AC through Pure L and C

PHYAlternating Current — Basics

A pure inductor and a pure capacitor each oppose AC through a frequency-dependent reactance — $X_L = \omega L$ for the inductor and $X_C = 1/\omega C$ for the capacitor — and neither dissipates power. Students compute the reactance, then the RMS current $I_{rms} = V_{rms}/X$ , and must track the $90^{\circ}$ phase relation.

The opposite frequency dependence of $X_L$ and $X_C$ is heavily examined.

Inductive reactance

X_L = \omega L = 2\pi f L

X_L

\Omega

L

in H,

f

in Hz; increases with frequency (an inductor opposes high

f

Capacitive reactance

X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}

X_C

\Omega

C

in F,

f

in Hz; decreases with frequency (a capacitor blocks DC/low

f

RMS current

I_{rms} = \dfrac{V_{rms}}{X}

X = X_L

for a pure inductor,

X = X_C

for a pure capacitor;

V_{rms}

= RMS source voltage.

In a pure inductor the current LAGS the voltage by $90^{\circ}$ ; in a pure capacitor the current LEADS by $90^{\circ}$ .
Always include $2\pi$ in $\omega = 2\pi f$ ; for $f = 50\,\text{Hz}$ , $\omega \approx 314\,\text{rad/s}$ .
$X_L$ rises with frequency while $X_C$ falls — at very high $f$ an inductor is nearly an open circuit and a capacitor nearly a short.
A capacitor blocks DC entirely ( $f = 0 \Rightarrow X_C = \infty$ ); a pure inductor passes DC freely ( $f = 0 \Rightarrow X_L = 0$ ).
Convert $\mu\text{F}$ to F: $10\,\mu\text{F} = 10\times10^{-6}\,\text{F}$ before using $X_C$ .
Average power in either pure element is zero because the phase angle is exactly $90^{\circ}$ ( $\cos 90^{\circ} = 0$ ).
Reactance plays the role of 'resistance' for the current magnitude, but unlike $R$ it stores and returns energy rather than dissipating it.

Where the marks go

Writing $X_C = \omega C$ or $X_L = 1/\omega L$ — the inductor's reactance is $\omega L$ and the capacitor's is $1/\omega C$ .
Dropping the $2\pi$ and using $f$ in place of $\omega$ .
Forgetting to convert $\mu$ F to F, giving $X_C$ off by $10^6$ .
Claiming a pure $L$ or $C$ consumes power; it is wattless.

How the board asks it

Numericalcapacitive reactance and rms current
A $5\,\mu\text{F}$ capacitor is connected across an AC supply of $220\,\text{V}$ , $50\,\text{Hz}$ . Calculate the capacitive reactance $X_C$ and hence the RMS current drawn from the source.
Derive / provephase relation in a pure inductor
An alternating voltage $V = V_0\sin\omega t$ is applied to a pure inductor of inductance $L$ . Obtain an expression for the instantaneous current and hence show that the current lags the voltage by $90^{\circ}$ .
Diagram / graphopposite frequency dependence of reactances
Draw, on the same axes, the variation of inductive reactance $X_L$ and capacitive reactance $X_C$ with the frequency $f$ of the applied AC, and explain the nature of each curve.
Give reasonsdc blocking and wattless power
Give reasons: (a) a capacitor blocks DC but allows AC to pass through it; (b) the average power consumed by a pure inductor over a complete cycle is zero.
Applicationcurrent response to changing frequency
An inductor and a capacitor are each connected, in turn, to the same AC source. State and explain how the current in each changes when the frequency of the source is increased.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.