10-Alternating Current -THEORY

Voltages and currents that vary symmetrically in magnitude and direction with time are very com- mon. 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 through the appliances is called the alternating current (ac current) Alternating voltage is generated by rotating a coil of conducting wire in a strong magnetic field. The magnetic flux linked with the coil changes with time and an alternating emf is thus induced. Instantaneous flux linked with coil is → →   ( A.B)n  ABn cos(t  0 ) where A = area of the coil (in m2) B = magnetic field ( in tesla) n = number of turns  = angular frequency  2  2f T (in rad s-1) f = frequency (in hertz) 0 = initial phase angle. With the change of time Farady’s law cost  0  changes consequently an emf V is induced. According to V   d dt   d [ ABn cos(t   )] dt 0  ABn sin(t  0 ) V  Vm sin(t  0 ) Here Vm = voltage amplitude of sinusoidal volt- age or the peak value of ac voltage where Vm  ABn The time variation of magnetic flux and the corre- V( sponding induced voltage are shown in the following dia- gram. Note: that (i) induced, voltage is maximum where the flux linked with the coil is zero and vice versa. This means that the slope of  □ t graph is the induced voltage (Faraday’s law). (ii) Cyclic variation of  and V are identical, consisting of positive and negative half cycles. AVERAGE VALUES OF ac VOLTAGE AND ac CURRENT: AC voltage or currrents are commonly sinusoidal (sine or cosine function) and their mean values for complete cycle is zero. The average values for half cycles are equally positive and negative (i) Average value for one half cycle (or rectified average value): V  Vm sin t T / 2  V dt 2 T / 2 av T / 2 T  0  (V )  0   dt 0 V sin t dt 0  2 V  0.637V .  0 0 This is also known as the rectified average value of a sinusoidal voltage and is repre- sented asVav . (ii) Root Mean Square Value (Vrms or Irms ) : Since V or I are equally negative and positive, their squares will always be positive and the square root of the average of their square will give the rms values.  V  Vm sin t 1 T (V 2 )  V 2 sin 2 t dt 0 V 2 T V 2  m (1  cos 2t) dt  m Thus 2T 0 2 V   Vm and rms Irms   or RMS value = How do we measure sinusoidally varying voltage or current? Moving coil galvanometer measure steady currents but if we pass sinusoidal current through them, the needle may wiggle at low fequency but its average deflection is zero. Hot wire instruments are commonly used to measure the rms values (also known as virtual values). It must be emphasied here that meters used for ac voltage and current measurement are always calibrated to read rms values, not peak (or maximum) or rectified average values. The usual domestic power supply “220 - volt ac” has an rms voltage of 220 V. The voltage amplitude or peak value is Vm  Vrms  2 (220V )  311V . Illustration: If a domestic appliance draws 2.5 A from a 220-V, 60- Hz power supply, find (a) the average current (b) the average of the square of the current (c) the current amplitude (d) the supply voltage amplitude. Solution: (a) The average of sinusoidal AC values over any whole number of cycles is zero. (b) RMS value of current = Irms = 2.5 A  (I 2 )  (I )2  6.25A2 (c) Irms  m  Current amplitude  = 3.5 A 2Irms  2(2.5 A) (d) Vrms  220V  Vm 2  Supply voltage amplitude Vm  2 (Vrms )  2(220V ) = 311 V. AC VOLTAGE APPLIED TO A RESISTOR : A resistor connected to a source  of ac voltage as shown in the ciruit digram . The symbol for an ac source in a circuit diagram is . For simplicity, we consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by V  Vm sin t …[i] where Vm is the amplitude of the sinusoidal voltage and  is its angular frequency.  R AC voltage applied to a resistor The instantaneous potential drop across the resistor R is Vm sin t  IR or I  Vm sin t R Since R is a constant, we can wire this equation as I  Im sin t …[ii] where I is the instantaneous current and the current amplitude Im is given by I  Vm m R …[iii] Equation [iii] is just Ohm’s law which for resistors work equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by equation [i] and [ii] are plotted as a function of time in figure. Note, in particular that both V and I reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase for a circuit containing pure resistance. Vm Im In a pure resistor, the voltage and current are in phase. The minima with zero and maxima occur at the same respective times. We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values during each cycle. Thus, the sum of the instantaneous current 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 is zero and that there is no dissipation of electrical energy. As you know, joule heating is given by I2 R and depends on I2 (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  I2R  I2 R sin2 t The average value of P over a cycle is …[iv] P  I2R  I2 R sin2 t  …[v (a)] where the bar over a letter(here, P) denotes its average value and <……> denotes taking average of the quantity inside the bracket. Since, 2 and R are constants, P  I2 R  sin2 t > … [v (b)] Using the trigonometric identity, sin2 t   (1 cos 2t) , we have  sin2 t  Thus,  1 (1  cos 2t ) 2  sin2 t  1 2 P  1 I2 R and since  cos 2t  0 . we have …[v (c)] 2 m To express ac power in the same form as dc power (P  I2 R) , as special value of current is used. It is called, root mean square (rms) or effective current and is denoted by Irms . It is defined by Irms    Im  0.707 Im … [vi] In terms of Irms , the average power is P  1 I2 R  I2 R …[vii] 2 m rms I Im Irms -Im The rms current Irms is related to the peak current I by I = Im = 0.707 I . m rms 2 m Similarly, we define the rms voltage or effective voltage by V  Vm  0.707 V …[viii] rms 2 m From equation [iii], we have Vm  Im R or Vm  Im R or Vrms  Irms R …[ix] Equation [ix] gives the relation between ac current and voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of Vm  Vrms  (1.414)(220 V)  311V In fact, the Irms or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation [vii] can also be written as P  V2 / R  I V (since V  I R ) rms rms rms rms rms Illustration: A light bulb is rated at 200 W for a 220 V supply. Find (a) 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 = 220 V. The resistance of the bulb is V2 (220V)2 R  rms   242 P 200W (b) The peak voltage of the source is Vm  2Vrms  311V (c) Since, P  Irms Vrms I  P  200W  0.90A rms V 220V . rms REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS – PHASORS: In the previous section, we saw 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 combination of these circuit elements. In order to show phase relationship between voltage and current in an ac circuit. we use the motion 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 figure. The vertical components of phasors V and I represent the sinusoidally varying quantities V and I. The magnitudes of phasors V and I represent the amplitudes or the peak values Vm and Im of these oscillating quantities. Figure (a) shows the voltage and their relationship at time t1 for the case of an ac source connected to a resistor i.e., corresponding to the circuit show in figure. The projection of voltage and current phasors on vertical axis, i.e., Vm sin t and Im sin t , respectively represent the instaneous value of voltage and current at that instant. As they rotate with frequency  , curves in figure(b) are generated which respresent the sinusoidal variation of voltage and current with time. (a) (b) (a) A phasor diagram for the circuit in figure (b) Graph of V and I versus t. From figure (a) we see that phasors V and I for the case of a resistor are in the same phase. This is so for all times. This means that the phase angle between the voltage and the current is zero. AC VOLTAGE APPLIED TO AN INDUCTOR: An ac source connected to an inductor as shown in the circuit below. Usually, inductors have appre- ciable resistance in their windings, but we shall assume that this ideal inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be Kirchhoff’s loop rule. (t)  0 , and since there is no resistor in the circuit. V  Vm sin t . Using the V  L dI  0 dt …[x]  L An AC source connected to an inductor where the second term is the self-induced emf in the inductor; and L is the self-inductance of the coil. The negative sign follows from Lenz’s law. Combining equation [i] and [x], we have dI  V  Vm sin t dt L L …[xi] Equation [xi] implies that the equation for I(t), the current as a function of time, must be such that its slope dI/dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by Vm . To obtain the current, we integrate dI with respect to time: L dt dI dt  Vm sin(t) dt dt L we get, I   Vm cos(t) L + 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(t)  sin t    , we have  2  i  I sin t      …[xii] m  2    where Im  Vm L is the amplitude of the current. The quantity L is analogous to the resistance and is called inductive reactance, denoted by XL : XL  L  2fL …[xiii] The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm () . The inductive reactance limits the current in a pure inductive circuit in the same way as does the resistance. In a pure resistive circuit. The inductive reactance is directly proportional to the inductance and in the fre- quency of the voltage source. A comparison of equation [i] and [ii] for the source voltage and the current in an inductor shows that  the current lags the voltage by 2 or one-quarter  1      cycle. Figure shows the voltage and the current  phasors in the present case at instant t1 . 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 equation [i] and [xii], respectively and as shown in figure (b) I (a) (b) (a) A phasor diagram for the circuit in figure (b) Graph of V and I versus t. We see that the current reaches its maximum value later than the voltage by one-fourth of a period T  / 2  . You have seen that an inductor has reactance that limits current similar to resistance in a dc  4   circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is P  IV  I sint   V sin(t) L m  2  m    Im Vm cos(t).sin(t) So, the average power over a complete cycle is   Im Vm sin(2t) 2 P   Im Vm sin(2t) L 2   Im Vm 2 sin(2t) = 0. since the average of sin(2t) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Physically, this result means the follows. During the first quarter of each current cycle, the flux through the inductor builds up and sets up a magnetic field and energy is stored in the inductor. In the next quarter of cycle, as the current decrease, the flux decreases and the stored energy is returned to the source. Thus, in each half cycle, the energy which is withdrawn from the source is returned to it without any dissipa- tion of power. Illustration: A pure inductor of 50.0 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  2fL  2  3.14  50  50 103   15.7 The rms current in the circuit is Irms  Vrms  220V  14.01A XL 15.7 AC VOLTAGE APPLIED TO A CAPACITOR: An ac source  connected to a capacitor only, a purely capacitive ac circuit is as shown. A B An ac source connected to a capacitor When a capacitor is connected to a voltage source in a dc circuit, transient 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 figure, 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(t) be the charge on the capacitor at any time t. The instantaneous voltage V(t) across the capacitor is V(t)  q(t) C …[xv] From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal, V sin t  q m C To find the current, we use the relation I  dq dt I  d (V Csin t)  CV cos(t) dt m m Using the relation, cos(t)  sin t    , we have  2    I  I sin t    …[xvi] m  2    where the amplitude of the oscillating current is Im  Vm (1/ C) Comparing it to Im  Vm R for a purely resistive circuit, we find that (1/ C) plays the role of resis- tance. It is called capacitive reactance and is denoted by Xc  1/ C  1/ 2fC so that the amplitude of the current is Xc , …[xvii] I  Vm Xc …[xviii] The dimension of capacitive reactance is the same as that of resistance and its SI unit is Ohm() . The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as does the resistance in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. I (a) (b) (a) Phasor diagram of V and I. (b) Graphs of V and I versus t for a series LCR circuit where XC > XL. A comparison of equation [xvii] with the equation of source voltage equation [i] shows that the current in a capacitor leads the voltage by / 2 . Figure (a) shows the phasor diagram at an instant t1 . Here  the current phasor I is 2 rad ahead of the voltage phasor V as they rotate counter clockwise. Figure (b) shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is PC  IV  I m cos(t).Vm sin(t) = Im Vm cos(t) sin(t)  Im Vm sin(2t) 2 …[xix] So, as in the case of an inductor, the average power P  sin(2t)  sin(2t)  0 since sin(2t)  0 over a complete cycle. As discussed in the case of an inductor, the energy stored by a capacitor in each quarter period is returned to the source in the next quarter period. Thus, we see that in the case of an inductor. the current lags the voltage by 90º and in the case of a capacitor, the current leads the voltage by 90º. Illustration: 30.0 F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reac- tance 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 XC  1 2fC  106 The rms current is Irms  Vrms  2.08A X C The peak current is Im  2Irms  2.96A This current oscillates between 2.96A and – 2.96A and is ahead of the voltage by 90º. If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. 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; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. Figure shows a series LCR circuit connected to an ac source . As usual, we take the voltage of the source to be V  Vm sin t . R A series LCR circuit connected to an ac source If q is the charge on the capacitor and I the current, at time t, we have, from Kirchhoff’s loop rule: L dI  IR  q  V dt C …[xx] We want to determine the instantaneous current I and its phase relationship to the applied alternating voltage V. We shall solve this problem by two methods. First, we use the technique of phasors and in the second method, we solve equation [xx] analytically to obtain the time –dependence of I. PHASOR-DIAGRAM SOLUTION: From the circuit shown in figure we see that the resistor, inductor and capacitor are in series. there- fore, the ac current in each element is the same, having the same amplitude and phase. Let it be I  Im sin[t  ) … [xxi] where  is the phase difference between the voltage across the source and the current in the circuit. On the basis of that we construct a phasor diagram for the present case. Let I be the phasor representing the current in the circuit as given by equation [xxi]. Further, let VL , VR , VC , and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previ- ous section, we know that VR  is in phase with I, VC is 2  rad behind I and VL is 2 rad ahead of I. VR , VC and I are shown in figure (a) with appropriate phase-relations. The length of these phasors or the amplitude of VR , VC VRm  Im R, VCm  Im XC , VLm  Im XL and VL are : … [xxii] The voltage equation (xx) for the circuit can be written as V = VL + VR + VC … [xxiii] The phasor relation whose vertical component gives the above equation is V  VL  VR  VC … [xxiv] (a) (b) (a) Relation between the phasors VL, VR, VC and I. (b) Relation between the phasors VL, VR and (VL + VC) for the circuit in → This relation is represented in figure (b). Since VC → and VL are always along the same line and in → → opposite direction, they can be combined vectorilly into a single phasor (VC  VL ) which has a magnitude → → → → → → | VCm  VLm | . Since, V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC  VL ) , the Pythagorean theorem gives: V2  V2  (V – V )2 m Rm Cm Lm Substituting the values of VRm , VCm , and VLm from equation [xxii] into the above equation, we have V2  (I R)2  (I X – I X )2  I2 [R 2  (X – X )2 ] or Im  m … [xxv (a)] By analogy to the resistance in a circuit, we introduce the impedance Z in ac circuit : I  Vm m Z … [xxv (b)] where Z  … [xxvi] Since phasor I is always parallel to phasor and can be determined from figure: tan   VCm – VLm VRm from the impedance triangle tan   XC – XL R VR , the phase angle  is the angle between VR … [xxvii] and V impedance triangle Equations [xxvi] and [xxvii] are shown in figure. This is called Impedance triangle which is a right triangle with Z as its hypotenuse. Equation [xxv (a)] gives the amplitude of the current and figure gives the phase angle. With these, equation [xxi] is completely specified. If XC  XL ,  is positive and the circuit is capacitive. Consequently, the voltage across the source lags the current. If XC  XL , is negative and the circuit is inductive. Consequently, the voltage across the source leads the current. Figure shows the phasor diagram and variation of V and I with t for the case XC  XL . Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analyzing ac circuits suffers from certain disadvantages. First, the phasor diagram states nothing about the initial condition. One can take any arbitrary value of t and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution. I (a) (b) (a) Phasor diagram of V and I. (b) Graphs of V and I versus t for a series LCR circuit where XC > XL. ANALYTICAL SOLUTION: The voltage equation for the circuit is L dI  RI  q  V dt C  Vm sin t We know that I  dq . Therefore, dt dI  d2q dt dt2 Thus, in terms of q, the voltage equation becomes d2q dq q L dt2 R dt C Vm sin t Let us assume a solution q  qm sin(t  ) … [xxviii] … [xxix (a)] so that dq  q dt m d2q  cos(t  ) 2 … [xxix (b)] and dt2 –qm sin(t  ) … [xxix (c)] Substituting these values in equation [xxviii], we get qm[R cos(t  )  (XC – XL ) sin(t  )]  Vm sin t … [xxx] where we have used the relation XC  C , XL  L . Multiplying and dividing equation [xxx] by Z  , we have = q Z R cos(t  )  (XC – XL ) sin(t  )  V sin t … [xxxi] m  Z Z  m Now, let R  cos Z And (XC – XL )  sin  Z So that   tan–1 XC – XL R Substituting this in equation [xxxi] and simplifying, we get: qmZcos(t   – )  Vm sin t Comparing the two sides of this equation, we see that Vm  qmZ  Im Z … [xxxii] … [xxxiii] where Im  qm … [xxxiii (a)] and  –   –  or   –    … [xxxiii (b)] 2 2 therefore, the current in the circuit is I  dq  q dt mcos(t  )  Im cos(t  ) or I  Im sin(t  ) … [xxxiv] where I  Vm  Vm m Z … [xxxiv (a)] and   tan–1 XC – XL R Thus, the analytical solution for the amplitude and phase of the current in the circuit agrees with the obtained by the technique of phasors. Illustration: A resistor of 200 and a capacitor of 15.0 F 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 voltage more than the source voltage? If yes, resolve the paradox. Solution: Given R  200, C  15.0F  15.0 106 F Vrms  220 V, f  50Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z      291.5 . Therefore, the current in the circuit is I  Vrms  220V  0.755A rms Z 291.5 (b) Since the current is the same throughout the circuit, we have VR  Irms R  (0.755A)(200)  151V VC  Irms XC  (0.755A)(212.3)  160.3V The algebraic sum of the two voltage, VR and VC is 311.3V 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 in 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: VR C  = 220 V Thus, if the phase difference between two voltage is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source. RESONANCE: An interesting characteristic of the series RLC circuit is the phenomenon of resonance. The phe- nomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is 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 child on the swing has a natural frequency for swinging back and forth. 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 Vm and frequency  , we found that the current amplitude is given by I  Vm  Vm m Z with X  1 and X  L . So if  is varied, then at a particular frequency  , X  X , and the C C L 0 C L impendance is minimum ( Z   R) . This frequency is called the resonant frequency : then 1 0C  0 L or 0  … [xxxv] At resonant frequency, the current amplitude is maximum; I  Vm m R The variation of Im with  in a RLC series circuit with L = 1.00 mH, C = 1.00 nF for two values of R: (i) R = 100  and (ii) R = 200  are shown in the figure. For the source applied Vm  100 V . 0 for this  case is 1   1.00 106 rad/s. 1.0 0.5 0.0 0.5  . M rad/s 1.5 2.0 Variation of Im with  for two cases: (i) R = 100, (ii) R = 200    mH. C = 1.00 nF and Vm = 100V for both cases. We see that the current amplitude is maximum at the resonant frequency. Since Im  Vm R at resonance We have seen that a voltage V  Vm sin t applied to a series RLC circuit drives a current in the circuit given by I  Im sin(t  ) where I  Vm and   tan1  XC  XL  m Z  R    Therefore, the instantaneous power P supplied to the source is P  VI  (Vm sin t) [Im sin(t  )]  Vm Im cos  cos(2t  ) 2 …[xxxvii] The average power over a cycle is given by the average of the two terms in R.H.S. of equation [xxxvii]. 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, P  Vm Im cos 2  Vm Im cos  Vrms Irms cos …[xxxviii (a)] This can also be written as, 2 rms Zcos …[xxxviii (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 cos is called the power factor. Let us discuss the following cases: If the circuit contains only pure R, it is called resistive. In that case   0, cos  1 . There is maxi- mum power dissipation. If the circuit contains only an inductor or capacitor, we know that the phase difference between  voltage and current is 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. In an LCR series circuit, power dissipated is given by equation [xxxviii] where   tan1 XC  XL . R  So,  may be non-zero (except 2 ) in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. At resonance X C  XL  0 , and   0 . Therefore, cos  1, and P  I2 z  I2 R . That is, maximum power is dissipated in a circuit (through R) at resonance. Illustration: A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R  3 , L = 25.48 mH, and C = 796 F . 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 XL and XC . XL  2fL  2  3.14  50  25.48103   8 XC   1 2fC 1  4 2  3.14  50  796 106 Therefore, Z   5 (b) Phase difference,    tan1 XC  XL R  tan1  4  8   53.1o  3    since  is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is 2 rms Now, Im rms 1  283   40A 2 5 Therefore,   P  (40A)2  3  4800W (d) Power factor  cos  cos 53.1o  0.6 . Illustration: Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which reasonance occurs? (b) Calculate the imped- ance, the current, and the power dissipated at the resonant condition. Solution: (a) The frequency at which the resonance occurs is 0   LC = 222.1 rad/s f  0  r 2 221.1 2  3.14 Hz  35.4 Hz (b) The impedance Z at resonant condition is equal to the resistance: Z  R  3 The rms current at resonance is  Vrms  Vrms   283  1  66.7A   Z R   The power dissipated at resonance is 2 rms  R  (66.7)2  3  13.35 kW You can see that in the present case, power dissipated at resonance is more than the power dissipated in Previous Illustration. For long-distance power transmission high voltage with small current is desirable so as to minmise heat loss (I2R) in the transmission line. This is accomplished with a device called “TRANSFORMER”. A transformer consists of two of coils or windings insulated from each other. They are wound on the same soft iron core, either one on top the other as in figure (a) or on separate arms of the core as in figure (b). One of the coils called the primary coil has NP turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. Soft iron-core (a) (b) 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 without 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 Vp is applied to it. Then the inducted emf or voltage S . in the secondary with Ns turns is   N d …[xxxxv] s s dt The alternating flux  also induces an emf. called back emf in the primary. This is   N d …[xxxxvi] But p p dt p  Vp . If this were not so, the primary current would be infinite since the primary has zero resistance. If the secondary is an open circuit or the current taken from it is small, then to a good approxima- tion S  VS where Vs written as is the voltage across the secondary. Therefore, equation [xxxxv] and [xxxxvi] can be V  N d …[xxxxv (a)] S S dt V  N d …[xxxxvi (a)] p p dt From equation [xxxxv (a)] and [xxxxv (a)], we have Vs  Ns Vp Np …[xxxxvii] 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 (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since P  IV , I p Vp  Is Vs …[xxxxviii] 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 equation [xxxxvii] and [xxxxviii], we have Ip  Vs  Ns Is Vp Np …[xxxxix] Now, we can see how a transformer affects the voltage and current. We have: V   Ns V I   Np  I s    p  p and s   p  s  …[xxxxx] That is, if the secondary coil has a greater number of turns than the primary (Ns  Np ) , the voltage is stepped up (Vs  Vp ). This type of arrangement is called a step-up transformer. However, in this arrange- ment, there is less current in the secondary than in the primary (Np / Ns  1 and Is  Ip ) . For example, if the primary coil of a transformer has 100 turns ad the secondary has 200 turns, the turn ratio Np  1 . Thus, a 220 V input at 10A will step-up to 440 V output at 5.0 A. Ns  2 Np and Ns 2 If the secondary coil has less number of turns than the primary (Ns  Np ) , we have a step-down transformer. In this case, current is increased. Vs  Vp and Is  Ip . That is, the voltage is stepped down, or reduced, and the The equations obtained above apply to an ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: (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. 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 (I2 R) . In high current, low voltage windings, these are minimized by using thick wire. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a laminated core. (vi) Hysteresis: The magnetization 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 to a minimum by using a magnetic material which has a low hysteresis loss. The large scale transmission and distribution of electrical energy over long 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 I2 R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers the voltage sub-station near the consumers. There the voltage at the is stepped down. It is further stepped down at distributing sub-stations power supplyed at 240 V. Angular frequency variation with power in LCR series circuit. R2 P  Pm   1 2  R2   L      C   Graph between P □  as shown in the figure. 1/ 2 1/ 2    R  2 R2     R  2 R2  1 2L  r  4L2  and 2 2L  r  4L2  Now,     R 1 2 L or (  )  (  )  R or 2  R . r r L L The frequency interval between half maximum power points is known as band width. The ratio resonance frequency and band width is known as quality factor (Q).  Q  r 2  r L . R Q factor is a measure of the sharpness of resonance. Resonance will be sharp if the value of bandwidth () around   r . is small. This is of course possible only when the power-frequency curve fall steeply Illustration: Obtain the resonant freqency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27F and R  7.4 . It is desired to improve the sharpness of the resonance of the circuit by reducing its full width at half maximum by a factor of 2. Suggest a suitable way. Solution: L = 3 henry, C  27 106 F, R  7.4 Resonant angular frequency is given by 0   rad s1  1 9 103 rad s1  1000 rad s1  111 rad s1 9 Q  0 L  111 3  45 R 7.4 To double Q without changing 0 , reeduce R to 3.7 .