Reactance of Capacitors and Inductors

A resistor obeys $V = IR$ at every instant. A capacitor or an inductor in an AC circuit also limits the current, but in its own way: by an amount that depends on the frequency, and with its current out of step with its voltage. The "resistance" of each is called its reactance.

Key idea

$$X_C = \frac{1}{\omega C}, \qquad X_L = \omega L, \qquad I_{\text{rms}} = \frac{V_{\text{rms}}}{X}.$$ In a capacitor the current leads the voltage by $90^\circ$; in an inductor it lags by $90^\circ$. Neither uses any average power.

Why it works

Capacitor: $q = CV$, so $I = C\dfrac{dV}{dt}$. With $V = V_0\sin\omega t$, $I = \omega CV_0\cos\omega t$: the peak current is $\omega CV_0 = V_0/X_C$. The cosine peaks a quarter cycle before the sine, so the current leads. At high frequency the capacitor is charged and discharged so quickly that it never fills up, so it passes current easily: $X_C$ falls.
Inductor: $V = L\dfrac{dI}{dt}$. With $I = I_0\sin\omega t$, $V = \omega LI_0\cos\omega t$: peak voltage $\omega LI_0 = X_LI_0$, and the voltage leads, so the current lags. The faster the current tries to change, the harder the inductor fights: $X_L$ rises with frequency.

A capacitor (with a tiny resistance) on an AC source. Change the frequency: the current grows as the frequency rises, and the phasors show the current a quarter turn ahead of the capacitor's voltage.

Worked examples

Worked example: A capacitor at two frequencies

Find the reactance of a $10\ \mu\text{F}$ capacitor at $60\ \text{Hz}$ and at $6.0\ \text{kHz}$.

$X_C = \dfrac{1}{2\pi(60)(10 \times 10^{-6})} = 265\ \text{Ω}$.
At $100$ times the frequency, $100$ times smaller: $2.65\ \text{Ω}$.

Answer: $265\ \text{Ω}$ and $2.65\ \text{Ω}$

Worked example: An inductor on the mains

A $50\ \text{mH}$ inductor is connected across $120\ \text{V}$ rms at $60\ \text{Hz}$. Find the rms current.

$X_L = 2\pi(60)(0.050) = 18.8\ \text{Ω}$.
$I_{\text{rms}} = 120/18.8 = 6.37\ \text{A}$, a quarter cycle behind the voltage.

Answer: $I_{\text{rms}} \approx 6.37\ \text{A}$

Worked example: Where they match

At what frequency do that inductor and that capacitor have equal reactances?

$\omega L = \dfrac{1}{\omega C}$ gives $\omega = \dfrac{1}{\sqrt{LC}}$.
$f = \dfrac{1}{2\pi\sqrt{(0.050)(10 \times 10^{-6})}} = 225\ \text{Hz}$. This is the resonant frequency you will meet soon.

Answer: $f \approx 225\ \text{Hz}$

Because the two reactances move in opposite directions, capacitors and inductors sort signals by frequency. In a loudspeaker crossover, a capacitor in series with the small tweeter lets the high notes through and holds back the low ones, while an inductor in series with the large woofer does the reverse.

Common mistake

"A capacitor's reactance grows with frequency." It shrinks: $X_C = 1/(\omega C)$. At high frequency a capacitor behaves almost like a wire, and at zero frequency (steady DC) it blocks completely, once charged. It is the inductor whose reactance grows. Check with the extremes: DC through a capacitor is blocked ($X_C \to \infty$ as $\omega \to 0$); DC through an inductor flows freely ($X_L \to 0$).

Reactance of Capacitors and Inductors

Worked examples

Practice

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Sources and further reading