Revise Characteristics of alternating currents for Cambridge 9702 (21.1)

Alternating current is an electrical current that continuously changes its magnitude and periodically reverses its direction of flow. This is fundamentally different from direct current (DC), which flows in a single, constant direction.

Because AC varies, we need ways to describe its value at any moment, its maximum strength, and its overall "effectiveness". Let's start with a snapshot in time. For any sinusoidal AC quantity (current or voltage), its instantaneous value can be represented by:

$X = x_0\sin(\omega t)$

Here, $X$ is the instantaneous value (voltage or current) at time $t$. $x_0$ is the peak value (or amplitude) – the absolute maximum magnitude reached. $\omega$ (omega) is the angular frequency, which tells us how quickly the AC oscillates.

$\omega = 2\pi f$

While the peak value tells us the maximum, it doesn't represent the average power delivered. The average value of a sinusoidal current over a full cycle is zero. For practical purposes, like calculating power, we use the Root Mean Square (RMS) value.

The RMS value of an AC voltage or current is the equivalent DC value that would dissipate the same average power in a resistor. It's found by squaring all the instantaneous values, finding the mean (average) of those squares, and then taking the square root of that mean.

$V_{rms} = \frac{V_{peak}}{\sqrt{2}}$

$I_{rms} = \frac{I_{peak}}{\sqrt{2}}$

The instantaneous power in an AC circuit continuously varies because voltage and current are changing. However, we are usually interested in the mean power dissipated over a full cycle.

To understand why mean power is half the maximum power, let's look at the instantaneous power in a resistor. Power is given by $P = I^2R$. Since the current is sinusoidal, $I = I_0 \sin(\omega t)$, the instantaneous power is:

$P(t) = (I_0 \sin(\omega t))^2 R = I_0^2 R \sin^2(\omega t)$

From this equation, we can see that the maximum power, $P_{max}$, occurs when $\sin^2(\omega t) = 1$. So, $P_{max} = I_0^2 R$. The power varies as $\sin^2(\omega t)$, which is a function that oscillates between 0 and 1. Over a full cycle, the average value of $\sin^2(\omega t)$ is exactly $\frac{1}{2}$.

Therefore, the mean power $\left< P \right>$ is the average of the instantaneous power:

$\left< P \right> = \text{average of } (I_0^2 R \sin^2(\omega t)) = I_0^2 R \times (\text{average of } \sin^2(\omega t)) = I_0^2 R \times \frac{1}{2}$

Since $P_{max} = I_0^2 R$, we can conclude that:

$\left< P \right> = \frac{1}{2} P_{max} = \frac{1}{2} I_0^2 R = \frac{1}{2} \frac{V_0^2}{R}$

A key advantage of using RMS values is that they simplify mean power calculations, making them identical to DC power calculations. This is precisely why RMS values are so useful!

$\left< P \right> = I_{rms}^2 R = \frac{V_{rms}^2}{R} = V_{rms} I_{rms}$