AC ELECTRICITY AIMS INTRODUCTION

2 AC ELECTRICITY AIMS The aims of this unit are to describe and quantify the behaviour of resistors, inductors and capacitors in AC circuits. INTRODUCTION In Unit 1 we studied direct current or DC electricity. In DC circuits, the voltage remains constant with respect to time and the current flows in one direction only through the circuit. In this unit we will be studying another form of electricity, called alternating current or AC. In AC circuits, the value of the voltage is constantly changing with time, as is the direction of current flow through the circuit. In particular, we will be looking at the behaviour of simple AC circuits containing resistances, inductances and capacitances. For the purpose of this course, this unit will complete our study of electricity. Why use AC? Why build circuits in which the magnitude and direction of the voltage is constantly changing, and in which the current is oscillating back and forth, rather than flowing steadily in one direction? DC circuits seem so straightforward in comparison: all that are needed in analysing them are Ohm s and Kirchoff s laws. So why not use DC in all electric and electronic applications? The answer is that AC is the only form in which electrical power can be economically transmitted over long distances. The following example demonstrates this point very clearly. Suppose we wish to supply a city with ten million watts of electrical power. Our generating station is located some distance away, and we wish to minimise the power dissipation along the transmission lines. Remember that electrical power (in watts) equals the line voltage (in volts) multiplied by the current (in amps). To supply our city with power, we can use any combination of voltage and current, so long as the product of the two gives us the figure of ten million watts. Shall we send the power out from our generating station in the form of high-voltage and low-current? Or as low-voltage and high-current? If we set the voltage at 100V, then the current must be 100,000A to satisfy the electricity consumers in the city. Alternatively, if we increase the voltage to 100,000V, then 100A of current will suffice. However, even the best conducting transmission lines will present some resistance to the flow of current, and accordingly, some power will be dissipated. Assuming line resistance is 20Ω, the power lost in transmission is given by the equation, P = I 2 R. If we generate 100,000A at 100V, then the power loss equals 200,000 million watts. This is some 20 thousand times the power required by the city! Unit 2: AC Electricity 2-1

On the other hand, if the power is sent out along the lines at 100,000V and only 100A, then the lost power is just 200,000 watts or 2% of the load required by the city. The reason for these hugely varying results is that power dissipated in the transmission of electricity increases not just with the current, but with the square of the current. It is for this reason that electrical power is distributed at high-voltages and low currents. But could we not transmit the high-voltage, lowcurrent power in DC form? Yes, we could; but then we would be unable to step-down the line voltage with transformers before passing it on to the power consumers. Only AC can be manipulated in this way. Today, most power stations produce electricity at 100,000-400,000V (100-400kV) AC, which transformer stations near the point of usage stepdown to the mains line voltage of 240V AC. Radio waves are of their nature also constantly changing in voltage and current. Their transmission, detection, processing, amplification and translation into audible sound are all other important areas in which an understanding of AC is essential. In summary, AC is a vital element in our electric and electronic world. SINUSOIDAL WAVEFORMS OBJECTIVES After studying this sub-unit, you should be able to: 1. Define or explain the meaning of the following terms:. alternating current and voltage. amplitude. angular frequency. cycle. frequency. hertz. instantaneous current and voltage. leading and lagging phase angles. peak value. periodic time. phase angle and phase difference. sinusoidal waveform 2. Write the mathematical equation for the instantaneous value of a voltage or current sinusoidal waveform, given the appropriate data. We learnt in Unit 1 that DC voltage and current remains constant with respect to time, as shown in Figure 2.1. As indicated above, very often in electric and electronic devices voltages and currents are used which are not constant with respect to time. The variation of these voltages or currents with respect to time are known as AC (alternating current ) wavefoms. By far the most important type of AC waveform is the sinusoid as shown in Figure 2.2. Note that although the abbreviation AC literally stands for alternating current, it is used generally to describe both current and voltage waveforms. 2-2 Communications Technology 1

A sinusoidal waveform is continuously changing in direction from positive to negative to positive and so on, ie the magnitude and polarity (direction) both change with time. It consists of a basic pattern, repeated every T seconds. We refer to a single and complete basic pattern as a cycle of the waveform. The duration of the cycle is called the period, denoted by the capital letter T. The sinusoid is therefore said to be periodic. Voltage V Figure 2.1 Time variation of a DC voltage or current Current I time t time t voltage or current + T 0 - t Figure 2.2 Sinusoidal Waveforms Sinusoidal Wave The frequency, f, is defined as the number of complete cycles it goes through every second. Frequency was formerly expressed in cycles per second, but the unit hertz (Hz) is now universally used. Thus, when the frequency of the mains electricity supply is said to be 50Hz, this means that 50 complete cycles of the sinusoidal voltage waveform take place every second. The relationship between frequency, f, in hertz and period, T, in seconds, is given by the expression: f = 1/T Mathematically, we can represent the sinusoid as a function of time using the equation vt () = Vˆ sin 2π ft = Vˆ sin θ, θ = 2π ft where ˆV is the peak value of the voltage waveform and f is its frequency in Hz.. Unit 2: AC Electricity 2-3

This is simply the sine function of an angle θ measured in radians and where the angle varies with time in accordance with the frequency of the voltage ω, ie θ = 2πft. If this voltage is connected across a resistor R the current flowing in the resistor will follow the same pattern of periodic variation as the emf, and its value at any instant is expressed as follows: it () = Iˆ sin 2π ft, where I ˆ the circuit. Vˆ = is the peak value of the current flowing in R It is important to note that this equation for instantaneous current applies only in what is termed a pure resistive circuit. A pure resistive circuit is one in which there is no capacitance or inductance, only resistance. Another way of expressing how fast the sinusoidal waveform is changing is in relation to its angular frequency in radians per second or Hz. Since one complete cycle corresponds to an angle of 2π radians (360 o ), a frequency of f Hz (cycles per second) corresponds to an angular frequency of 2πf radians per second. Angular frequency is denoted by the symbol ω. Thus: ω = 2πf or f = ω/2π As f = 1/T, where T is the time for one period, we can write: ω = 2π/T or T = 2π/ω We now have two equivalent expressions for both the instantaneous voltage and instantaneous current of an AC sinusoidal waveform: vt () = Vˆ sin 2π ft = Vˆ sinω t it () = Iˆ sin 2π ft = Iˆ sinω t where ˆV and Î are the peak voltage and peak current respectively, and ω is the angular frequency in radians per second and f is the waveform frequency in cycles per second (hertz). Voltage and current in DC circuits are represented usually by the capital letters, V and I. AC voltage and current are represented by the same letters, but in lower case form, v and i. When instantaneous values of AC voltage and current are discussed, the symbols v(t) and i(t) are most commonly used. The peak value (amplitude) of an AC voltage or current waveform is shown by the symbols. ˆV or Î 2-4 Communications Technology 1

t EXAMPLE An emf, e, is given by the expression e(t) = 20 sin (2000πt) Volts Determine: (a) The amplitude of the waveform. (b) The angular frequency of the waveform. (c) The frequency of the waveform in Hz. (d) The time for one complete period of the wave. e (t) +20V Figure 2.3 Graph of the emf waveform -20V 0.5 1 1.5 T = 1ms t (ms) SOLUTION The waveform is illustrated in Figure 2.3. Comparing the equation e(t) = 20 sin (2000πt) to the general equation: vt () = Vˆ sin 2π ft = Vˆ sinω t we can deduce that: (a) The amplitude of the waveform is 20V (b) The angular frequency, ω = 2000π rad/s (c) The frequency, f = ω/2π = 2000π/2π = 1000Hz (d) The periodic time, T = 1/f = 1/1000 s = 1ms SAQ 1 In a domestic light bulb connected to the mains electricity supply, the instantaneous current is zero twice in each cycle of the current. Why doesn t the light go out during these times of zero current? Unit 2: AC Electricity 2-5

SAQ 2 A sinusoidal current, i(t), has a maximum value of 10A and a frequency of 100Hz. Calculate: (a) The angular frequency in radians per second. (b) The time for one complete cycle of the current. (b) Write down an expression for the instantaneous current, i(t), given that i(t) = 0 when t = 0 and sketch the waveform. SAQ 3 A sinusoidal voltage v(t) is described by the equation v(t) = 100 sin (100πt) Volts. Calculate: (a) The instantaneous value of the voltage after 6ms. (b) The first time after t = 0 that the instantaneous voltage is 50V. SUMMARY 1. A voltage or current which varies with some particular pattern is called a waveform. Waveforms which vary in magnitude and which change from positive to negative and vice versa are known as alternating or AC waveforms. 2. A periodic waveform is composed of the repetition of a single basic waveform pattern. 3. The following parameters may be defined for any periodic waveform: amplitude - the maximum value of the waveform cycle - a complete basic wave pattern period - the time T for one cycle frequency - the number of cycles in one second The unit of frequency is the hertz (Hz), where f = 1/T and T is the periodic time. 4. The instantaneous current and voltage of an AC sinusoidal waveform are given by the expressions: vt () = Vˆ sin 2π ft = Vˆ sinω t it () = Iˆ sin 2π ft = Iˆ sinω t 5. Two sinusoidal waveforms are said to be in phase when they have their maximum and minimum values at the same instant of time; otherwise, they are described as out of phase. 2-6 Communications Technology 1

ANSWERS TO SAQS SAQ 1 One of the reasons that the frequency of the mains electricity was chosen at 50 Hz throughout Europe was that this frequency was high enough to stop electric lights from flickering. In other words, the current does not stay at zero long enough to allow the light to diminish. Other electric devices for which the frequency of the mains supply is important are those driven by synchronous motors, such as record players and clocks. 50Hz is also convenient for them. In the United States, the mains frequency is slightly higher at 60 Hz. SAQ 2 Peak value or amplitude of current, I =10A frequency, f = 100Hz (a) angular frequency, ω (b) periodic time, T = 1/f = 1/100 = 0.01s = 10ms = 2πf = 2π 100 = 628 rads/s (c) instantaneous current, i(t) = Î sin ωt i(t) = 10 sin (200πt) SAQ 3 (a) At t = 6ms = 6 10-3 s and measuring angles in radians, v(t) = 100 sin (100π 6 10-3 ) = 100 sin (1.88) = 100 0.951 = 95.1V (b) Let t 1 = time at which v(t) is 50V 50 = 100 sin (100πt 1 ) 1 = sin( 100πt 1 ) 2 sin 1 1 2 = 100πt 1 0.52 = 100πt 1 giving t 1 = 0.52/100π = 1.67 ms. Unit 2: AC Electricity 2-7

POWER DISSIPATION AND RMS VALUES OBJECTIVES After studying this sub-unit, you should be able to: 1. Define or explain the meaning of the following terms:. average power. effective value of AC current or voltage. pure resistance. rms voltage and current 2. Calculate rms and peak values of sinusoidal AC waveforms, given the appropriate data. AVERAGE POWER DISSIPATION IN AN AC CIRCUIT If a DC current, I, flows in a pure resistance, R, and is constant with respect to time, as shown in Figure 2.4 then the power dissipation, P, is given by the expression P = I 2R Current I t (a) Current variation through a resistance R in a DC circuit Figure 2.4 Power dissipation in a DC circuit Power ( I 2 R) t (b) Power dissipation in a resistance R in a DC circuit However, consider a sinusoidal AC current, i(t), flowing through a pure resistance, R. The resulting instantaneous power dissipation, P(t), is given by the expression: P(t) = i 2 (t)r In the case of an AC current, the instantaneous power, P(t), is clearly not constant with respect to time. Figure 2.5 shows how both the instantaneous current, i(t), and the square of the instantaneous current i 2 (t), continually vary with respect to time. Current Figure 2.5 Power dissipation in an AC circuit ^ 2 I ^ 2 1/2 I ^ I π ω 2π ω 3π ω 2 i (t) i (t) Time, t i 2 2-8 Communications Technology 1

Note that squaring the instantaneous current, i(t), has the effect of removing the negative sign and making the result permanently positive; that is, the graph of the square of the instantaneous current, i 2 (t), never passes beneath the horizontal axis. Thus, the instantaneous power dissipation, P(t), is always positive, regardless of whether the instantaneous current, i(t), is flowing in the positive or negative direction. In this situation, it is more convenient to deal with average power dissipation, P av, rather than with the constantly changing instantaneous power, P(t). The average power dissipation is defined as the average value of the instantaneous power dissipation over one complete period. P av = average value of P(t) over one period, T = 2π/ω To find P av we perform the following, four-step operation: 1. We take the various positive and negative values of the instantaneous current, i(t), over one complete cycle. 2. We square all these values, making them all positive. 3. We then calculate the average of the squared values, i 2. 4. Finally, we multiply the average of the squared values over one complete cycle by the value of the resistance, R, thus finding the average power dissipated in that resistance. Therefore P av = i 2 R where i 2 is the average of the squares of the instantaneous current values over one complete period. For a sinusoidal AC current, such as that in Figure 2.5, we can draw a straight, horizontal line which is proportional to the constant value of the average power, given by the expression P av = i 2 R It can be shown that: the average value of the square of the instantaneous current is equal to half the square of the peak value of that current: i 1 = Iˆ 2 2 2 Thus, the average power dissipation, P av, can be written as: P av = i 2 2 R I = ˆ 2 R The average power dissipation of a pure resistance in an AC circuit is therefore equal to half the square of the peak current multiplied by the resistance. Unit 2: AC Electricity 2-9

ROOT-MEAN SQUARE (RMS) VALUES OF AC CURRENTS AND VOLTAGES We now have an expression for the average power dissipation for a sinusoidal AC current flowing in a pure resistance. Is this expression related to the average current? Reference to Figure 2.5 reveals that the answer to this question clearly is no. The area enclosed by the i(t) graph above the t-axis is equal to the area under the t-axis enclosed by the negative part of the i(t) graph. The average value of i(t) is therefore zero. Instead of average values for AC currents (or voltages), we use so-called effective values, where: An effective value of an AC current or voltage is that value which, if in DC form, would produce the same average power dissipation. The equation for average power dissipation in a purely resistive AC circuit contained the expression for the average value of the square of the instantaneous current, i 2. If we find the square root of this, we then have the effective current: effective current = i 2 = I ˆ 2 / 2 = 0707. I ˆ Note how we obtained this effective value: we found the square root of the average of the square. Another word for average is mean. The effective value of an AC current is therefore the square root of the mean of the square; or, as it is commonly known, the root-mean square (rms) value. We can follow the same reasoning and derive a similar expression for the rms value of an AC voltage. These expressions for the rms values of AC current and voltage are important ones and you should memorise them: i = 0. 707Iˆ v = 0. 707Vˆ RMS values are almost universally used to specify AC voltages and currents. For example, the voltage of the mains supply is quoted as 240V. By this is meant that the rms value of the voltage is 240V. Since the AC mains voltage is sinusoidal, is peak value is: 240 2 = 339V SAQ 4 Calculate the rms values of a sinusoidal voltage with a peak value 10V. SAQ 5 Calculate the peak values of a sinusoidal voltage waveform with an rms value of 20V. 2-10 Communications Technology 1

SUMMARY 1. When an AC current, i(t), is applied across a pure resistance, R, the power dissipation at any instant, P(t), continually varies as the instantaneous current varies. Mathematically: P(t) = i 2 (t)r 2. The average power dissipation of an AC current in a pure resistance is equal to half the square of the maximum value of that current, multiplied by the resistance: PAV = 1 I R 2 ˆ2 3. The effective value or root-mean square (rms) value of a sinusoidal AC current or voltage is that value, which if in DC form, would produce the same average power dissipation. 4. The rms values of a sinusoidal AC current and voltageare related to its peak amplitude values by the expressions: i = 0. 707Iˆ v = 0. 707Vˆ Unit 2: AC Electricity 2-11

ANSWERS TO SAQS SAQ 4 For a sinusoidal voltage, the rms value, v, is 0.707 times the peak value. That is: v = vˆ / 2 v = 0.707 ˆV = 0.707 10 = 707V SAQ 5 For a sinusoidal waveform: Vˆ = 2v = 1. 14v = 1.414 20 = 28.3V 2-12 Communications Technology 1

COMPONENT BEHAVIOUR IN AC CIRCUITS OBJECTIVES After studying this sub-unit, you should be able to: 1. Define or explain the meaning of the following terms:. capacitive reactance. impedance. inductive reactance. reactance. phasor diagram 2. Calculate the rms values of current, voltage and average power in simple resistive AC circuits. 3. Describe and calculate the response of a pure resistance to an AC voltage. 4. Describe and calculate the response of a pure capacitance to an AC voltage. 5. Describe and calculate the response of a pure inductance to an AC voltage. RESISTANCE IN AC CIRCUITS When an AC voltage, v, is applied across a resistance R, as shown in Figure 2.6, an alternating current, i, flows. voltage or current V ^ ^ I v i t Figure 2.6 AC current and voltage relationship in a pure resistance (a) graphical relationship of voltage and current i v (b) phasor relationship of voltage and current If the applied voltage is sinusoidal, then the voltage at any instant, v(t), is given by the expression: vt () = Vˆ sinω t Unit 2: AC Electricity 2-13

From Ohm s law, the instantaneous current will be: vt () Vˆ it () = = sin t Iˆ sin t R R ω = ω where ˆ Vˆ I = R Therefore the current and voltage are in phase as shown in Figure 2.6(a). Generally phase relationshipscan be conveniently indicated on what s called a phasor diagram In a phasor diagram, each sinusoidal waveform is represented by a single line called a phasor. The length of the phasor is proportional to the amplitude of the wave; the greater the amplitude, the longer the line. All lines are drawn from a single point, which may be imagined to be the centre of a clock face. The phase difference in radians between each waveform is represented by a corresponding angle between the lines. Leading phase angles are drawn in an anti-clockwise sense from the reference phasor, while lagging phase angles are drawn in a clockwise direction. In this case the voltage and current are in phase and therefore the phase angle between the phasors is zero as shown in Figure 2.6(b). For one half-cycle, the current is flowing in one direction through the resistor and for the next half-cycle, it is flowing in the opposite direction and so on. Furthermore, at two instants in time during each cycle, the current is zero. Multiplying both sides of the equation ˆV = ÎR by 0.707 converts the peak values of voltage and current to their rms values: 0. 707Vˆ = 0. 707IR ˆ v = ir which is the same basic form of Ohm s law as used in DC circuits. Thus, if rms values of voltage and current are used, a resistor behaves the same for AC as it does for DC. The techniques already encountered in our study of DC circuits - Ohm s law, Kirchoff s laws - are equally applicable to AC resistor networks. From the definition of the root-mean square (rms) value of AC waveforms, it follows that the average power, P av, dissipated in a resistance, R, is given by the expression: P av = v i = i 2 R = v 2 /R where v is the rms value of the potential difference across the resistor and i is the rms value of the current through it. So: power calculations in AC resistive circuits are exactly the same as in DC circuits, when the rms values of voltages and currents are used. For this reason we will drop the bar over the symbols for voltage v and current i, and proceed on the basis that when we use these lower case symbols in ac circuits, we are As we shall see, AC circuits which contain capacitors or inductors behave quite differently assuming that they are rms values. 2-14 Communications Technology 1

In conclusion: in AC resistive circuits, the current and voltage are in phase. If rms values of current and voltage are used, Ohm s and Kirchoff s laws can be applied, and the calculations of power dissipation are the same as those in DC circuits. EXAMPLE For the circuit shown in Figure 2.7, calculate: (a) The rms current supplied by the generator. (b) The rms voltage across the 2.2kΩ resistor. (c) The average power dissipated in the 6.8kΩ resistor. 18V rms 1kHz R 2 = 2.2 kω Figure 2.7 AC circuit with two resistances V R 1 = 6.8kΩ SOLUTION (a) The total circuit resistance, R T = 6.8 + 2.2 = 9kΩ The rms current, i = v/r = 18/(9 10 3 ) = 2mA (b) Let v 2 = rms voltage across R 2 = 2.2kΩ v 2 = ir 2 = 2 10-3 2.2 10 3 = 4.4V (c) Let P = average power dissipation in R 1 = 6.8kΩ P = i 2 R i = (2 10-3 ) 2 6.8 10 3 = 27.2 10-3 = 27.2mW SAQ 6 In the circuit shown in Figure 2.8, calculate: (a) The rms current supplied by the generator. (b) The average power dissipated in the 5.6 kω resistor. (c) The frequency of the AC current in Hz through the 2.7 kω resistor. Unit 2: AC Electricity 2-15

2.7kΩ 3.3kΩ 5.6kΩ Figure 2.8 Circuit for SAQ 6 v(t) = 30 sin (400πt) volts CAPACITANCE IN AC CIRCUITS The effect of placing a capacitor in a DC circuit is to stop the flow of electric current. Electrons cannot travel across the gap between the capacitor plates and so the continuous conducting path required for a current to flow is interrupted. For example, if a capacitor is connected in series with an electric light bulb and a DC power supply, the bulb will not illuminate as the circuit is open at the capacitor. If we remove the DC power supply, however, and replace it with an AC one, then the electric bulb does light up! If a variable capacitor is used, it can also be seen that the brightness of the bulb is directly proportional to the value of the capacitance and to the waveform frequency of the applied voltage. In this section, we shall examine in some detail the behaviour of a capacitance in an AC circuit. In our study of capacitance in Unit 1, we learnt that a capacitor of capacitance C, charged to a DC voltage V, stores an amount of charge, Q, given by the expression: Q = CV If an alternating voltage v is applied across a pure capacitance C, as shown in Figure 2.9, the capacitor will be continually charged and discharged. i Figure 2.9 AC voltage applied across a capacitor C V ^ sin ωt X Y C v c 2-16 Communications Technology 1

During the positive half-cycle of the voltage waveform, plate X of the capacitor becomes positively charged and plate Y negatively charged. During the negative half-cycle, X receives a negative charge and Y a positive one. There is therefore an alternating flow of charge or alternating current, i, through the capacitor Thus, unlike the case for a dc voltage where the capacitor eventually charges up to the value of the dc supply voltage and no more current flows, a capacitor continually conducts an ac current. Thus we may say that a capacitor passes ac and blocks dc. It may be shown that the alternating current is also sinusoidal but leads the supply voltage by a phase angle of π/2 radians (90 o ). The expression for the current i(t) maybe shown to be it () = Iˆ sin( ωt+ π / 2 ) = Iˆ cos ωt, where Iˆ = ωcvˆ Thus the graph of the instantaneous voltage across the capacitor is a sine function and the graph of the instaneous current through it is a cosine function. Figure 2.10 shows the plots of v and i as functions of time, illustrating the current/voltage relationship in a capacitor. voltage or current i C v C Figure 2.10 AC current and voltage relationship in a pure capacitance t The equations for instantaneous current and voltage are: vt () = Vˆ sinω t and i(t) = ω CVˆ sin( ω t + π / 2) These equations, together with their corresponding graphs, indicate very clearly that the instantaneous voltage and current in a capacitor are out of phase by a quarter of a cycle - π/2 radians or 90 o - with the current leading the voltage. Thus, the phasor diagram is as shown in Figure 2.11. The peak value of the current, Î, is related to the peak value of the voltage, ˆV, by the expression. Iˆ = ωcvˆ Vˆ 1 = Iˆ ωc Unit 2: AC Electricity 2-17

i C Figure 2.11 Phasor diagram showing the relationship between the instantaneous voltage and current in a capacitor V C 90 Converting to rms values gives: i = ωcv v 1 1 = = i ωc 2πfC where f is the waveform frequency in hertz. Note that this expression is in the form of Ohm s law, V/I = R. The quantity 1/ωC or 1/2πfC is the factor resisting the flow of current through the capacitor and is called capacitive reactance. The term resistance is not used to denote opposition in this case, for the following reason: power is dissipated in a resistance, but not in a capacitor. However, the units of capacitive reactance are ohms. The symbol for capacitive reactance is X C, where: Xc = Vˆ c / Iˆ c = vc / ic = 1/ωC = 1/2πfC ohms Note that X C is inversely proportional to both the capacitance, C, and the waveform frequency, f. Hence, if either (or both) f or C increase, the capacitive reactance decreases and vice versa. In other words, the greater the waveform frequency or the capacitance, the smaller the reactance and the larger will be the current flowing across the capacitor. The relationship between capacitive reactance and waveform frequency is shown in Figure 2.12. capacitive reactance X c X = c 1 2πfC Figure 2.12 Variation of capacitive reactance with waveform frequency frequency f(hz) 2-18 Communications Technology 1

EXAMPLE A 1µF capacitor has an rms current of 2mA flowing through it at a frequency of 1000Hz. Calculate the rms voltage across the capacitor. SOLUTION Capacitive reactance, X C = 1/2πfC = 1/(2π 1000 10-6 ) = 159Ω rms voltage across capacitor, v= ix c = 2 10-3 159 = 0.32V = 320mV SAQ 7 Why do capacitive reactances become small in high-frequency circuits, such as those in FM radios or TV sets? SAQ 8 A sinusoidal source of emf of 10V rms and frequency 2kHz is applied across a 0.1µF capacitor. Calculate the rms value of the current flowing through the capacitor. In conclusion we can state the following points concerning the behaviour of a capacitance in an AC circuit: If a sinusoidal waveform is applied across a capacitor, the current will lead the voltage by 90 o. Capacitive reactance is the opposition which every capacitor presents to the flow of current. Its units are ohms. INDUCTANCE IN AC CIRCUITS A piece of wire wound in the form of a coil possesses an electrical property known as inductance. The property arises from the observable phenomenon that if a current flowing through the coil changes for some reason then an emf e is induced in the coil which tries to oppose the current change. The magnitude of the induced emf is proportional to the rate of change of current. The constant of proportionality is known as the inductance of the coil L and is measured in a unit called the henry. Mathematically we can summarise this with the equation e = L di dt Where di/dt denotes the charge of current with respect to time. Consider a sinusoidal voltage, v, applied across a lossless (zero resistance) inductor of inductance L, as shown in Figure 2.13. Figure 2.13 Instantaneous voltage and current relationship in a pure inductance i L v L voltage or current v (t) L (t) i L Unit 2: AC Electricity 2-19

This results in a sinusoidal current flowing through the inductor, i, which is taken as the reference waveform with equation it () = Iˆ sinω t It may be shown that the voltage across the inductor is given by vt () = Vˆ cos ωt= Vˆ sin( ωt+ π / 2 ), where V ˆ = ωliˆ Figure 2.13 plots v and i as functions of time. Again, the equations for v and i indicate that the instantaneous voltage and current in an inductor are out of phase by an angle of 90 o (π/2 radians), with the voltage leading the current. The phasor diagram will be as shown in Figure 2.14. Figure 2.14 Phasor diagram showing the relationship between the instantaneous voltage and current in an inductor v L 90 i L Since Vˆ = ωliˆ Vˆ / Iˆ = v/ i = ωl = 2πfL where f is the waveform frequency in hertz. The quantity ωl = 2πfL is the factor resisting the flow of current and is called the inductive reactance, X L, measured in ohms. X L = v/i = ωl = 2πfL X L is directly proportional to frequency; the higher the frequency, the greater the inductive reactance. See Figure 2.15. inductive reactance Figure 2.15 Variation of inductive reactance with waveform frequency X L X L = 2 πfl frequency (Hz) 2-20 Communications Technology 1

Like capacitive reactance, X C, inductive reactance, X L, cannot be added directly to resistance in an electric circuit. No power is dissipated in a pure inductance; like a pure capacitance, it presents no resistance to the flow of current. Energy is repeatedly stored in the magnetic field surrounding the inductor and released back to the supply. A device that behaves in this way is called a reactor. Lossless capacitors and inductors are pure reactors. Practical capacitors and inductors are never quite lossless and contain some resistance, which dissipates some of the supplied power. SAQ 9 Manufacturers of resistors, those circuit components in circuits which offer specified resistances to the flow of current, always show the resistance value of their component by means of colour-coded bands on the outer surface of each resistor. In this way, the user can readily distinguish between - say - a 30kΩ resistor and a 2Ω one. Why don t capacitor and inductor manufacturers similarly display the reactance of their components? SAQ 10 When an AC voltage is applied across a pure inductance of 0.1H, a sinusoidal current, i = 300 sin (100πt) ma, flows. Calculate: (a) the rms current (b) the rms value of the applied voltage (c) the peak value of the supply voltage Write a mathematical expression for the instantaneous value of the supply voltage. PHASE DIFFERENCE BETWEEN SINUSOIDAL WAVEFORMS In circuits containing resistors and capacitors and/or inductors the phase difference between the voltages and currents may vary in a general way and do not necessarily have a phase difference of just 90, or one quarter of a cycle, (lag or lead), as is the case for a pure inductance or capacitance. Consider the two sinusoidal waveforms, v 1 and v 2, sketched in Figure 2.16(a). Unit 2: AC Electricity 2-21

v 2 v 1 π π ω 2π π ω 3 π θ = ωt 3 π ω = θ ω 2 t (a) in phase v1 v 2 Figure 2.16 Illustration of phase difference φ t d π π ω 2π (b) out of phase 3π θ= ωt 2π ω 3π t = θ ω ω We can say that v 1 and v 2 are in phase because v 1 has its maximum and minimum values at exactly the same instants of time as v 2. For the waveform in Figure 2.16 (b), however, this is not the case. The waveforms are displaced relative to each other and are said to be out of phase. In other words, there is a phase difference between v 1 and v 2. The amount of phase difference is expressed in terms of the phase angle φ, measured in radians on the ωt-axis. If we define v 1 as the reference waveform, then v 2 lags behind v 1 (since v 2 reaches it maximum value after v 1 ) by a phase angle φ. Thus: v1() t = Vˆ 1 sin ωt v2() t = Vˆ 2 sin( ωt φ ) A phase lag φ on the θ-axis corresponds to a time lag t d on the t-axis. Both voltages are sinusoidal waveforms with v 1 = 0 when t = 0 and v 2 = 0 when t = t d. That is: v 2 () t = Vˆ 2 sin( ω t ω t d ) v 2 (t d ) = 0 and thus ωt d - φ = 0 ωt d = φ t d = φ/ω We can therefore conclude that: the time delay between two out-of-phase sinusoidal waveforms is equal to the phase angle between them divided by their angular frequency. 2-22 Communications Technology 1

EXAMPLE A voltage v 1 in an electric circuit is sinusoidal, and has a peak value of 4V and a frequency of 200Hz. Another voltage, v 2, has the same frequency as v 1 and a peak value of 2V. Also it reaches its peak 1ms before v 1 does. (a) Determine the phase angle between v 1 and v 2. (b) Write mathematical expressions for v 1 and v 2 using v 1 as the reference waveform. SOLUTION (a) Time lead of v 2 over v 1 = 1ms angular frequency of v 2 and v 1, ω = 2πf = 2π 200 = 400π phase lead of v 2 over v 1, φ = ωt d = 400π 10-3 = 0.4π radians (b) If v 1 is the reference voltage, then: v () t = Vˆ sinω t 1 1 where ˆV 1 = 4 V and ω = 400π rads/s thus v 1 (t) = 4 sin (400πt)V Since v 2 leads v 1 by a phase angle: v () t = Vˆ sin( ω t + φ) 2 2 where ˆV 2 = 2 V. ω = 400π rads/s φ = 0.4π rads v 2 (t) = 2 sin (400πt + 0.4π)V = 2 sin π(400t + 0.4)V SAQ 11 A current in an AC circuit, i 1, is sinusoidal and has a peak value of 100mA and a frequency of 500Hz. Another current in the same circuit, i 2, has the same frequency but is twice as large as i 1 and lags it by a phase angle of π/4 radians (45 ). (a) Write the mathematical expressions for i 1 and i 2. (b) Calculate the time interval between a peak in the waveform of i 1 and a peak in the waveform of i 2. (c) Sketch roughly i 1 and i 2 on the same axis. Unit 2: AC Electricity 2-23

SUMMARY 1. In a purely resistive AC circuit, the current and voltage are in phase with one another. The rms values of voltage and current are related to resistance by Ohm s law: v = ir i = v/r R = v/i 2. The average power dissipation in a purely resistive circuit is given by the expressions: P = vi or P = i 2 R or P = v 2 /R where v and i are rms values. 3. In a purely capacitive AC circuit, the current leads the voltage by 90 o or π/2 radians. 4. Capacitive reactance, X C, measured in ohms, is the opposition that a pure capacitance, C, presents to the flow of AC current. It is inversely proportional to the angular or waveform frequency of the applied voltage. X C = 1/ωC or X C = 1/2πfC 5. The voltage and current in a purely capacitive circuit are related by the expression: v = ix C where v and i are rms. values. 6. In a purely inductive AC circuit, the current lags the voltage by 90 o or π/2 radians. 7. Inductive reactance, X L, measured in ohms, is the opposition that a pure inductance, L, presents to the flow of AC current. It is directly proportional to the angular or waveform frequency of the applied voltage. X L = ωl or X L = 2πfL 8. The voltage and current in a purely inductive circuit are related by the expression: v = ix L where v and i are rms values. 9. Capacitive and inductive reactances cannot be added directly to resistance, as a component which offers reactance does not dissipate power. 10. The phase difference between two out-of-phase sinusoidal waveforms is expressed in terms of the phase angle φ between them. The phase angle is equal to the time difference, t d, between the two waveforms multiplied by their angular frequency: φ = ωt d 2-24 Communications Technology 1

11. When comparing two out-of-phase waveforms, one is described as the reference waveform and the other waveform as lagging or leading the reference one. Unit 2: AC Electricity 2-25

ANSWERS TO SAQS SAQ 6 (a) The equivalent resistance, R p, of 5.6kΩ and 2.7kΩ in parallel is: R p = (5.6 2.7)/(5.6 + 2.7) = 1.82kΩ Total circuit resistance, R T = 3.3 + 1.82 = 5.12kΩ RMS voltage of generator, C = 30 0.707 = 21.2V RMS current of generator, i = v/r = 21.2/(5.12 10 3 ) = 4.14 10-3 = 4.14mA (b) Let v 1 be the rms voltage across R 1, the 3.3kΩ resistor and let v 2 be the rms voltage across R p, the parallel combination of the 5.6k Ω and the 2.7kΩ resistor. v 1 = ir 1 = 4.14 10-3 3.3 10 3 = 13.7V v 2 = v - v 1 = 21.2-13.7 = 7.5 The average power dissipation in the 5.6kΩ resistor is given by: = (7.5) 2 /(5.6 10 3 ) = 0.01W = 10mW (c) The frequency of the AC current through the 2.7kΩ resistor is the same as the frequency of the supply voltage. angular frequency, ω = 400πt waveform frequency, f = ω/2π = 400π/2π = 200Hz 2-26 Communications Technology 1

SAQ 7 Because capacitive reactance decreases as the frequency across the capacitor increases; the higher the frequency, the lower the reactance. SAQ 8 Capacitive reactance, X C = 1/2πfC = 1/(2π 2000 0.1 10-6 ) = 796Ω rms current through capacitor, i = v/x c = 10/796 = 0.0126A = 12.6mA SAQ 9 The resistive value of a resistor remains constant, regardless of the frequency of the AC current passing through it. However, the reactances of capacitors and inductors vary according to the AC frequency applied across them. As a result, manufacturers of capacitors and inductors cannot possibly predict their reactances in every circumstance; that depends on how the components are used. Unit 2: AC Electricity 2-27

SAQ 10 (a) The rms current = 0.707 Î = 0.707 300 = 212mA (b) Frequency of supply, f = ω/2π = 100π /2π = 50Hz inductive reactance, X L = 2πfL = 2π 50 0.1 = 31.4Ω rms value of supply voltage, v= ix L = 212 10-3 31.4 = 6.7V (c) Peak value of supply voltage, ˆV = v 2 = 6.7 1.414 = 6.7V The voltage across a pure inductance leads the current through it by π/2 radians or by 90 o. v(t) = 9.5 sin (100πt + π/2) volts 2-28 Communications Technology 1

SAQ 11 (a) i () t = Iˆ sinω t = Iˆ sin π ft 1 1 1 2 = 100 sin (2π 500 t) = 100 sin (1000πt) ma If i 2 is twice as large as i 1, then: Iˆ = 2Iˆ = 2 100 = 200mA 2 1 i 2 lags i 1 by a phase angle of φ = π/4 radians (45 o ). Therefore i 2 is of the form: i 2 = 200 sin (1000πt - π/4) ma (b) Time delay between waveforms, t d = φ/ω = (π/4)/1000π = 2.5 10-4 = 0.25ms (c) Your sketch of i 1 and i 2 should look like that shown in Figure 2.17. current (ma) 200 i 2 100 Figure 2.17 Waveforms in SAQ 11 0-100 - 200 0.25ms 1 2 3 i i t (ms) Unit 2: AC Electricity 2-29