ALTERNATING CURRENT. Chapter Seven 7.1 INTRODUCTION

Transcription

1 Chapter Seven ALTENATING CUENT 7. INTODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with tie. But voltages and currents that vary with tie are very coon. The electric ains supply in our hoes and offices is a voltage that varies like a sine function with tie. 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, ost of the electrical devices we use require ac voltage. This is ainly because ost of the electrical energy sold by power copanies is transitted and distributed as alternating current. The ain reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted fro one voltage to the other by eans of transforers. Further, electrical energy can also be transitted econoically over long distances. AC circuits exhibit characteristics which are exploited in any devices of daily use. For exaple, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits one of any that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they ean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying siple haronic tie dependance has becoe so universally accepted that we follow others in its use. Further, voltage another phrase coonly used eans potential difference between two points.

2 Physics 7. AC VOLTAGE APPLIED TO A ESISTO Figure 7. shows a resistor connected to a source ε of ac voltage. The sybol for an ac source in a circuit diagra is ~. We consider a source which produces sinusoidally varying potential difference across its terinals. Let this potential difference, also called ac voltage, be given by v = v sin ωt (7.) where v is the aplitude of the oscillating potential difference and ω is its angular frequency. NICOLA TESLA ( ) 34 Nicola Tesla ( ) Yugoslov scientist, inventor and genius. He conceived the idea of the rotating agnetic field, which is the basis of practically all alternating current achinery, and which helped usher in the age of electric power. He also invented aong other things the induction otor, the polyphase syste of ac power, and the high frequency induction coil (the Tesla coil) used in radio and television sets and other electronic equipent. The SI unit of agnetic field is naed in his honour. FIGUE 7. In a pure resistor, the voltage and current are in phase. The inia, zero and axia occur at the sae respective ties. To find the value of current through the resistor, we apply Kirchhoff s loop rule ε ( t ) =, to the circuit shown in Fig. 7. to get v sin ω t = i v or i = sin ω t Since is a constant, we can write this equation as i = i sin ω t (7.) where the current aplitude i is given by i v FIGUE 7. AC voltage applied to a resistor. = (7.3) Equation (7.3) is just Oh s law which for resistors works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.) and (7.) are plotted as a function of tie in Fig. 7.. Note, in particular that both v and i reach zero, iniu and axiu values at the sae tie. Clearly, the voltage and current are in phase with each other. We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values during each cycle. Thus, the su of the instantaneous current values over one coplete cycle is zero, and the average current is zero. The fact that the average current is zero, however, does

3 Alternating Current not ean that the average power consued is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i and depends on i (which is always positive whether i 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 = i = i sin ω t (7.4) The average value of p over a cycle is* p = < i > = < i sin ωt > [7.5(a)] where the bar over a letter(here, p) denotes its average value and <...> denotes taking average of the quantity inside the bracket. Since, i and are constants, p = i < sin ωt > [7.5(b)] Using the trigonoetric identity, sin ωt = / ( cos ωt), we have < sin ωt > = (/) ( < cos ωt >) and since < cosωt > = **, we have, < sin ω t > = Thus, p = i [7.5(c)] To express ac power in the sae for as dc power (P = I ), a special value of current is defined and used. It is called, root ean square (rs) or effective current (Fig. 7.3) and is denoted by I rs or I. George Westinghouse (846 94) A leading proponent of the use of alternating current over direct current. Thus, he cae into conflict with Thoas Alva Edison, an advocate of direct current. Westinghouse was convinced that the technology of alternating current was the key to the electrical future. He founded the faous Copany naed after hi and enlisted the services of Nicola Tesla and other inventors in the developent of alternating current otors and apparatus for the transission of high tension current, pioneering in large scale lighting. GEOGE WESTINGHOUSE (846 94) FIGUE 7.3 The rs current I is related to the peak current i by I = i / =.77 i. * The average value of a function F (t) over a period T is given by F( t) = F( t) T T T sin ω t < cos > = cos = sin T T = = ω ωt T ** ω t ωt [ ω T ] 35

4 Physics It is defined by i I = i = i = =.77 i (7.6) In ters of I, the average power, denoted by P is P = p = i = I (7.7) Siilarly, we define the rs voltage or effective voltage by v V = =.77 v (7.8) Fro Eq. (7.3), we have v = i v i or, = or, V = I (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is siilar to that in the dc case. This shows the advantage of introducing the concept of rs values. In ters of rs values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the sae as those for the dc case. It is custoary to easure and specify rs values for ac quantities. For exaple, the household line voltage of V is an rs value with a peak voltage of v = V = (.44)( V) = 3 V In fact, the I or rs current is the equivalent dc current that would produce the sae average power loss as the alternating current. Equation (7.7) can also be written as P = V / = I V (since V = I ) Exaple 7. A light bulb is rated at W for a V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rs current through the bulb. Solution (a) We are given P = W and V = V. The resistance of the bulb is 36 EXAMPLE 7. ( ) V V = = = 484Ω P W (b) The peak voltage of the source is v = V = 3V (c) Since, P = I V P W I = = =.45A V V

5 7.3 EPESENTATION OF AC CUENT AND VOLTAGE BY OTATING VECTOS PHASOS In the previous section, we learnt that the current through a resistor is in phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a cobination of these circuit eleents. In order 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 diagra. A phasor* is a vector which rotates about the origin with angular speed ω, as shown in Fig The vertical coponents of phasors V and I represent the sinusoidally varying quantities v and i. The agnitudes of phasors V and I represent the aplitudes or the peak values v and i of these oscillating quantities. Figure 7.4(a) shows the voltage and current phasors and their relationship at tie t for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.. The projection of voltage and current phasors on vertical axis, i.e., v sinωt and i sinωt, respectively represent the value of voltage and current at that instant. As they rotate with frequency ω, curves in Fig. 7.4(b) are generated. Fro Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the sae direction. This is so for all ties. This eans that the phase angle between the voltage and the current is zero. 7.4 AC VOLTAGE APPLIED TO AN INDUCTO Figure 7.5 shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assue that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v = v sinωt. Using the Kirchhoff s loop rule, ε ( t) =, and since there is no resistor in the circuit, di v L = (7.) d t where the second ter is the self-induced Faraday ef in the inductor; and L is the self-inductance of Alternating Current FIGUE 7.4 (a) A phasor diagra for the circuit in Fig 7.. (b) Graph of v and i versus ωt. FIGUE 7.5 An ac source connected to an inductor. * Though voltage and current in ac circuit are represented by phasors rotating vectors, they are not vectors theselves. They are scalar quantities. It so happens that the aplitudes and phases of haronically varying scalars cobine atheatically in the sae way as do the projections of rotating vectors of corresponding agnitudes and directions. The rotating vectors that represent haronically varying scalar quantities are introduced only to provide us with a siple way of adding these quantities using a rule that we already know. 37

6 38 Interactive aniation on Phasor diagras of ac circuits containing,, L, C and LC series circuits: Physics the inductor. The negative sign follows fro Lenz s law (Chapter 6). Cobining Eqs. (7.) and (7.), we have v di v = = sin ω t (7.) L L Equation (7.) iplies that the equation for i(t), the current as a function of tie, ust be such that its slope di/ is a sinusoidally varying quantity, with the sae phase as the source voltage and an aplitude given by v /L. To obtain the current, we integrate di/ with respect to tie: v di d t = sin( ωt )d t L and get, v i = cos( ω t) + constant ωl The integration constant has the diension of current and is tieindependent. Since the source has an ef which oscillates syetrically about zero, the current it sustains also oscillates syetrically about zero, so that no constant or tie-independent coponent of the current exists. Therefore, the integration constant is zero. Using π cos( ωt) = sin ω t, we have π i = i sin ωt (7.) v where i L analogous to the resistance and is called inductive reactance, denoted by X L : = is the aplitude of the current. The quantity ω L is ω X L = ω L (7.3) The aplitude of the current is, then i v = X (7.4) L The diension of inductive reactance is the sae as that of resistance and its SI unit is oh (Ω). The inductive reactance liits the current in a purely inductive circuit in the sae way as the resistance liits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A coparison of Eqs. (7.) and (7.) for the source voltage and the current in an inductor shows that the current lags the voltage by π/ or one-quarter (/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t. The current phasor I is π/ behind the voltage phasor V. When rotated with frequency ω counterclockwise, they generate the voltage and current given by Eqs. (7.) and (7.), respectively and as shown in Fig. 7.6(b).

7 Alternating Current FIGUE 7.6 (a) A Phasor diagra for the circuit in Fig (b) Graph of v and i versus ωt. We see that the current reaches its axiu value later than the T π/ voltage by one-fourth of a period = 4 ω. You have seen that an inductor has reactance that liits current siilar to resistance in a dc circuit. Does it also consue power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is π pl = i v = i t v t sin ω sin ( ω ) ( ω ) sin( ω ) = i v cos t t iv = sin ( ωt ) So, the average power over a coplete cycle is iv PL = sin ( ω t ) iv = sin ( ω t ) =, since the average of sin (ωt) over a coplete cycle is zero. Thus, the average power supplied to an inductor over one coplete cycle is zero. Figure 7.7 explains it in detail. Exaple 7. A pure inductor of 5. H is connected to a source of V. Find the inductive reactance and rs current in the circuit if the frequency of the source is 5 Hz. Solution The inductive reactance, X L =. 3 π ν L = W = 7.85Ω The rs current in the circuit is V V I = = = 8A X 7.85 Ω L EXAMPLE 7. 39

8 Physics - Current i through the coil entering at A increase fro zero to a axiu value. Flux lines are set up i.e., the core gets agnetised. With the polarity shown voltage and current are both positive. So their product p is positive. ENEGY IS ABSOBED FOM THE SOUCE. - Current in the coil is still positive but is decreasing. The core gets deagnetised and the net flux becoes zero at the end of a half cycle. The voltage v is negative (since di/ is negative). The product of voltage and current is negative, and ENEGY IS BEING ETUNED TO SOUCE. One coplete cycle of voltage/current. Note that the current lags the voltage. -3 Current i becoes negative i.e., it enters at B and coes out of A. Since the direction of current has changed, the polarity of the agnet changes. The current and voltage are both negative. So their product p is positive. ENEGY IS ABSOBED. 3-4 Current i decreases and reaches its zero value at 4 when core is deagnetised and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. ENEGY ABSOBED DUING THE /4 CYCLE -3 IS ETUNED TO THE SOUCE. 4 FIGUE 7.7 Magnetisation and deagnetisation of an inductor.

9 7.5 AC VOLTAGE APPLIED TO A CAPACITO Figure 7.8 shows an ac source ε generating ac voltage v = v sin ωt connected to a capacitor only, a purely capacitive ac circuit. When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short tie required to charge the capacitor. As charge accuulates on the capacitor plates, the voltage across the increases, opposing the current. That is, a capacitor in a dc circuit will liit 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. 7.8, it liits or regulates the current, but does not copletely 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 tie t. The instantaneous voltage v across the capacitor is q v = (7.5) C Fro the Kirchhoff s loop rule, the voltage across the source and the capacitor are equal, v q sin ω t = C dq To find the current, we use the relation i = d i = ( vc sin ω t ) = ω C v cos( ω t) Using the relation, π cos( ω t) = sin ω t +, we have Alternating Current FIGUE 7.8 An ac source connected to a capacitor. π i = i sin ω t + (7.6) where the aplitude of the oscillating current is i = ωcv. We can rewrite it as v i = (/ ωc) Coparing it to i = v / for a purely resistive circuit, we find that (/ωc) plays the role of resistance. It is called capacitive reactance and is denoted by X c, X c = /ωc (7.7) so that the aplitude of the current is i v = X (7.8) C 4

10 Physics The diension of capacitive reactance is the sae as that of resistance and its SI unit is oh (Ω). The capacitive reactance liits the aplitude of the current in a purely capacitive circuit in the sae way as the resistance liits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. A coparison of Eq. (7.6) with the FIGUE 7.9 (a) A Phasor diagra for the circuit in Fig (b) Graph of v and i versus ωt. equation of source voltage, Eq. (7.) shows that the current is π/ ahead of voltage. Figure 7.9(a) shows the phasor diagra at an instant t. Here the current phasor I is π/ ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.9(b) shows the variation of voltage and current with tie. We see that the current reaches its axiu value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is p c = i v = i cos(ωt)v sin(ωt) = i v cos(ωt) sin(ωt) iv = sin( ωt) (7.9) So, as in the case of an inductor, the average power iv iv PC = sin( ωt) = sin( ωt) = since <sin (ωt)> = over a coplete cycle. Figure 7. explains it in detail. Thus, we see that in the case of an inductor, the current lags the voltage by π/ and in the case of a capacitor, the current leads the voltage by π/. 4 EXAMPLE 7.4 EXAMPLE 7.3 Exaple 7.3 A lap is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Solution When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lap will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (/ωc) and the current flows in the circuit. Consequently, the lap will shine. educing C will increase reactance and the lap will shine less brightly than before. Exaple 7.4 A 5. µf capacitor is connected to a V, 5 Hz source. Find the capacitive reactance and the current (rs and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Solution The capacitive reactance is X C = = = Ω πνc π (5Hz)(5. 6 F) The rs current is

11 Alternating Current - The current i flows as shown and fro the axiu at, reaches a zero value at. The plate A is charged to positive polarity while negative charge q builds up in B reaching a axiu at until the current becoes zero. The voltage v c = q/c is in phase with q and reaches axiu value at. Current and voltage are both positive. So p = v c i is positive. ENEGY IS ABSOBED FOM THE SOUCE DUING THIS QUATE CYCLE AS THE CAPACITO IS CHAGED. - The current i reverses its direction. The accuulated charge is depleted i.e., the capacitor is discharged during this quarter cycle.the voltage gets reduced but is still positive. The current is negative. Their product, the power is negative. THE ENEGY ABSOBED DUING THE /4 CYCLE - IS ETUNED DUING THIS QUATE. One coplete cycle of voltage/current. Note that the current leads the voltage. -3 As i continues to flow fro A to B, the capacitor is charged to reversed polarity i.e., the plate B acquires positive and A acquires negative charge. Both the current and the voltage are negative. Their product p is positive. The capacitor ABSOBS ENEGY during this /4 cycle. 3-4 The current i reverses its direction at 3 and flows fro B to A. The accuulated charge is depleted and the agnitude of the voltage v c is reduced. v c becoes zero at 4 when the capacitor is fully discharged. The power is negative.enegy ABSOBED DUING -3 IS ETUNED TO THE SOUCE. NET ENEGY ABSOBED IS ZEO. FIGUE 7. Charging and discharging of a capacitor. 43

12 Physics V V I = = =.4 A X Ω C The peak current is EXAMPLE 7.5 EXAMPLE 7.4 i = I = (.4)(.4 A) =.47A This current oscillates between +.47A and.47 A, and is ahead of the voltage by π/. If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. Exaple 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.. FIGUE 7. The switch is closed and after soetie, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. Solution As the iron rod is inserted, the agnetic field inside the coil agnetizes the iron increasing the agnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil 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. FIGUE 7. A series LC circuit connected to an ac source AC VOLTAGE APPLIED TO A SEIES LC CICUIT Figure 7. shows a series LC circuit connected to an ac source ε. As usual, we take the voltage of the source to be v = v sin ωt. If q is the charge on the capacitor and i the current, at tie t, we have, fro Kirchhoff s loop rule: di q L + i + = v (7.) C We want to deterine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this proble by two ethods. First, we use the technique of phasors and in the second ethod, we solve Eq. (7.) analytically to obtain the tie dependence of i.

13 7.6. Phasor-diagra solution Fro the circuit shown in Fig. 7., we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each eleent is the sae at any tie, having the sae aplitude and phase. Let it be i = i sin(ωt+φ) (7.) where φ is the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous sections, we shall construct a phasor diagra for the present case. Let I be the phasor representing the current in the circuit as given by Eq. (7.). Further, let V L, V, V C, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. Fro previous section, we know that V is parallel to I, V C is π/ behind I and V L is π/ ahead of I. V L, V, V C and I are shown in Fig. 7.3(a) with apppropriate phaserelations. The length of these phasors or the aplitude of V, V C and V L are: v = i, v C = i X C, v L = i X L (7.) The voltage Equation (7.) for the circuit can be written as v L + v + v C = v (7.3) The phasor relation whose vertical coponent gives the above equation is V L + V + V C = V (7.4) This relation is represented in Fig. 7.3(b). Since V C and V L are always along the sae line and in opposite directions, they can be cobined into a single phasor (V C + V L ) which has a agnitude v C v L. Since V is represented as the hypotenuse of a right-traingle whose sides are V and (V C + V L ), the pythagorean theore gives: ( ) v = v + v v C L Substituting the values of v, v C, and v L fro Eq. (7.) into the above equation, we have v = ( i ) + ( i X i X ) C L Alternating Current FIGUE 7.3 (a) elation between the phasors V L, V, V C, and I, (b) elation between the phasors V L, V, and (V L + V C ) for the circuit in Fig. 7.. = i + ( XC X L ) v or, i = + ( X X ) C L [7.5(a)] By analogy to the resistance in a circuit, we introduce the ipedance Z in an ac circuit: i where v Z = [7.5(b)] Z = + ( X X ) (7.6) C L 45

14 Physics FIGUE 7.4 Ipedance diagra. Since phasor I is always parallel to phasor V, the phase angle φ is the angle between V and V and can be deterined fro Fig. 7.4: v tan φ = C v v L Using Eq. (7.), we have X C X L tan φ = (7.7) Equations (7.6) and (7.7) are graphically shown in Fig. (7.4). This is called Ipedance diagra which is a right-triangle with Z as its hypotenuse. Equation 7.5(a) gives the aplitude of the current and Eq. (7.7) gives the phase angle. With these, Eq. (7.) is copletely specified. If X C > X L, φ is positive and the circuit is predoinantly capacitive. Consequently, the current in the circuit leads the source voltage. If X C < X L, φ is negative and the circuit is predoinantly inductive. Consequently, the current in the circuit lags the source voltage. Figure 7.5 shows the phasor diagra and variation of v and i with ω t for the case X C > X L. Thus, we have obtained the aplitude and phase of current for an LC series circuit using the technique of phasors. But this ethod of analysing ac circuits suffers fro certain disadvantages. First, the phasor diagra say nothing about the initial condition. One can take any arbitrary value of t (say, t, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution. This is not a general solution. Additionally, we do have a transient solution which exists even for v =. The general solution is the su of the transient solution and the steady-state solution. After a sufficiently long tie, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. FIGUE 7.5 (a) Phasor diagra of V and I. (b) Graphs of v and i versus ω t for a series LC circuit where X C > X L Analytical solution The voltage equation for the circuit is di q L + i + = v C = v sin ωt 46 We know that i = dq/. Therefore, di/ = d q/. Thus, in ters of q, the voltage equation becoes

15 Alternating Current d q dq q L + + = v sin ω t (7.8) C This is like the equation for a forced, daped oscillator, [see Eq. {4.37(b)} in Class XI Physics Textbook]. Let us assue a solution q = q sin (ω t + θ) [7.9(a)] so that d q = qω cos( ω t + θ) [7.9(b)] d q and = q sin( ) ω ω t + θ [7.9(c)] Substituting these values in Eq. (7.8), we get [ cos( ) ( )sin( )] q ω ω t + θ + X X ωt + θ = v sin ω t (7.3) C L where we have used the relation X c = /ωc, X L = ω L. Multiplying and dividing Eq. (7.3) by ( ) Z = + X X, we have c L ( XC X L ) q ω Z cos( t ) sin( t ) Z ω θ Z ω θ = v sin ωt (7.3) Now, let and Z = cosφ ( XC X L ) = sin φ Z X X so that φ Substituting this in Eq. (7.3) and siplifying, we get: C L = tan (7.3) q ω Z cos( ω t + θ φ) = v sin ωt (7.33) Coparing the two sides of this equation, we see that where v = q ω Z = i Z i = q ω [7.33(a)] π π and θ φ = or θ = + φ [7.33(b)] Therefore, the current in the circuit is dq i = = q ω cos( ωt + θ) = i cos(ωt + θ) or i = i sin(ωt + φ) (7.34) v v where i = = Z + ( X X ) X C X L and φ = tan C L [7.34(a)] 47

16 Physics Thus, the analytical solution for the aplitude and phase of the current in the circuit agrees with that obtained by the technique of phasors esonance An interesting characteristic of the series LC circuit is the phenoenon of resonance. The phenoenon of resonance is coon aong systes that have a tendency to oscillate at a particular frequency. This frequency is called the syste s natural frequency. If such a syste is driven by an energy source at a frequency that is near the natural frequency, the aplitude of oscillation is found to be large. A failiar exaple of this phenoenon is a child on a swing. The the swing has a natural frequency for swinging back and forth like a pendulu. If the child pulls on the rope at regular intervals and the frequency of the pulls is alost the sae as the frequency of swinging, the aplitude of the swinging will be large (Chapter 4, Class XI). For an LC circuit driven with voltage of aplitude v and frequency ω, we found that the current aplitude is given by i v v = = Z + ( X X ) C L with X c = /ωc and X L = ω L. So if ω is varied, then at a particular frequency ω, X c = X L, and the ipedance is iniu ( Z ) frequency is called the resonant frequency: X c = X L or = L ω C ω = + =. This or ω = (7.35) LC At resonant frequency, the current aplitude is axiu; i = v /. Figure 7.6 shows the variation of i with ω in a LC series circuit with L =. H, C =. nf for two values of : (i) = Ω and (ii) = Ω. For the source applied v = V. ω for this case is LC =. 6 rad/s. We see that the current aplitude is axiu at the resonant frequency. Since i = v / at resonance, the current aplitude for case (i) is twice to that for case (ii). FIGUE 7.6 Variation of i esonant circuits have a variety of with ω for two cases: (i) = Ω, (ii) = Ω, applications, for exaple, in the tuning L =. H. echanis of a radio or a TV set. The antenna of a radio accepts signals fro any broadcasting stations. The signals picked up in the antenna acts as a source in the 48 tuning circuit of the radio, so the circuit can be driven at any frequencies.

17 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 becoes nearly equal to the frequency of the radio signal received. When this happens, the aplitude of the current with the frequency of the signal of the particular radio station in the circuit is axiu. It is iportant to note that resonance phenoenon is exhibited by a circuit only if both L and 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 aplitude is v /, the total source voltage appearing across. This eans that we cannot have resonance in a L or C circuit. Sharpness of resonance The aplitude of the current in the series LC circuit is given by i = v + ω L ω C and is axiu when ω = ω = / L C. The axiu value is i = v. ax / For values of ω other than ω, the aplitude of the current is less than the axiu value. Suppose we choose a value of ω for which the current aplitude is / ties its axiu value. At this value, the power dissipated by the circuit becoes half. Fro the curve in Fig. (7.6), we see that there are two such values of ω, say, ω and ω, one greater and the other saller than ω and syetrical about ω. We ay write ω = ω + ω ω = ω ω The difference ω ω = ω is often called the bandwih of the circuit. The quantity (ω / ω) is regarded as a easure of the sharpness of resonance. The saller the ω, the sharper or narrower is the resonance. To get an expression for ω, we note that the current aplitude i is ax ( / ) i for ω = ω + ω. Therefore, Alternating Current at ω, i = v + ω L ω C ax i v = = 49

18 Physics or or + ω L = ω C + ω = ω C L ω L = ω C which ay be written as, ( ω + ω) L ( ω + ω) C = ω L ω + = ω ω ωc + ω Using ω = in the second ter on the left hand side, we get L C ω ω L ωl ω ω + ω + = We can approxiate ω + ω as ω ω since ω ω <<. Therefore, ω ω ω L + ω L = ω ω ω or ωl = ω 5 ω = [7.36(a)] L The sharpness of resonance is given by, ω ωl ω = ωl The ratio is also called the quality factor, Q of the circuit. [7.36(b)] ω Q = L [7.36(c)] ω Fro Eqs. [7.36 (b)] and [7.36 (c)], we see that ω =. So, larger the Q

19 value of Q, the saller is the value of ω or the bandwih and sharper is the resonance. Using ω = /L C, Eq. [7.36(c)] can be equivalently expressed as Q = /ω C. We see fro Fig. 7.5, that if the resonance is less sharp, not only is the axiu current less, the circuit is close to resonance for a larger range ω of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit or vice versa. Fro Eq. (7.36), we see that if quality factor is large, i.e., is low or L is large, the circuit is ore selective. Alternating Current Exaple 7.6 A resistor of Ω and a capacitor of 5. µf are connected in series to a V, 5 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rs) across the resistor and the capacitor. Is the algebraic su of these voltages ore than the source voltage? If yes, resolve the paradox. Solution Given 6 = Ω, C = 5. µ F = 5. F V = V, ν = 5Hz (a) In order to calculate the current, we need the ipedance of the circuit. It is Z = + X = + ( π νc) C = ( Ω ) + ( F) 6 = ( Ω ) + ( Ω) = 9.5 Ω Therefore, the current in the circuit is (b) V V I = = =.755 A Z 9.5 Ω Since the current is the sae throughout the circuit, we have V V C = I = (.755 A)( Ω ) = 5 V = I X = (.755 A)(.3 Ω ) = 6.3 V C The algebraic su of the two voltages, V and V C is 3.3 V which is ore than the source voltage of V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the sae phase. Therefore, they cannot be added like ordinary nubers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages ust be obtained using the Pythagorean theore: V+ C = V + VC = V 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. EXAMPLE 7.6 5

20 Physics 7.7 POWE IN AC CICUIT: THE POWE FACTO We have seen that a voltage v = v sinωt applied to a series LC circuit drives a current in the circuit given by i = i sin(ωt + φ) where i v Z = and φ tan X X C L = Therefore, the instantaneous power p supplied by the source is ( sin ω ) [ sin( ω φ) ] p = v i = v t i t + vi = [ cos φ cos( ω t + φ) ] (7.37) The average power over a cycle is given by the average of the two ters in.h.s. of Eq. (7.37). It is only the second ter which is tie-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, vi v i P = cosφ = cos φ = V I cosφ [7.38(a)] This can also be written as, P I Z cosφ = [7.38(b)] So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between the. The quantity cosφ is called the power factor. Let us discuss the following cases: Case (i) esistive circuit: If the circuit contains only pure, it is called resistive. In that case φ =, cos φ =. There is axiu 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 π/. Therefore, cos φ =, and no power is dissipated even though a current is flowing in the circuit. This current is soeties referred to as wattless current. Case (iii) LC series circuit: In an LC series circuit, power dissipated is given by Eq. (7.38) where φ = tan (X c X L )/. So, φ ay be non-zero in a L or C or CL circuit. Even in such cases, power is dissipated only in the resistor. Case (iv) Power dissipated at resonance in LC circuit: At resonance X c X L =, and φ =. Therefore, cosφ = and P = I Z = I. That is, axiu power is dissipated in a circuit (through ) at resonance. 5 EXAMPLE 7.7 Exaple7.7 (a) For circuits used for transporting electric power, a low power factor iplies large power loss in transission. Explain. (b) Power factor can often be iproved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

21 Alternating Current Solution (a) We know that P = I V cosφ where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is sall, we have to increase current accordingly. But this will lead to large power loss (I ) in transission. (b)suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =/Z. We can iprove the power factor (tending to ) by aking Z tend to. Let us understand, with the help of a phasor diagra (Fig. 7.7) how this can be achieved. Let us resolve I into two coponents. I p along FIGUE 7.7 the applied voltage V and I q perpendicular to the applied voltage. I q as you have learnt in Section 7.7, is called the wattless coponent since corresponding to this coponent of current, there is no power loss. I P is known as the power coponent because it is in phase with the voltage and corresponds to power loss in the circuit. It s clear fro this analysis that if we want to iprove power factor, we ust copletely neutralize the lagging wattless current I q by an equal leading wattless current I q. This can be done by connecting a capacitor of appropriate value in parallel so that I q and I q cancel each other and P is effectively I p V. EXAMPLE 7.7 Exaple 7.8 A sinusoidal voltage of peak value 83 V and frequency 5 Hz is applied to a series LC circuit in which = 3 Ω, L = 5.48 H, and C = 796 µf. Find (a) the ipedance 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 ipedance of the circuit, we first calculate X L and X C. X L = πνl = Ω = 8 Ω X C = πν C EXAMPLE

22 Physics = = 4Ω Therefore, Z = + ( X X ) = 3 + (8 4) L C = 5 Ω (b) Phase difference, φ = tan X X C L EXAMPLE = tan = Since φ is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = Now, I i 83 I = = = 4A 5 Therefore, P = (4 A) 3 Ω = 48 W (d) Power factor =cosφ = cos53. =.6 Exaple 7.9 Suppose the frequency of the source in the previous exaple can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the ipedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is ω = = LC =.rad/s 3 6 ω. ν r = = Hz = 35.4Hz π 3.4 (b) The ipedance Z at resonant condition is equal to the resistance: Z = = 3 Ω 54 EXAMPLE 7.9 The rs current at resonance is V V 83 = = = 66.7A Z = 3 The power dissipated at resonance is P I = = = (66.7) kw You can see that in the present case, power dissipated at resonance is ore than the power dissipated in Exaple 7.8.

23 Alternating Current Exaple 7. At an airport, a person is ade to walk through the doorway of a etal detector, for security reasons. If she/he is carrying anything ade of etal, the etal detector eits a sound. On what principle does this detector work? Solution The etal detector works on the principle of resonance in ac circuits. When you walk through a etal detector, you are, in fact, walking through a coil of any turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with etal in your pocket, the ipedance 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 eitted as an alar. EXAMPLE LC OSCILLATIONS We know that a capacitor and an inductor can store electrical and agnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenoenon of electrical oscillations siilar to oscillations in echanical systes (Chapter 4, Class XI). Let a capacitor be charged q (at t = ) and connected to an inductor as shown in Fig The oent the circuit is copleted, the charge on the capacitor starts decreasing, giving rise to current in the circuit. Let q and i be the charge and current in the circuit at tie t. Since di/ is positive, the induced ef in L will have polarity as shown, i.e., v b < v a. According to Kirchhoff s loop rule, q di L = (7.39) C i = (dq/) in the present case (as q decreases, i increases). Therefore, Eq. (7.39) becoes: d q + q = (7.4) LC FIGUE 7.8 At the instant shown, the current is increasing so the polarity of induced ef in the inductor is as shown. d x This equation has the for + ω x = for a siple haronic oscillator. The charge, therefore, oscillates with a natural frequency ω = (7.4) LC and varies sinusoidally with tie as ( ω φ) q = q cos t + (7.4) where q is the axiu value of q and φ is a phase constant. Since q = q at t =, we have cos φ = or φ =. Therefore, in the present case, 55

24 Physics q = q cos( ω t) (7.43) d The current q i = is given by i = i sin( ω t) (7.44) where i = ωq Let us now try to visualise how this oscillation takes place in the circuit. Figure 7.9(a) shows a capacitor with initial charge q connected to an ideal inductor. The electrical energy stored in the charged capacitor is q U E =. Since, there is no current in the circuit, energy in the inductor C is zero. Thus, the total energy of LC circuit is, U = U = E q C FIGUE 7.9 The oscillations in an LC circuit are analogous to the oscillation of a block at the end of a spring. The figure depicts one-half of a cycle. 56 At t =, the switch is closed and the capacitor starts to discharge [Fig. 7.9(b)]. As the current increases, it sets up a agnetic field in the inductor and thereby, soe energy gets stored in the inductor in the for of agnetic energy: U B = (/) Li. As the current reaches its axiu value i, (at t = T/4) as in Fig. 7.9(c), all the energy is stored in the agnetic field: U B = (/) Li. You can easily check that the axiu electrical energy equals the axiu agnetic energy. The capacitor now has no charge and hence no energy. The current now starts charging the capacitor, as in Fig. 7.9(d). This process continues till the capacitor is fully charged (at t = T/) [Fig. 7.9(e)]. But it is charged with a polarity opposite to its initial state in Fig. 7.9(a). The whole process just described will now repeat itself till the syste reverts to its original state. Thus, the energy in the syste oscillates between the capacitor and the inductor.

25 The LC oscillation is siilar to the echanical oscillation of a block attached to a spring. The lower part of each figure in Fig. 7.9 depicts the corresponding stage of a echanical syste (a block attached to a spring). As noted earlier, for a block of a ass oscillating with frequency ω, the equation is d x ω + x = Here, ω = k /, and k is the spring constant. So, x corresponds to q. In case of a echanical syste F = a = (dv/) = (d x/ ). For an electrical syste, ε = L (di/) = L (d q/ ). Coparing these two equations, we see that L is analogous to ass : L is a easure of resistance to change in current. In case of LC circuit, ω = / LC and for ass on a spring, ω = k /. So, /C is analogous to k. The constant k (=F/x) tells us the (external) force required to produce a unit displaceent whereas /C (=V/q) tells us the potential difference required to store a unit charge. Table 7. gives the analogy between echanical and electrical quantities. Alternating Current TABLE 7. ANALOGIES BETWEEN MECHANICAL AND ELECTICAL QUANTITIES Mechanical syste Mass Force constant k Displaceent x Velocity v = dx/ Mechanical energy Slectrical syste Inductance L eciprocal capacitance /C Charge q Current i = dq/ Electroagnetic energy E = kx + v U q = + L i C Note that the above discussion of LC oscillations is not realistic for two reasons: (i) Every inductor has soe resistance. The effect of this resistance is to introduce a daping effect on the charge and current in the circuit and the oscillations finally die away. (ii) Even if the resistance were zero, the total energy of the syste would not reain constant. It is radiated away fro the syste in the for of electroagnetic waves (discussed in the next chapter). In fact, radio and TV transitters depend on this radiation. 57

26 Physics TWO DIFFEENT PHENOMENA, SAME MATHEMATICAL TEATMENT You ay like to copare the treatent of a forced daped oscillator discussed in Section 4. of Class XI physics textbook, with that of an LC circuit when an ac voltage is applied in it. We have already rearked that Eq. [4.37(b)] of Class XI Textbook is exactly siilar to Eq. (7.8) here, although they use different sybols and paraeters. Let us therefore list the equivalence between different quantities in the two situations: Forced oscillations Driven LC circuit d x dx d q dq q + b + kx = F cos ω d t L + + = v sin ω t C Displaceent, x Charge on capacitor, q Tie, t Tie, t Mass, Self inductance, L Daping constant, b esistance, Spring constant, k Inverse capacitance, /C Driving frequency, ω d Driving frequency, ω Natural frequency of oscillations, ω Natural frequency of LC circuit, ω Aplitude of forced oscillations, A Maxiu charge stored, q Aplitude of driving force, F Aplitude of applied voltage, v You ust note that since x corresponds to q, the aplitude A (axiu displaceent) will correspond to the axiu charge stored, q. Equation [4.39 (a)] of Class XI gives the aplitude of oscillations in ters of other paraeters, which we reproduce here for convenience: A = F { ( ω ω ) d + ωdb } / eplace each paraeter in the above equation by the corresponding electrical quantity, and see what happens. Eliinate L, C, ω, and ω, using X L = ωl, X C = /ωc, and ω = /LC. When you use Eqs. (7.33) and (7.34), you will see that there is a perfect atch. You will coe across nuerous such situations in physics where diverse physical phenoena are represented by the sae atheatical equation. If you have dealt with one of the, and you coe across another situation, you ay siply replace the corresponding quantities and interpret the result in the new context. We suggest that you ay try to find ore such parallel situations fro different areas of physics. One ust, of course, be aware of the differences too. 58

27 Alternating Current Exaple 7. Show that in the free oscillations of an LC circuit, the su of energies stored in the capacitor and the inductor is constant in tie. Solution Let q be the initial charge on a capacitor. Let the charged capacitor be connected to an inductor of inductance L. As you have studied in Section 7.8, this LC circuit will sustain an oscillation with frquency ω = π ν = LC At an instant t, charge q on the capacitor and the current i are given by: q (t) = q cos ωt i (t) = q ω sin ωt Energy stored in the capacitor at tie t is q q U E = C V = = cos ( ωt) C C Energy stored in the inductor at tie t is U M = = L i L q ω ωt sin ( ) q = ωt = C ( ω LC ) sin ( ) / Su of energies q U E + U M = cos ωt + sin ωt C q = C This su is constant in tie as q o and C, both are tie-independent. Note that it is equal to the initial energy of the capacitor. Why it is so? Think! EXAMPLE TANSFOMES For any purposes, it is necessary to change (or transfor) an alternating voltage fro one to another of greater or saller value. This is done with a device called transforer using the principle of utual induction. A transforer consists of two sets of coils, insulated fro each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.(a) or on separate libs of the core as in Fig. 7.(b). One of the coils called the priary coil has N p turns. The other coil is called the secondary coil; it has N s turns. Often the priary coil is the input coil and the secondary coil is the output coil of the transforer. 59

28 Physics FIGUE 7. Two arrangeents for winding of priary and secondary coil in a transforer: (a) two coils on top of each other, (b) two coils on separate libs of the core. When an alternating voltage is applied to the priary, the resulting current produces an alternating agnetic flux which links the secondary and induces an ef in it. The value of this ef depends on the nuber of turns in the secondary. We consider an ideal transforer in which the priary has negligible resistance and all the flux in the core links both priary and secondary windings. Let φ be the flux in each turn in the core at tie t due to current in the priary when a voltage v p is applied to it. Then the induced ef or voltage ε s, in the secondary with N s turns is dφ ε s = Ns (7.45) The alternating flux φ also induces an ef, called back ef in the priary. This is dφ ε p = N p (7.46) But ε p = v p. If this were not so, the priary current would be infinite since the priary has zero resistance(as assued). If the secondary is an open circuit or the current taken fro it is sall, then to a good approxiation ε s = v s where v s is the voltage across the secondary. Therefore, Eqs. (7.45) and (7.46) can be written as dφ = [7.45(a)] vs Ns d t v p dφ N d t = p [7.46(a)] 6 Fro Eqs. [7.45 (a)] and [7.46 (a)], we have v v s p N = N s p (7.47)

29 Alternating Current Note that the above relation has been obtained using three assuptions: (i) the priary resistance and current are sall; (ii) the sae flux links both the priary and the secondary as very little flux escapes fro the core, and (iii) the secondary current is sall. If the transforer is assued to be % efficient (no energy losses), the power input is equal to the power output, and since p = i v, i p v p = i s v s (7.48) Although soe energy is always lost, this is a good approxiation, since a well designed transforer ay have an efficiency of ore than 95%. Cobining Eqs. (7.47) and (7.48), we have i p vs Ns = = i v N (7.49) s p p Since i and v both oscillate with the sae frequency as the ac source, Eq. (7.49) also gives the ratio of the aplitudes or rs values of corresponding quantities. Now, we can see how a transforer affects the voltage and current. We have: N s N p Vs = Vp N and Is = I p p N (7.5) s That is, if the secondary coil has a greater nuber of turns than the priary (N s > N p ), the voltage is stepped up(v s > V p ). This type of arrangeent is called a step-up transforer. However, in this arrangeent, there is less current in the secondary than in the priary (N p /N s < and I s < I p ). For exaple, if the priary coil of a transforer has turns and the secondary has turns, N s /N p = and N p /N s =/. Thus, a V input at A will step-up to 44 V output at 5. A. If the secondary coil has less turns than the priary(n s < N p ), we have a step-down transforer. In this case, V s < V p and I s > I p. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transforers (without any energy losses). But in actual transforers, sall energy losses do occur due to the following reasons: (i) Flux Leakage: There is always soe flux leakage; that is, not all of the flux due to priary 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 priary and secondary coils one over the other. (ii) esistance of the windings: The wire used for the windings has soe resistance and so, energy is lost due to heat produced in the wire (I ). In high current, low voltage windings, these are iniised by using thick wire. (iii) Eddy currents: The alternating agnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a lainated core. (iv) Hysteresis: The agnetisation of the core is repeatedly reversed by the alternating agnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a iniu by using a agnetic aterial which has a low hysteresis loss. 6

30 Physics The large scale transission and distribution of electrical energy over long distances is done with the use of transforers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I loss is cut down). It is then transitted over long distances to an area sub-station near the consuers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 4 V reaches our hoes. SUMMAY. An alternating voltage v = v sin ω t applied to a resistor drives a current i = i sinωt in the resistor, i the applied voltage. v =. The current is in phase with. For an alternating current i = i sinωt passing through a resistor, the average power loss P (averaged over a cycle) due to joule heating is (/ )i. To express it in the sae for as the dc power (P = I ), a special value of current is used. It is called root ean square (rs) current and is donoted by I: i I = =.77i Siilarly, the rs voltage is defined by v V = =.77v We have P = IV = I 3. An ac voltage v = v sinωt applied to a pure inductor L, drives a current in the inductor i = i sin (ωt π/), where i = v /X L. X L = ωl is called inductive reactance. The current in the inductor lags the voltage by π/. The average power supplied to an inductor over one coplete cycle is zero. 4. An ac voltage v = v sinωt applied to a capacitor drives a current in the capacitor: i = i sin (ωt + π/). Here, i = v, XC X = ωc is called capacitive reactance. C The current through the capacitor is π/ ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one coplete cycle is zero. 5. For a series LC circuit driven by voltage v = v sinωt, the current is given by i = i sin (ωt + φ) where i = and φ = tan v ( ) + XC X L X C X L 6 ( ) Z XC X L = + is called the ipedance of the circuit.