Alternating Currents

1. Introduction

An alternating current (AC) is a current whose magnitude changes continuously with time and whose direction reverses periodically. It is usually sinusoidal:

$$ \begin{aligned} I &= I_0\sin\omega t \ \qquad E &= E_0\sin\omega t \end{aligned} $$

where $I_0$ and $E_0$ are the peak (maximum) values or amplitudes, and $\omega = 2\pi f$ is the angular frequency. AC is used for power distribution because it can be stepped up or down efficiently with transformers. This chapter covers the peak and rms values, the behaviour of AC through a resistor, an inductor, and a capacitor, series RL, RC, and LCR circuits, resonance, the quality factor, and power in AC circuits.

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2. Peak and RMS Value of AC (Subtopic 19.1)

2.1 RMS value

The root-mean-square (rms) value of an alternating current is that steady (direct) current which, flowing through a given resistance for a given time, produces the same amount of heat as the alternating current does in the same resistance in the same time. It is also called the effective or virtual value.

Derivation. Let $I = I_0\sin\omega t$ flow through a resistance $R$. In a small time $dt$ the heat produced is (Joule's law):

$$dH = I^2 R,dt = I_0^2 R\sin^2\omega t,dt$$

The heat produced in one full cycle (from $t = 0$ to $t = T$) is:

$$ \begin{aligned} H &= I_0^2 R\int_0^T \sin^2\omega t\ \ dt &= I_0^2 R\int_0^T \frac{1 - \cos 2\omega t}{2}\ \ dt &= \frac{I_0^2 R T}{2} \end{aligned} $$

If $I_{rms}$ is the equivalent steady current producing the same heat, $H = I_{rms}^2 R T$. Equating:

$$ \begin{aligned} & I_{rms}^2 R T = \frac{I_0^2 R T}{2} \ \ & \Rightarrow\ \ & \boxed{I_{rms} = \frac{I_0}{\sqrt{2}} = 0.707,I_0} \end{aligned} $$

Similarly, for the alternating EMF:

$$E_{rms} = \frac{E_0}{\sqrt{2}} = 0.707,E_0$$

So the rms value of AC is $0.707$ times (that is, $70.7%$ of) the peak value. AC meters read rms values; the "$220\ \text{V}$" mains is an rms value.

2.2 Mean (average) value

Over one complete cycle the average value of a sinusoidal AC is zero, because it is positive for one half-cycle and equally negative for the next. The average is therefore defined over a half-cycle.

The mean value over a half-cycle is that steady current which sends the same charge through the circuit in that half-period as the AC does. Integrating $I = I_0\sin\omega t$ over the half-cycle:

$$ \begin{aligned} q &= \int_0^{T/2} I_0\sin\omega t\ \ dt &= \frac{2I_0}{\omega} \ &= \frac{I_0 T}{\pi} \end{aligned} $$

Since $q = I_m\dfrac{T}{2}$, the mean value is:

$$\boxed{I_m = \frac{2I_0}{\pi} = 0.637,I_0}$$

that is $63.7%$ of the peak value. Likewise $E_m = \dfrac{2E_0}{\pi} = 0.637,E_0$.

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