ALTERNATING CURRENT (Contents: Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, Transformer.) 1. INTRODUCTION: * We have so far considered direct current (dc) sources and circuits with de sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. + The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current). * Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. * The mainreason for preferring use of ac voltage over de voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. + AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits.2. AC VOLTAGE APPLIED TO A RESISTOR: « Figure.1 shows a resistor connected to a source (€) of ac voltage. + The symbol for an ac source in a circuit diagram is ©). * We consider a source which produces sinusoidally varying potential difference across its terminals. * Let this potential difference, also called ac voltage, be given ay 1 v= vm sinot Where, vm is the amplitude of the oscillating potential difference and o is its angular frequency. i > Ew’ R FIGURE .1: AC voltage applied to a resistor. + Tofind the value of current through the resistor, we apply Kirchhoff’s loop rule Daa=o to the circuit shown in Fig. above to get, vm sin@t =iR or i=**sinot R Since Ris a constant, we can write this equation as, iH insinot (2) where the current amplitude /,, is given by on (3) ‘+ Equation (3) is Ohm’s law, which for resistors, works equallywell for both ac and de voltages. * The voltage across a pure resistorand the current through it, given by Eqs. (1) and (2) are plotted as a function of time in Fig.2. * Note, in particular that both v and / reach zero, minimum and maximum values at thesame time. Clearly, the voltage and current are in phase with FIGURE .2: Ina pure resistor, the voltage and current are in phase. The minima, zero and maxima * Like the applied voltage, the current varies Peemet teas sinusoidally and has corresponding positive and ieee ne negative valuesduring each cycle. + Thus, the sum of the instantaneous current each other.values over one complete cycle is zero, and the average current is zero. ‘+The fact that the average current is zero, however, does not mean that the average power consumed is zero andthat there is no dissipation of electrical energy. + As youknow, Joule heating is given by ®R and depends on i? (which is always positive whether /is positive or negative}and not on i. ‘+ Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. «The instantaneous power dissipated in the resistor is p=PR=/Rsin*ot The average value of p over a cycle is P== 5a] where the bar over a letter (here, p) denotes its average value and <..... > denotes taking average of the quantity inside the bracket. Since, ?,and R are constants, P=/R .5(b)] Using the trigonometric identity, sin? wt =1/2 (1—cos 20t), we have = (1/2) (1-< cos 2t>}and since < cos2ut > = 0, we have, =+ Thus, + To express ac power in the same form as de power (P = PR), a special value of current is defined and used.It is called, root mean square (rms) or effective current (Fig.3) and is denoted bv! ori. FIGURE .3: The rms current is related to the peak current jy By 1 = in/ 2 =0.707 inItis defined by 0.707 i, vm (6) In terms of /, the average power, denoted by P is p=P=4?R=7R (7) 2” Similarly, we define the rms voltage or effective voltage by V= JF 0.707 v,, ws (8) From €q, (7.3), we have Vp = ig (2) Equation (9) gives the relation between ac current and ac voltage and is similar to that in the de case. This shows the advantage of introducing the concept of rms values In terms of rms values, the equation for power [Eq. (7)] and relation between current and voltage in ac circuitsare essentially the same as those for the dc case. * It is customary to measure and specify rms values for ac quantities. Forexample, the household line voltage of 220 V is an rms value with a peak voltage of v,= V2 V= (1.414)(220 v) = 311.V * In fact, the / or rms current is the equivalent de current that would produce the same average power loss as the alternating current. Equation(7) can also be written as P=V/R=IV (since V=1R) Example. 1: A light bulb is rated at 100W for a 220 V supply. Find {@) the resistance of the bulb; (b) the peak voltage of the source; and {c) the rms current through the bulb. Solution (a) We are given P = 100 W and V bulb is 220 V. The resistance of the v2_@20vy | P 100W (b) The peak voltage of the source is. v, =v anv {c) Since, P p _100w Vv R0v 1 4540,+ REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS: © The current through a resistor isin phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a combination of these circuit elements. © Inorder to show phase relationship between voltage and current in an ac circuit, we use the notion of phasors. © The analysis of an ac circuit is facilitated by the use of a phasor diagram. A phasor is a vector which rotates about the origin with angular speed ©, as shown in Fig. * The vertical components of phasors V and | represent the sinusoidally varying quantities v and /. * The magnitudes of phasors V and 1 represent the amplitudes or the peak values v,, and i,, of these oscillating quantities. : i,sin wt, v,,8in wt, @ © FIGURE: (a) A phasor diagram for thecircuit in Fig. (b) Graph of v and i versus et. + Figure (a) shows the voltage and current phasors and their relationship at time ¢, for the case of an ac source connected toa resistori.e., corresponding to the circuit shown in Fig. * The projection of voltage and current phasors on vertical axis, i.e., v,, sino t and im Sino t, respectively represent the value of voltage and current at that instant. As they rotate with frequency «, curves in Fig. (b) are generated From Fig. (a) we see that phasors V and I for the case of a resistor arein the same direction. This is so for all times, ‘+ This means that the phase angle between the voltage and the currents zero.3. AC VOLTAGE APPLIED TO AN INDUCTOR: FIGURE: An ac source connected to an inductor. + Figure shows an ac source connected to an inductor. ‘+ Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. © Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v= v,, sino t. oo () ‘© Using the Kirchhoff’s loop rule, >) £(¢, 0, and since thereis no resistor in the circuit, (2) where the second term is the self- induced Faraday emf in the inductor; and L is the self-inductance of the inductor. + The negative sign follows from Lenz’s law. * Combining Eqs. (1) and (2), we have ML? Pm sin wt ove (3) aL Equation (3) implies that the equation for i(9, the current as a function of time, must be such that its slope di/dtis a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by v,,/L. To obtain the current, we integrate di/dt with respect to Tr founteoonte and get, (= —2e cos(ut) +constant (The integration constant has the dimension of current and is time independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time- independent component of the current exists. Therefore, the integration constant is zero.)Using ~cos(eo) = sin{ ot ln sin oe (4) », where f, =—"js the amplitude of the current. The quantity wl is analogous to the resistance and is called inductive reactance, denotedby x,: X=0L The amplitude of the current is, then (5) in OE ve (6) The dimension of inductive reactance is the same as that of resistanceand its SI unit is ‘ohm (Q). The inductive reactance limits the current in apurely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A.comparison of Eqs. (1) and (4) for the source voltage and thecurrent in an inductor shows that the current lags the voltage by x/2 orone-quarter (1/4) cycle. Figure. (a) shows the voltage and the current phasors in the present case at instant t . The current phasor I is /2 behind the voltage phasor V. When rotated with frequency counter- clockwise, they generate the voltage and current given by Eqs. (1) and (5), respectively and as shown in Fig. (b). rs v,8in wt, i,sin(@t- 2/2) @ ©) FIGURE: (a) A Phasor diagram for the circuit. (b) Graph of v and i versus wt.We see that the current reaches its m: T _n/3 4 0 | ‘An inductor has reactance that limits current simi The instantaneous power supplied to the inductor i F to resistance in a de circ PB =iv =i, sin[ ot -3] Xv, sin(wt) vp 608 (A) stn ( at) {a0 sn (2ut) the average power over a complete cycle 1s, {a sun(2u0)) inte (sin(2eut)) since the average of sin (208) over a complete cycle is. zero. ‘Thus, the average power supplied to an inductor over one complete cycle is zero. Example. 2: A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. Solution The inductive reactance, X, =2xVL=2x 3.14 x 50% 25%10°O 850 ‘The rms current in the circuit is v _ 220V X 7.850 1um value later than the voltage by one-fourth of a period4. AC VOLTAGE APPLIED TO A CAPACITOR: * Figure shows an ac source € generating ac voltage v=vm sin ot wom (1) connected to a capacitor only, a purely capacitive ac circuit. E~O =e FIGURE: An ac sourceconnected to a capacitor. ‘When a capacitor is connected to a voltage sourcein a de circuit, current will flow for the short time required to charge the capacitor. + As charge accumulates on the capacitor plates, the voltage across them increases, ‘opposing the current. That is,a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, as in Fig., it limits or regulates the current, but does not completely prevent the flow of charge. ‘+The capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor 4 one 2) From the Kirchhof’s loop rule, the voltage across the source and the capacitor are equal, a », e-4 rm sie ct = ‘To find the current, we use the relation IqCSIN WE} = W/CDy COSfw) Using the relation, costwt) = sin{ wt +5) . we have in sin{ oe +3) 8) where the amplitude of the oscillating current is 4,, = @Cv,, We can rewrite tas fe 170 Comparing 1 tof, (1/@C) plays the Is denoted by X,. X= 1/oc so that the amplitude of the current ts. bn i»/R for a purely resistive circuit, we find that resistance. ICs called capacitive reactance and @ Xe 8)‘* | The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (2). ‘+ The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limitsthe current in a purely resistive circuit. But itis inversely proportional to the frequency andthe capacitance. ‘+A comparison of Eq. (3) with the equation of source voltage, Eq. (1) shows that the current is x/2 ahead of voltage. i,sin( @t,+7/2) fe) ow FIGURE: (a) A Phasor diagram for the circuitin Fig. 7.8, (b) Graph of v and i versus wt. + Figure (a) shows the phasor diagram at an instant t,. Here the current phasor 1 is x/2 ahead of the voltage phasor V as they rotate counterclockwise. * Figure (b) shows the variation of voltage and currentwith time. We see that the current reaches its maximum value earlier thanthe voltage by one-fourth of a period. * The instantaneous power supplied to the capacitor is- = 10= §, COS(OD ep, SIN(O1) = 4,0, costo) sin(ot) Pe * {nf sinus) ‘So, as in the ease of an inductor, the average power id bP . ( : sint2at)) AS (sin(2at)) =0 since = 0 over a complete eycle. «Thus, we see that in the case of an inductor, the current lags the voltageby x/2 and in the case of a capacitor, the current leads the voltage by 2/2. 10Example. 3: A lamp is connected in series with a capacitor. Predict your observations for de and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Solution When a de source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if Cis reduced. ‘With ac source, the capacitor offers capacitative reactance (1/wC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly ‘than before. Example. 4: A 15.0 j1F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Solution The capacitive reactance Is ee © 2nve — 2m(5SOHz)(15.0 x 10° F) =2120 ‘The rms current is Vv _20V x 2120 The peak current is fg = V1 = (1.41)(2.08 A) = 1.470 ‘This current oscillates between +1.47A and ~1.47 A, and the voltage by 7/2. If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled Example. 5: A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. oa The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (¢) is unchanged, as the iron rod is inserted. Give your answer with reasons. Solution As the iron rod is inserted, the magnetic field inside the coil ‘magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coll increases. Consequently, the inductive reactance of the coll increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases. ahead of 115.AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT: figure reas ‘A series LCR circuit connected to on ac source Figure shows a series LCR circuit connected to an ac source £. As usual, we take the voltage of the source to be vevpsin wt a If q is the charge on the capacitor and / the current, at time t, we have, from Kirchhoff’s loop rule: ruins ae c We want to determine the instantaneous current /and its phase relationship to the applied alternating voltage v. We can solve this problem by two methods. First, use the technique of phasors and in the second method, we can solve Eq. (2) analytically to obtain the time— dependence of i v (2) 12Phasor-diagram solution From the circuit shown in Fig., we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element isthe same at anytime, having the same amplitude and phase. Let it be i, sinfoted) (3) where @ is the phase difference between the voltage across the source and the currentin the circuit. Construction of a phasor diagram for the present case, Let I be the phasor representing the current in the circuit as given by £q.(3). Further, let V,, Vq, Vo. and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that V, is parallel tol, V, is /2 behind and V is /2 ahead of |. V/V Vand! are shown in Fig.(a) with appropriate phase- relations. The length of these phasors or the amplitude Of Vy, Ve and V, are: Van = bon Re Wem = don Ker Vem = bm Mi (a) The voltage Equation (7.20) for the circuit can be written as Wt vetveeY (s) w o) The phasor relation whose vertical component FIGURE . (2) Relation between the gives the above equation is Nese vol idea yg ommend between the phasors V,, Vy, and (VY, + i) VitVatVeaV (6) for the circuit in Fig. This relation is represented in Fig. (b).Since V, and V, are always along the same line and in opposite directions, they can be combined into a single phasor (V, + V,) which has a magnitude | v,., ~ v,,| . Since V is represented as the hypotenuse of a right-triangie whose sides are V,and (V, + V,), the pythagorean theorem gives: veev 4(y -v ) Substituting the values Of Ven Vem ANG V,, from Eq. (4) into the above equation, we have wai RP +X -1 XP 2 [a ox, -x P) (Pome (a) By analogy to the resistance in a circult, we introduce the impedance Z inanaccircut or, a (7(o)1 where Z= fA +0. —X,) (8) 13FIGURE (¢) Impedance diagram, Since phasor T is always parallel to phasor V _, the phase angle is the angle between V, and V and can be determined from Fig. 7.12: tang = Yen Min Using Eq. (7.22), we have Xe =X R (9) Equations (8) and (9) are graphically shown in Fig. (c). This is called Impedance diagram which is a right-triangle withZ as its hypotenuse. tand Equation 7(a) gives the amplitude of the current and Eq. (9) gives the phase angle. With these, Eq. (3) is completely specified If X,>X, @ is positive and the circuit is predominantly capaci e. Consequently, the current in the circuit leads the source voltage. If X_X,. @) Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of tand draw different phasors which show the relative angle between phasors. » FIGURE (4) Phasor diagram of V and I. and Graphs of v and i versus @ ¢ for a sei eR circuit where X,> X,. 146. RESONANCE: An interesting characteristic of the series ALC circuit is the phenomenon of resonance. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequencyis called the system's natural frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large. A familiar example of this phenomenon is a child on a swing. The swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging, the amplitude of the swinging will be large. For an RLC circuit driven with voltage of amplitude v,, and frequency ©, we found that the current amplitude is given by At resonant frequency, the current amplitude is maximum; /,, = v,/R- Figure 7.16 shows the variation of j,, with in a RLC series circuit with L = 1.00 mH, C= 1.00 nF for two values of R: (i) R = 100 2 and (ii) R = 200 ©. For the source applied v,,» 1 100 V. @, for this case is 77E = 1.00105 og alg, 5 Meads rad/s. FIGURE . Variation of i with @ for two cases: (i) R = 100 © (ii) R = 200 0, (00 mH, We see that the current amplitude is maximum at the resonant frequency. Since i, v,,/ Rat resonance, the current amplitude for case (i)s twice to that for case (i). Resonant circuits have a variety of applications, for example, in the tuning mechanism ofa radio ora TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. in tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximurn. 15It is important to note that resonance phenomenon is exhibited by a circuit only if both Land C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is v,/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. 16Example .6 A resistor of 200 @ and a capacitor of 15.0 41F are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Solution Given R= 2000, C =15.0nF =15.0x 10°F V =220V, v = SOHz (2) inorder to calculate the current, we need the impedance of the circuit. It is 2=|POXE= (RV RRO = 200 + (2x 3:14 x 5015.05 10°F)= =/2000F + @1230" = 201670 Therefore, the current in the circuft is <0 = 2Y Lassa z isa {b) Since the current is the same throughout the circuit, we have Vp = 1R = (0.755 A)(2000) = 151¥ Ve= 1X = (0.755 A)(212.30) = 160.3 V The algebraic sum of the two voltages, V_and V-is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not jn the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. ‘Therefore, the total of these voltages must be obtained using the Pythagorean theorem: vo = verve =220V ‘Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source. POWER IN AC CIRCUIT and THE POWER FACTOR: We have seen that a voltage v= v,, sinovt applied to a series RLC circuit drives a current in the circuit given by i= i, sinfot-+4) where . (4) ly “2 and ¢ = tan” Therefore, the instantaneous power p supplied by the source is 17p=vi=(vmsinot)x[in sinfot+@)) = “Teo = cos(2ot +6)] (a) ‘The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.37). It is only the second term which is time-dependent, Its average is zero (the positive half of the cosine cancels the negative half). Therefore, Pe cosh = “0 In 2 v2 V2 =Vicosh (2(@)] This can also be written as, cos Pp 2Z cos [2(b)) So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle between them. The quantity cosé is called the power factor. Let us discuss the following cases: Case (i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case $= 0, cos § =1. There is maximum power dissipation. Case (Ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is x/2. Therefore, cos @ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. Case (Ili) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq, (2) where 6 = tan (X —X )/R. So, 4 may be non-zero in @ RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. Case (iv) Power dissipated at resonance in LCR circuit: At resonance X,-X,= 0, and = 0. Therefore, cos = 1 and P = 122 = /? R. That is, maximum power is dissipated in a circuit (through A) at resonance. Example .7 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. {b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. Solution (a) We know that P = / V cos} where cos is the power factor. To supply a given power at a given voltage, if cos@ is small, we have to increase current accordingly. But this will lead to large power loss (FR) in transmission. (b)Suppose in a circuit, current / lags the voltage by an angle 6. Then power factor cos =R/2. We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram (Fig. below) 18ae ° 7 FIGURE how this can be achieved. Let us resolve I into two components. I, along the applied voltage V and |, perpendicular to the applied voltage. |, as you have learnt in Section 7.7, is called the wattless ‘component since corresponding to this component of current, there is no power loss. |, is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current |, by an equal leading wattless current 1, This can be done by connecting 2 capacitor of appropriate value in parallel so that I, and I’, cancel each other and P is effectively J, V. Example .8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which Q, L= 25.48 mH, and C = 796 pF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor. Solution {a) To find the impedance of the circuit, we first calculate X,and X,. X,=2 vl 23.14 x 50x 25.48 x 10° 0= 8 1 x ve 1 ta, 2x 3:14 x 50 x 796 x10 Therefore, 2=\R a0, xX =F 0-4 =52 (b) Phase difference, § = tan? <= R =tan[4=) - ar 19Since ¢ 15 negative, the current in the cireutt lags the voltage across the source. (c) The power dissipated in the circuit 1s P=PR wn an ‘Therefore, P = (40A)’ x30 =4800W (d) Power factor =c0s¢ = cos(-53.1°) =0.6 Example .9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is 1 a eo Jee Y25.48 x10 796 x10 = 222.1rad/s y, 92 22d, "an 2x36 Hz = 35.4Hz (b) The impedance Z at resonant condition is equal to the resistance: Z=R=30 ‘The rms current at resonance is Foran ‘The power dissipated at resonance is 687A, P =P? xR = (66.7)? x3 =13.35 kW You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example .8. Example -10 At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. if she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes — resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. 208. TRANSFORMERS: For many purposes, it is necessary to change (or transform) an alternating, voltage from one to another of greater or smaller value. This is done with a device called transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig.(a) or on separate limbs of the core as in Fig.(b). One of the coils called the primary coil has N, turns. The other coil is called the secondary coil; it has N, turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. Soft iron-core . + ° a ¢ —H cr Z iP 8 7 +1 3 == 5 3 . at . a ® ” FIGURE . Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other, (b) two coils on separate limbs of the core. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let @ be the flux in each turn in the core at time t due to current in the primary when a voltage v, is applied to it, Then the induced emf or voltage &., in the secondary with IV turns is a eo @ a a ‘dt The alternating flux @ also induces an emf, called back emf in the primary. This is 4 5 =v 2) € Pde Bute, =v, If this were not so, the primary current would be infinite since the prifnary hes zero resistance (as assumed). Ifthe secondary is an open circuit or the current taker from it is small, then toa good approximation oY,where v, is the voltage across the secondary. Therefore, Eqs. (1) and(2) can be written as hy a a) at a - a) ae From Eqs. [1 (a)] and [2 (a)], we have vs (3) Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (il) the secondary current is small. Ifthe transformer is assumed to be 100% efficient (no energy losses), ‘the power input is equal to the power output, and since p =v, ivy = hv. (4) Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (3) and (4), we have fs Me (5) i 0, 0, » Np Since / and v both oscillate with the same frequency as the ac source, Eq. (5) also gives the ratio of the amplitudes or rms values of corresponding quantities. Now, we can see how a transformer affects the voltage and current. We have: v-(# Vp and Is (6) N, P That is, if the secondary coil has a greater number of turns than the primary (N, > W,), the voltage is stepped up (V, > V,). This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (N,/N,< 1. and J, /,. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: 22(i) Flux Leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire(/?R). in high current, low voltage windings, these are minimised byusing thick wire. (ii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using alaminated core. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept toa minimum by using a magnetic material which has a low hysteresis loss. The large scale transmission and distribution of electrical energy overlong distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the / 7R loss is cut down). Itis then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. 23SUMMARY 1, An alternating voltage V= vv, sin @t applied to a resistor R drives a current =/, sinwt inthe resistor, jy = 2 the current isin phase with the applied voltage. 2. For an alternating current i= /,sin ot passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is (172 Wap. To express It in the same form as the de power (P= 1?R), a special "value of current is used. It is called rect mean square (rms) current and is donoted by f: 12% 07071, 2 ‘Similarly, the rms voltage is defined by va¥ 2070, 2 We have P= IV=7°R 3. An ac voltage v = vq sin oot applied to a pure inductor t, drives a current in the inductor 7 = ip sin (at ~ 2/2), where in = Val. X= Ol 15 called inductive reactance. The current in the inductor lags the voltage by ‘n/2. The average power supplied to an inductor over one complete cycle is zero. ure v v 1 A z z Me o 24An ac voltage v = v, sinat applied to a capacitor drives a current in the ‘capacitor: / = jp sin (ot + 7/2). Here, v, 1 i= x oh Meg Ie called capacitive reactance ‘The current through the capacitor is =/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor ‘over one complete cycle Is zero. For a series RLC circuit driven by voltage v= v,, sin wt, the current is given by / =/,, sin (ot + 4) where: = Ya Cray ochre EME fae +(X_—X,} |s called the impedance of the circuit The average power loss over a complete cycle is given by P= V1 cos ‘The term cosh is called the power factor. In a purely inductive or capacitive circuit, cos = 0 and no power Is dissipated even though a current is flowing in the circuit. In such cases, current Is referred to as a wattless current. ‘The phase relationship between current and voltage in an ac circult can be shown conveniently by representing voltage and current by rotating vectors called phasors. A phasor is a vector which rotates about the origin with angular speed @. The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. ‘The analysis of an ac circuit is facilitated by the use of a phasor diagram, A transformer consists of an iron core on which are bound a primary of N, turns and a secondary coil of N, turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by vat) + |—+ Ve N wy) and the currents are related by (mY WJ If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (V, > V,). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer. 25