($p > 0$).
2. Quarter cycle 2: the capacitor discharges back through the source ($p < 0$), returning every joule.
3. $p_C = \frac{i_mv_m}{2}\sin 2\omega t$ averages to zero — borrowing, never spending.

Example 10: The R-L-C phase scoreboard

Summarise the phase of current relative to voltage for pure R, L and C on AC, with the power verdict for each.

Solution:

1. R: in phase ($\phi = 0$) — average power $I^2R$ (the only dissipator).
2. L: current lags by 90 degrees — average power zero.
3. C: current leads by 90 degrees — average power zero. Memory hook: 'L lags, C leads, R pays the bills.'