Electricity and Magnetism

ALTERNATING CURRENTS

THE ALTERNATOR

Recall the alternator previously described: Also, recall that the emf  'E' produced by an alternator can be expressed as: This is called an alternating emf since the current it produces in a conductor alternates in direction. In this equation:  In the above case: RMS VALUES AND POWER

Consider: With the switch in position (1), the current produced by the ac supply is adjusted to make the bulb reasonably bright. Then, with the switch in position (2), the current produced by the battery is varied until the bulb is at the same brightness. The switch can be flicked left and right to compare the brightnesses. The reading on the ammeter gives the value of the direct current ('dc'), and this is equal to the average or mean or effective value of the ac current.

In the above set-up, when the bulb is equally bright for both supplies, then: Definition:

• The root mean square value of an alternating current is that value of direct current which has the same heating effect in a given resistor
In the above arrangement, the reading on the ammeter is Idc, which equals Irms.
Using P = V2/R, steps similar to the above can be followed to infer that: (we often do not bother with the ‘dc’ subscripts)

In the ac case, the power is not constant, but its mean value, <P>, in the above set-up, equals the steady power of the dc supply, so: Sinusoidal ac

For a sinusoidal ac supply:   We can see this visually in the above graph - if we ‘chop off’ the top half of a ‘hill’ it will exactly fill the adjacent ‘valley’.   Note that:
• We have the same equation for power for both dc and ac, P = VI, so long as it is remembered that V and I are rms values in the ac case
• For a sinusoidal supply we can determine Vrms and Irms directly from the peak values as V0/Ö2 , and I0/Ö2 respectively
• Ammeters and voltmeters that measure ac values usually give rms values
• When we specify an input or an output of, say, 200V for a transformer, the value is an rms value

AC THROUGH A RESISTOR Suppose that the applied voltage is given by:  V and I reach their peak values at the same instant and are said to be in phase. AC THOUGH A CAPACITOR

Observation With switch (1) closed and switch (2) open the lamp does not light, because there is only a current flowing for the brief time that it takes the capacitor to charge.

With switch (1) open and switch (2) closed, the lamp does light, because the capacitor is being repeatedly charged and discharged, so there is always a current flowing. In effect, an alternating current can flow 'through' a capacitor, whereas a direct current cannot.

Phase difference between V and I If the p.d. applied to the capacitor is given by:  Notice that I has its maximum value ¼ cycle before V has its maximum value.

We say that the current leads the voltage by ¼ cycle. So, we can say that when ac is applied to a capacitor that: Reactance of a capacitor, XC

This is the opposition that a capacitor has to the flow of ac, and is defined by: Example  AC THROUGH AN INDUCTOR

Observation When a dc supply is used (see earlier notes on self induction) bulb (2) lights shortly after bulb (1), because the inductor opposes the rise of current, making it gradual.

When an ac supply is used, as in the above set up, bulb (1) lights up, but bulb (2) never does. This is because the current is repeatedly changing directions and the inductor prevents it rising enough to light the bulb. Phase difference between V and I We consider an inductor with zero resistance, called a pure inductance (in practice, the coil does have some resistance).

An ac supply is applied across the inductor, and if the current I flowing at time t is given by:    Reactance of an inductor, XL This is the opposition that an inductor has to the flow of ac, and is defined by: Example Calculate XL for an inductor of inductance L = 10H, when:  (a) f = 50Hz and (b) f = 1kHz POWER IN AC CIRCUITS

Resistor Notice that the power varies at twice the frequency of V or I (the above shows 2 cycles of P for one of V or I).

Inductor During the first ¼ cycle the current in the coil is rising, as is the energy stored in the magnetic field created. During the second ¼ cycle, the current falls to zero, as does the field, and the energy returns to the source. Thus:
• over a cycle, the power absorbed by a pure inductor is zero
In practice the coil of an inductor has some resistance, so some power is dissipated in the coil as heat.

Capacitor The power variation is similar to that of an inductor. During the first ¼ cycle the charge on the capacitor is rising, as is the energy stored in the electric field created. During the second ¼ cycle, the charge falls to zero, as does the field, and the energy returns to the source. Thus:

• over a cycle the power absorbed by a capacitor is zero

LCR SERIES CIRCUIT

We consider a circuit in which an inductor L, a capacitor C and a resistor R are in series, and across which there is an alternating p.d. V: At any instant: Since L, C and R are in series, at any instant there is the same current I through each of them.

From previous notes:

• VR is in phase with I VL leads I by 900 Þ VL leads VR by 900
• I leads VC by 900 Þ VR leads VC by 900
We can represent the relationships by: Impedance, Z

This is the opposition that the circuit as a whole has to the flow of ac, and is defined by: Again, the unit is the ohm, as it is for resistance and reactance. It can be shown that: Resonance in LCR circuits Both XL and XC depend upon frequency of the supply, and therefore so does Irms. The frequency f0 at which Irms is a maximum is called the resonant frequency of the circuit.

The equation above implies that Irms is a maximum, i.e. resonance occurs, when:    We define a quantity called the Q-factor (‘quality factor’), also called the magnification factor: A circuit with a high Q-factor gives rise to sharp resonance. We can ‘tune’ a circuit to a particular frequency by varying the value of C. Variable capacitors can be made by arranging the plates in such a way that the area of overlap can be varied, thus varying the capacitance. Example Calculate:

1. the value of capacitance at which resonance would occur the p.d.s across the inductor and capacitor at resonance
2. the Q factor at resonance   RECTIFICATION

While a.c. (alternating current) is very useful for power transmission, in many uses of electricity we want d.c. (direct current, such as is provided by a battery). ‘Rectification’ is the process of changing a.c. to d.c. Half-wave rectifier  Without the diode in the circuit, current would alternate in direction in the same way as the supply voltage, represented above.

However, the diode only lets current through R on the positive part of the cycle, but not on the negative half.
The p.d. across R = I R, so when the current is zero so is the voltage across it. Although the voltage, and therefore current, is going up and down, it is only ever in one direction, so this is now a direct current (since its direction does not change). Since only half of the original wave ‘survives’ we refer to this as half-wave rectification.

Smoothing Although the above produces a direct voltage and current, it is not the steady sort of dc we get from a battery. To ‘smooth’ the voltage we add a capacitor. The capacitor, labelled C, is placed in the half-wave rectifier circuit as indicated below: The effect of the capacitor on the voltage across R is represented below: Initially the voltage across the resistor rises to its maximum, and the capacitor charges to its maximum, the top plate becoming positive.

Without the capacitor, the voltage across R then drops to zero and stays there for half a cycle. But now, as soon as the voltage starts to fall, the capacitor start to discharge through the resistor – this maintains the voltage across it close to its maximum until the next cycle starts. Thus, the voltage across the resistor and the current though it are smoother than without the capacitor. Full-wave rectifier The following arrangement of four diodes is called a bridge rectifier circuit: If an a.c. supply is applied to XY, on the first half cycle, X is positive relative to Y, and on the second half cycle Y is positive relative to X.
• On the first half cycle, when X is positive, current follows the route XARCY
• On the second half cycle, when Y is positive, current follows the route YBRDX
In both half cycles, the current flows the same way through resistor R. So, on each half cycle, the current though R, and the voltage across it, rises and falls, but it never changes direction. Thus, the a.c. supply is now d.c. Since both halves of the original a.c. are rectified, this is called full-wave rectification. As with the previous circuit, a capacitor can be used to partially smooth the output voltage - in more complex circuits the conversion from a.c. to d.c. is practically perfect.

CATHODE RAY OSCILLOSCOPE ('CRO') The cathode-anode arrangement is called an ‘electron gun’. In a practical CRO the arrangement is more complex than represented above - it is designed to allow the electrons to be focused into a fine beam. A voltage applied to the x-plates can pull the electron beam side to side (along the x-axis), and a voltage applied to the y-plates can pull the electron beam up and down (along the y-axis). The x-plates are connected to a varying voltage which pulls the beam from left to right at a constant rate and then turns off to let the beam ‘fly back’ to the left again. It does this repeatedly, and on screen a horizontal line is seen. The rate at which it is pulled side to side is controlled by the 'time-base' setting on the CRO. Saw-tooth time-base voltage: Using a CRO:

1. To measure a steady voltage If a fixed voltage is applied to the y-plates, the horizontal line will move up or down. How much it does so depends on the size of the voltage and the ‘y-gain’ setting on the CRO. • The trace moves 2 squares, i.e. 2 cm upwards when the voltage V is applied. A y-gain of 4V/cm means that each vertical cm represents 4volts.
• Thus the value of V is 2*4V = 8 volts.
2. To measure frequency The time-base tells us how long each horizontal cm corresponds to. The time-base can be varied till at least one full wave appears on the screen.
• One wavelength means, for example, from crest to crest, and this occupies 4cm on the screen The time-base is set to 5ms/cm Thus, as one wave occupies 4cm, this is equivalent to 20ms = 20*10-3 seconds  = 0.02 sec This is the period, T, of the wave
• Frequency = 1/period = 1/0.02 = 50 waves per sec or 50 Hz (Hz = 'hertz'). This is the frequency of the mains voltage supply