Alternating current: peak and r.m.s. values: WASSCE Physics

The electricity from the wall socket is not a steady push in one direction: it reverses fifty times every second. Alternating current is easy to generate and, above all, easy to transform to different voltages — which is why every power grid uses it.

Learning objectives

By the end of this lesson you should be able to:

describe alternating current and its waveform
define peak value, frequency, period and r.m.s. value
convert between peak and r.m.s. values
compare a.c. and d.c.

What is alternating current?

Definition

Alternating current (a.c.) is a current that periodically reverses its direction, usually varying sinusoidally with time. Direct current (d.c.) flows in one direction only.

An alternating voltage varying as a sine wave between plus and minus the peak value; the r.m.s. value is 0.707 of the peak

Figure 1. An alternating voltage. WASSCE Tutor diagram.

V = V_0\sin\omega t = V_0\sin 2\pi f t

\begin{aligned}V &= V_0\sin\omega t \\ &= V_0\sin 2\pi f t\end{aligned}

Term	Meaning
Peak value $V_0$ or $I_0$	the maximum value of the voltage or current
Period $T$	time for one complete cycle
Frequency $f$	number of cycles per second, $f = 1/T$ (mains: 50 Hz in Sierra Leone)
Peak-to-peak value	$2V_0$

The r.m.s. value

Because an alternating current keeps changing, we need a single "effective" value to compare it with d.c.

Definition

The root-mean-square (r.m.s.) value of an alternating current is the value of the steady direct current that would produce the same heating effect in the same resistor:

I_{rms} = \dfrac{I_0}{\sqrt{2}} \approx 0.707\,I_0 \qquad V_{rms} = \dfrac{V_0}{\sqrt{2}} \approx 0.707\,V_0

\begin{gathered}\begin{aligned}I_{rms} &= \dfrac{I_0}{\sqrt{2}} \\ &\approx 0.707\,I_0\end{aligned} \\[6pt] \begin{aligned}V_{rms} &= \dfrac{V_0}{\sqrt{2}} \\ &\approx 0.707\,V_0\end{aligned}\end{gathered}

Mains voltages are always quoted as r.m.s. values. Sierra Leone's mains supply is about 230 V r.m.s., so its peak value is about $230 \times \sqrt{2} \approx 325$ V.

Worked example

An alternating voltage has a peak value of 340 V and a frequency of 50 Hz. Find (a) the r.m.s. voltage, (b) the period.

(a) $V_{rms} = \dfrac{340}{\sqrt{2}} = 240$ V. (b) $T = \dfrac{1}{50} = 0.02$ s.

Worked example

An a.c. supply of 220 V r.m.s. is connected to a 55 Ω heater. Find (a) the r.m.s. current, (b) the peak current, (c) the average power.

(a) $I_{rms} = \dfrac{220}{55} = 4$ A.

(b) $I_0 = 4\sqrt{2} = 5.66$ A.

Common mistake

Power calculations with a.c. use r.m.s. values ( $P = V_{rms}I_{rms}$ ). Using peak values gives twice the real average power.

Measuring a.c.

Ordinary moving-coil meters cannot read a.c. directly (the needle would try to swing back and forth 50 times a second and stay at zero). They are used with a rectifier.
Moving-iron and hot-wire meters work with both a.c. and d.c., and read r.m.s. values.
A cathode-ray oscilloscope (CRO) displays the waveform, so the peak value and period can be read directly.

a.c. versus d.c.

	a.c.	d.c.
direction	reverses periodically	one direction
sources	generators (alternators), the mains	cells, batteries, solar cells, rectified a.c.
voltage can be changed by a transformer	yes	no
transmission over long distances	efficient (step up with transformers)	less convenient
needed for	most household appliances	electroplating, charging batteries, electronics

Rectification converts a.c. into d.c. using diodes (covered with semiconductors). Phone chargers contain a transformer, a rectifier and a smoothing capacitor.

Key points

a.c. reverses direction periodically; mains frequency in Sierra Leone is 50 Hz.
$V_{rms} = 0.707\,V_0$ , $I_{rms} = 0.707\,I_0$ ; mains is about 230 V r.m.s. (about 325 V peak).
r.m.s. value = steady d.c. giving the same heating effect.
a.c. can be transformed to high or low voltages; d.c. cannot.

WASSCE examination focus

WAEC tests this

Define the r.m.s. value of an alternating current and use $I_{rms} = I_0/\sqrt{2}$ , read peak value and period from a waveform, and compare a.c. and d.c. (including why a.c. is used for transmission). Calculations of power with r.m.s. values are common.

Quick check

The peak value of an alternating current is 10 A. What is its r.m.s. value?
What is the period of a 50 Hz supply?
Why is a.c. rather than d.c. used in the national grid?

Answers: 1. $10 \times 0.707 = 7.07$ A. 2. 0.02 s. 3. Its voltage can be stepped up and down with transformers, so power can be transmitted at high voltage with small losses.

Alternating current: peak and r.m.s. values