Why a Capacitor Blocks Steady Current but Lets Alternating Current Through

Household current reverses direction many times a second, so its value keeps changing and simple averages give zero. Inductors and capacitors also push current out of step with voltage, which Ohm's law alone cannot describe.

This lesson covers peak, mean and RMS values, phasors for resistors, inductors and capacitors, and impedance and resonance in a series LCR circuit.

**For $I=I_0\sin \omega t$, the peak value is $I_0$, the mean over a half-cycle is $\frac{2I_0}{\pi }=0.637I_0$, and the RMS value is $\frac{I_0}{\sqrt{2}}=0.707I_0$.

Why RMS matters. The RMS value is the steady DC current that would produce the same heating in a resistor. Ammeters and voltmeters on AC read RMS values, and mains voltage is quoted as an RMS value.

Why the mean is over a half-cycle. Over a full cycle, the positive and negative halves cancel, so the mean is zero.

Worked example 1.** An alternating current has a peak of 5.0 A:

$$I_{\text{rms}}=\frac{5.0}{\sqrt{2}}=3.54\text{ A}{I}_{\text{mean}}=\frac{2\times 5.0}{\pi }=3.18\text{ A}$$

Worked example 2. For $V=311\sin (100\pi t)$ volts:

$$V_{\text{rms}}=\frac{311}{\sqrt{2}}=220\text{ V}f=\frac{100\pi }{2\pi }=50\text{ Hz}$$

Across a $44\Omega$ heater, $I_{\text{rms}}=5.0$ A and the power is $5.0^2\times 44=1100$ W.

An everyday example. An electric iron rated for mains supply heats exactly as it would on a steady DC supply equal to the RMS voltage.

The substance. Never use the mean value for heating — power depends on $I^2$, so only the RMS value gives the right answer.

**A phasor is a rotating arrow whose projection gives the instantaneous value; in a resistor current and voltage are in phase, in an inductor current lags voltage by $90^{\circ }$ with reactance $X_L=\omega L$, and in a capacitor current leads by $90^{\circ }$ with reactance $X_C=\frac{1}{\omega C}$.

The three elements:

- Resistor** — V and I in phase, $V=IR$
- Inductor — the back emf opposes changes in current, so I lags V, and $X_L$ rises with frequency
- Capacitor — charge must build before voltage appears, so I leads V, and $X_C$ falls with frequency

Worked example 1. A 0.10 H inductor on 220 V, 50 Hz:

$$X_L=2\pi \times 50\times 0.10=31.4\Omega I=\frac{220}{31.4}=7.0\text{ A}$$

Worked example 2. A 100 $\mu$F capacitor on the same supply:

$$X_C=\frac{1}{2\pi \times 50\times 100\times 10^{-6}}=31.8\Omega I=\frac{220}{31.8}=6.9\text{ A}$$

Steady DC has $\omega =0$, so $X_C$ becomes infinite and the capacitor blocks it, while $X_L$ becomes zero.

An everyday example. The capacitor in a ceiling fan passes alternating current to the second winding while shifting its phase, which gives the motor its starting turn.

The substance. A capacitor passes AC without any charge crossing the gap — the plates simply charge and discharge every half-cycle.

**In a series LCR circuit the impedance is $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ with phase angle $\tan \varphi =\frac{X_L-X_C}{R}$, and resonance occurs when $X_L=X_C$, at $f_0=\frac{1}{2\pi \sqrt{LC}}$, where the impedance is least and equals R.

Why voltages add as phasors.** $V_L$ and $V_C$ point in opposite directions, both at $90^{\circ }$ to $V_R$, so they partly cancel.

Worked example 1. With $R=30\Omega$, $X_L=80\Omega$ and $X_C=40\Omega$ on 220 V:

$$Z=\sqrt{30^2+(80-40{)}^2}=50\Omega I=\frac{220}{50}=4.4\text{ A}\tan \varphi =\frac{40}{30},\varphi =53^{\circ }$$

The current lags, because the circuit is mainly inductive.

Worked example 2 — resonance. With $L=0.20$ H and $C=50$ $\mu$F:

$$f_0=\frac{1}{2\pi \sqrt{0.20\times 50\times 10^{-6}}}=50.3\text{ Hz}$$

At resonance, a 30 $\Omega$ resistor and 220 V give the largest current, $220/30=7.3$ A. The quality factor $Q={\omega }_{0}L/R=2.1$ measures how sharp the peak is.

An everyday example. A wireless phone charger tunes its coil circuit to resonance so that energy passes efficiently to the phone.

The substance. **At resonance, $V_L$ and $V_C$ can each exceed the supply voltage** — they cancel each other, not vanish.