Feeds:
Posts

## Alternating current basics

I have recently been teaching my Physics students the difference between Altnernating Current (AC) and Direct Current (DC). I thought I would try and explain some of what I have been saying here.

## Direct Current

The kind of current we get from a battery is direct current. This means that the current remains constant with time. Many electrical devices such as radios, computers and electronic devices run on DC.

A direct current of 4 Amps. The current stays constant with time.

The power dissipated in a device which has a resistance $R$ when we have a direct current of value $I$ is simply $P=I^{2}R$. So, for example, if our device has a resistance of $100 \Omega$ and we have a current of $4 A$ then the power disippated will be $4^{2} \times 100 = 16 \times 100 = 1600$ Watts or 1.6kW.

## Alternating Current

An alternating current is constantly varying with time. A plot of the current against time looks like this:

An alternating current. This particular plot shows an alternating current which has a peak value of 4 A and a frequency of 0.25 Hz.

The peak current, 4A in this case, can be denoted by $I_{0}$. Mathematically this alternating current can then be described by the equation $I=I_{0}sin(2\pi f t)$ where $f$ is the frequency. The power disippated can be determined by looking at $I^{2}$. Below is a plot of $I$ and $I^{2}$.

A plot of current I and I squared (I x I) for an alternating current.

As we can see from the plot of $I^{2}$, it is also always varying with time. It varies between a maximum of $I_{0}^{2}$ and zero.

If we look closely at the plot of the square of alternating current over half of the AC cycle (so from a time of 0 to 2s in our example), we can see that it takes 1s to reach the value of 16, and a further 1s to come back down to a value of zero. How long does it take to reach a value of 8, half the peak value of 16? To work this out we need to calculate $t$ when $4 \sin\left(\frac{\pi}{2}t\right)= \sqrt{8}$ so when $\sin\left(\frac{\pi}{2}t\right)= \frac{\sqrt{8}}{4} = \sqrt{\frac{8}{16}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$. To find $t$ we do $\frac{\pi}{2}t=\arcsin\left(\frac{1}{\sqrt{2}}\right)$ so $\left(\frac{\pi}{2}t\right)=\frac{\pi}{4}$ which gives us $t=\frac{2}{4}=\frac{1}{2}=0.5s$ which is exactly half of the time it takes to reach its maximum value of 16. As the curve is symmetrical, it will be above 8 for exactly half of the time and below 8 for exactly half of the time, so the average value of $I^{2} = \frac{1}{2}I_{0}^{2}$.

We therefore say that the average power disippated in the device is $P_{av} = \frac{1}{2}I_{0}^{2}R$. This leads us to define a new quantity, $I_{rms}$, the root mean square current, which is defined as $I_{rms}^{2}=\frac{1}{2}I_{0}^{2}$ and so $I_{rms} = \frac{1}{\sqrt{2}}I_{0}$. This means we can write the average power dissipated in an AC device as $\boxed {P_{av}=I_{rms}^{2}R}$.

The root mean square current and its related root mean square voltage are often more useful to us than the peak current and voltage. When we say that the voltage from the mains in Europe is 240V, and in the USA is 110V, this is actually the root mean square voltage, not the peak voltage. This is quoted because the average power $P_{av}=I_{rms}^{2}R$ used by any device is also given by $P_{av}=V_{rms}I_{rms}$.