ALTERNATING CURRENTS

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Alternating currents Root mean square voltage and current Additional questions	Heating effects of electric currents Inductive circuits Electric machines Resistive, capacitive and inductive AC circuits Alternating currents and safety	Data Glossary
Learning Outcomes
After studying this section, you will (a) understand what is meant by an "alternating current", and (b), be familiar with the expressions for root mean square voltages and currents.

Alternating currents:

The types of generators in normal use do not produce direct currents (DC), whose voltage remains constant with time, but ALTERNATING CURRENTS (AC).

Alternating current sources are indicated in circuit diagrams as

.

Alternating currents continuously reverse the direction of electron flow through the circuit, and the electromotive force (voltage) E, causing such currents varies with the time, t, according to the formula

where ω is the angular velocity of the rotating coil(s) of the generator, measured in radians per second (2π radians = 360^o) and f is the frequency. Note that ω = 2πf.

The set of diagrams above (A - E) show the position of the plane of the rotating electric coil with respect to a fixed magnetic flux at various times, for one half a rotation, that is, one half of a current- generating cycle.

• When ωt = 0^o (A), the coil is parallel to the flux, and so does not cut the magnetic flux lines and no current is therefore induced in the coil.
• When ωt = 45^o (B), the rate at which flux lines are cut by the coil is maximal, but since the area of the coil presented to the flux is not yet at its maximum value, the current in the coil is still rising.
• When ωt = 90^o (C), the coil is perpendicular to the flux, and the induced current is at a maximum.
• When ωt = 135^o (D), the rate at which flux lines are cut by the coil decreases maximally, but since the area of the coil presented to the flux is not yet at its minimum value, the current in the coil is still decreasing.
• When ωt = 180^o (E), the coil is again parallel to the flux, and so does not cut the magnetic flux lines and no current is therefore induced in the coil.
• For the second half of the cycle, the induced current at various stages will be in the opposite direction to that which was found in the first half of the cycle.

It is clear from the above that the graph of the induced voltage against time follows a sine curve, that is, it is SINUSOIDAL.

When the angle of rotation of the coils is 0^o, 180^o, 360^o, 540^o..., E will be zero. When the angle of rotation is 90^o, 270^o, 450^o, 630^o..., E will be E_o, the VOLTAGE AMPLITUDE, also known as the PEAK VOLTAGE. Note that the average voltage (or current) over a whole number of cycles is zero.

Root mean square voltage and current:

The root mean square value of an alternating current produces the same heating effect as that of a direct current of the same value in identical resistances under the same conditions. Note that measuring instruments (ammeters and voltmeters) always report root mean square values of the current or the voltage.

The ROOT MEAN SQUARE value of the voltage is Erms = Eo / √2, and is the nominal value of the power supply. In South Africa, this is 220 V, at 50 cycles. This means that 50 times a second, the voltage varies between Emax and Emin, where Eo = √2 Erms = 1.41 x 220 = 311 V. (In the United States, nominal voltage is 120 V at 60 Hz, with a peak voltage of 169 V.) The current, I, in an alternating current circuit will also vary sinusoidally with time: I = Io sin ωt, where Io is the CURRENT AMPLITUDE. The root mean square value of the current is Irms = Io / √2 (See the derivation)

Why use alternating currents at all? (Click here for a discussion)

Additional questions

Derivation of root mean square current formula:

At any time, t, the voltage is given by

E = E_o sinωt

Squaring

E² = E_o² sin²ωt

But, from trigonometry, sin²α = ½(1 - cos2α)

Giving

E² = ½E_o² - ½ E_o² cos 2ωt

But the average of cos 2ωt over a complete cycle is zero. The average square of the voltage over a whole cycle will be E² = ½E_o².

Thus, the ROOT MEAN SQUARE VOLTAGE will be E_RMS = E_o/√2 = 0.707E_o

A similar argument will show that the ROOT MEAN SQUARE CURRENT will be I_o/√2 = 0.707I_o.