7.3 REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS 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 combination of these circuit elements. 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 diagram. A phasor* is a vector which rotates about the origin with angular speed w, as shown in Fig. 7.4. 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 7.4(a) shows the FIGURE 7.4 (a) A phasor diagram for the voltage and current phasors and their circuit in Fig 7.1. (b) Graph of v and i versus wt.relationship at time t1 for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1. The projection of voltage and current phasors on vertical axis, i.e., vm sinwt and im sinwt, respectively represent the value of voltage and current at that instant. As they rotate with frequency w, curves in Fig. 7.4(b) are generated. From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero.

1.00 nF for two values of R: (i) R = 100 W and (ii) R = 200 W. For the source applied vm = 1 100 V. w0 for this case is = 1.00×106 LC FIGURE 7.14 Variation of im with w for tworad/s. cases: (i) R = 100 W, (ii) R = 200 W, We see that the current amplitude is L = 1.00 mH. maximum at the resonant frequency. Since im = vm / R at resonance, the current amplitude for case (i) is twice to that for case (ii). Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a 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 maximum. It is important to note that resonance phenomenon 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 amplitude is vm/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. 189 Reprint 2025-26 Physics Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF 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 = 200 Ω, C = 15.0 µF = 15.0 × 10 −6 F V = 220 V, ν = 50Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z = R 2 + X C2 = R 2 + (2πνC )− 2 = (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10 −6 F )−2 = (200 Ω )2 + (212.3 Ω )2 = 291.67 Ω Therefore, the current in the circuit is V 220 V I = = = 0.755 A Z 291.5 Ω (b) Since the current is the same throughout the circuit, we have V R = I R = (0.755 A)(200 Ω=) 151V VC = I X C = (0.755 A)(212.3 Ω=) 160.3 V The algebraic sum of the two voltages, VR and VC 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 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: V R +C = V R2 + VC2 7.6 = 220 V Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor EXAMPLE is equal to the voltage of the source. 7.7 POWER IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinwt applied to a series RLC circuit drives a current in the circuit given by i = im sin(wt + f) where v m − 1  X C − X L  i m = Z and φ = tan  R  190 Therefore, the instantaneous power p supplied by the source is Reprint 2025-26 Alternating Current p = v i = (v m sin ωt ) × [ i m sin(ωt + φ)] v m i m = [ cosφ− cos(2ωt + φ)] (7.29) 2 The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.29). 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, v m i m v m i m P = cos φ = cos φ 2 2 2 = V I cos φ [7.30(a)] This can also be written as, P = I 2 Z cos φ [7.30(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 them. The quantity cosφ is called the power factor. Let us discuss the following cases: CaseCaseCaseCaseCase (i)(i)(i)(i)(i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case φ = 0, cos φ = 1. There is maximum power dissipation. CaseCaseCaseCaseCase (ii)(ii)(ii)(ii)(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/2. Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. CaseCaseCaseCaseCase (iii)(iii)(iii)(iii)(iii) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.30) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. CaseCaseCaseCaseCase (iv)(iv)(iv)(iv)(iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2 R. That is, maximum power is dissipated in a circuit (through R) at resonance. ExampleExampleExampleExampleExample 7.77.77.77.77.7 (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. SolutionSolutionSolutionSolutionSolution (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 small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission. EEEE (b)Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. EXAMPLEXAMPLEXAMPLEXAMPLEXAMPLE We can improve the power factor (tending to 1) by making Z tend to 7.77.77.77.77.7 us understand, with the help of a phasor diagram (Fig. 7.15) R. Let 191 Reprint 2025-26 Physics FIGURE 7.15 how this can be achieved. Let us resolve I into two components. Ip along the applied voltage V and Iq perpendicular to the applied voltage. Iq 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. IP 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, 7.7 we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I¢q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I¢q cancel EXAMPLE each other and P is effectively Ip V. Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 mF. 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 pnL = 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W 1 X C = 2 π νC 1 = 6 = 4 Ω 2 × 3.14 × 50 × 796 × 10− Therefore, Z = R 2 + ( X L − X C )2 = 3 2 + (8 − 4)2 = 5 W X C − X L 7.8 (b) Phase difference, f = tan–1 R −1  4 − 8  EXAMPLE = tan  3 = −53. 1°192 Reprint 2025-26 Alternating Current Since f is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I 2 R im 1  283  = Now, I = 2 2  5 = 40A EXAMPLE Therefore, P = (40 A )2 × 3 Ω= 4800 W (d) Power factor = cos cos –53.1 0.6 7.8 Example 7.9 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 1 ω0 = = LC 25.48 × 10 −3 × 796 × 10 −6 = 222.1rad/s ω0 221.1 νr = = Hz = 35.4Hz 2 π 2 × 3.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3 Ω The rms current at resonance is V V  283  1 = = = = 66.7 A Z R  2  3 The power dissipated at resonance is P = I 2 × R = (66.7)2 × 3 = 13.35 kW EXAMPLE You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. 7.9 Example 7.10 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 EXAMPLE 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. 7.10 193 Reprint 2025-26 Physics 7.8 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. 7.16(a) or on separate limbs of the core as in Fig. 7.16(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. FIGURE 7.16 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 f 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 induced emf or voltage es, in the secondary with Ns turns is dφ εs = − N s (7.31) d t The alternating flux f also induces an emf, called back emf in the primary. This is dφ εp = − N p (7.32) d t But ep = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation es = vs Reprint 2025-26 Alternating Current where vs is the voltage across the secondary. Therefore, Eqs. (7.31) and (7.32) can be written as dφ v s = − N s [7.31(a)] d t dφ v p = − N p [7.32(a)] d t From Eqs. [7.31 (a)] and [7.32 (a)], we have v s N s = (7.33) v p N p 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 = i v, ipvp = isvs (7.34) 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. (7.33) and (7.34), we have i p v s N s = = (7.35) i s v p N p Since i and v both oscillate with the same frequency as the ac source, Eq. (7.35) 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:  N s   N p  Vs = V p and I s = I p (7.36)  N p   N s  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 arrangement, 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 and the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A. If the secondary coil has less turns than the primary (Ns < Np), we have a step-down transformer. In this case, Vs < Vp and Is > Ip. 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: (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 195 Reprint 2025-26 Physics 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 (I 2R). In high current, low voltage windings, these are minimised 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 using a laminated 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 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 I 2R loss is cut down). It is 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. SUMMARY 1. An alternating voltage v = v m sin ω t applied to a resistor R drives a v m current i = im sinwt in the resistor, i m = . The current is in phase with R the applied voltage. 2. For an alternating current i = im sin wt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a special value of current is used. It is called root mean square (rms) current and is donoted by I: i m I = = 0.707 i m 2 Similarly, the rms voltage is defined by vm V = = 0.707 v m 2 We have P = IV = I 2R 3. An ac voltage v = vm sin wt applied to a pure inductor L, drives a current in the inductor i = im sin (wt – p/2), where im = vm/XL. XL = wL is called inductive reactance. The current in the inductor lags the voltage by p/2. The average power supplied to an inductor over one complete cycle is zero. Reprint 2025-26 Alternating Current 4. An ac voltage v = vm sinwt applied to a capacitor drives a current in the capacitor: i = im sin (wt + p/2). Here, v m 1 i m = , X C = is called capacitive reactance. X C ωC The current through the capacitor is p/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. 5. For a series RLC circuit driven by voltage v = vm sin wt, the current is given by i = im sin (wt + f) v m where i m = R 2 + ( X C − X L )2 −1 X C − X L and φ = tan R is called the impedance of the circuit. Z = R 2 + ( X C − X L )2 The average power loss over a complete cycle is given by P = V I cosf The term cosf is called the power factor. 6. In a purely inductive or capacitive circuit, cosf = 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. 7. The phase relationship between current and voltage in an ac circuit 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 w. 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. 8. A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by  N s  Vs =  N p  V p and the currents are related by  N p  Is = I p  N s  If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). 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. 197 Reprint 2025-26 Physics Physical quantity Symbol Dimensions Unit Remarks 2 –3 v m rms voltage V [M L T A –1] V V = , v m is the 2 amplitude of the ac voltage. i m rms current I [ A] A I = , im is the amplitude of 2 the ac current. Reactance: Inductive XL [M L2 T –3 A–2] XL = L Capacitive XC [M L2 T –3 A–2] XC = 1/C Impedance Z [M L2 T –3 A–2] Depends on elements present in the circuit. 1 Resonant wr or w0 [T –1] Hz w0  for a frequency LC series RLC circuit ω0 L 1 Quality factor Q Dimensionless Q = = for a series R ω0 C R RLC circuit. Power factor Dimensionless = cosf, f is the phase difference between voltage applied and current in the circuit. POINTS TO PONDER 1. When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is v m = 2V = 2(240) = 340 V 2. The power rating of an element used in ac circuits refers to its average power rating. 3. The power consumed in an ac circuit is never negative. 4. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac 198 currents, as the dc ampere is derived. An ac current changes direction Reprint 2025-26 Alternating Current with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. 5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is V RC = VR2 + VC2 and not VR + VC since VC is p/2 out of phase of VR. 6. Though in a phasor diagram, voltage and current are represented by vectors, these quantities are not really vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element. 8. In a RLC circuit, resonance phenomenon occur when XL = XC or 1 ω0 = . For resonance to occur, the presence of both L and C LC elements in the circuit is a must. With only one of these (L or C ) elements, there is no possibility of voltage cancellation and hence, no resonance is possible. 9. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. 10. In generators and motors, the roles of input and output are reversed. In a motor, electric energy is the input and mechanical energy is the output. In a generator, mechanical energy is the input and electric energy is the output. Both devices simply transform energy from one form to another. 11. A transformer (step-up) changes a low-voltage into a high-voltage. This does not violate the law of conservation of energy. The current is reduced by the same proportion. 199 Reprint 2025-26 Physics EXERCISES 7.1 A 100 W resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? 7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? 7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. 7.4 A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. 7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. 7.6 A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? 7.7 A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? 7.8 Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W. FIGURE 7.17 (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. Reprint 2025-26 Chapter Eight ELECTROMAGNETIC WAVES 8.1 INTRODUCTION In Chapter 4, we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency. Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities. These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time- varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to Reprint 2025-26 Physics the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and light. Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today. In this chapter, we first discuss the need for displacement current and its consequences. Then we present a descriptive account of electromagnetic waves. James Clerk Maxwell The broad spectrum of electromagnetic waves, (1831 – 1879) Born in stretching from g rays (wavelength ~10–12 m) to long Edinburgh, Scotland, radio waves (wavelength ~106 m) is described. was among the greatest physicists of the nineteenth century. He 8.2 DISPLACEMENT CURRENT derived the thermal We have seen in Chapter 4 that an electrical current velocity distribution of produces a magnetic field around it. Maxwell showed molecules in a gas and that for logical consistency, a changing electric field must was among the first to obtain reliable also produce a magnetic field. This effect is of great estimates of molecular importance because it explains the existence of radio parameters from waves, gamma rays and visible light, as well as all other measurable quantities forms of electromagnetic waves. like viscosity, etc. To see how a changing electric field gives rise to Maxwell’s greatest a magnetic field, let us consider the process of acheivement was the unification of the laws of charging of a capacitor and apply Ampere’s circuital electricity and law given by (Chapter 4) magnetism (discovered by Coulomb, Oersted, “B.dl = m0 i (t) (8.1)(1831–1879) Ampere and Faraday) to find magnetic field at a point outside the capacitor. into a consistent set of Figure 8.1(a) shows a parallel plate capacitor C which equations now called Maxwell’s equations. is a part of circuit through which a time-dependent From these he arrived at current i (t) flows . Let us find the magnetic field at a the most important point such as P, in a region outside the parallel plateMAXWELL conclusion that light is capacitor. For this, we consider a plane circular loop of an electromagnetic radius r whose plane is perpendicular to the direction wave. Interestingly, of the current-carrying wire, and which is centred Maxwell did not agree symmetrically with respect to the wire [Fig. 8.1(a)]. FromCLERK with the idea (strongly suggested by the symmetry, the magnetic field is directed along the Faraday’s laws of circumference of the circular loop and is the same in electrolysis) that magnitude at all points on the loop so that if B is theJAMES electricity was magnitude of the field, the left side of Eq. (8.1) is B (2p r). particulate in nature. So we have B (2pr) = m0i (t) (8 .2) 202 Reprint 2025-26 Electromagnetic Waves Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand side is zero and not m0i, since no current passes through the surface of Fig. 8.1(b) and (c). So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig. 8.1(c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field! If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/e0 (see Eq. 2.41). The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux FE through the surface S ? Using Gauss’s law, it is 1 Q Q ΦE = E A = A = (8.3) ε0 A ε0 Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using Eq. (8.3), we have dΦE d  Q  1 d Q = FIGURE 8.1 A d t d t  ε0 = ε0 d t parallel plate capacitor C, as part of This implies that for consistency, a circuit through which a time  dΦE  dependent current i (t) flows, (a) a loop of ε0  d t  = i (8.4) radius r, to determine This is the missing term in Ampere’s circuital law. If we generalise magnetic field at a this law by adding to the total current carried by conductors through point P on the loop; the surface, another term which is e0 times the rate of change of electric (b) a pot-shaped surface passingflux through the same surface, the total has the same value of current i through the interior for all surfaces. If this is done, there is no contradiction in the value of B between the capacitor obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop shown in (a) as itsis non-zero no matter which surface is used for calculating it. B at a rim; (c) a tiffin- point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with inside, as it should be. The current carried by conductors due to flow of the circular loop as charges is called conduction current. The current, given by Eq. (8.4), is a its rim and a flat circular bottom S new term, and is due to changing electric field (or electric displacement, between the capacitor an old term still used sometimes). It is, therefore, called displacement plates. The arrows show uniform electriccurrent or Maxwell’s displacement current. Figure 8.2 shows the electric field between the and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates. The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing 203 Reprint 2025-26 Physics charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ✒0 (d✂E/dt)). So we have dΦE i = i c + i d = i c + ε0 (8.5) d t In explicit terms, this means that outside the capacitor plates, we have only conduction current ic = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., ic = 0, and there is only displacement current, so that id = i. The generalised (and correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. The generalised law is dΦE B idl = µ0 i c + µ0 ε0 (8.6) Ñ∫ dt and is known as Ampere-Maxwell law. In all respects, the displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero since the electric field E does not change with time. In other FIGURE 8.2 (a) The cases, for example, the charging capacitor above, both conduction electric and magnetic and displacement currents may be present in different regions of fields E and B between space. In most of the cases, they both may be present in the same the capacitor plates, at region of space, as there exist no perfectly conducting or perfectly the point M. (b) A cross insulating medium. Most interestingly, there may be large regions sectional view of Fig. (a). of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P). The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field. So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is * They are still not perfectly symmetrical; there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of 204 electric field. Reprint 2025-26 Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Thus, time- dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5). One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section. MAXWELL’S EQUATIONS IN VACUUM 1. “E.dA = Q/✒0 (Gauss’s Law for electricity) 2. “B.dA = 0 (Gauss’s Law for magnetism) –dΦB 3. “E.dl == (Faraday’s Law) d t d ΦE + µ0 ε0 (Ampere – Maxwell Law) 4. “B.dl = µ0 i c d t

7.4 AC VOLTAGE APPLIED TO AN INDUCTOR Figure 7.5 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 = vm sinw t. Using the Kirchhoff’s loop rule, ε ()t = 0 , and since there ∑ is no resistor in the circuit, d i v − L = 0 (7.10) d t where the second term is the self-induced Faraday FIGURE 7.5 An ac source emf in the inductor; and L is the self-inductance of connected to an inductor. * Though voltage and current in ac circuit are represented by phasors – rotating vectors, they are not vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know. 181 Reprint 2025-26 Physics the inductor. The negative sign follows from Lenz’s law (Chapter 6). Combining Eqs. (7.1) and (7.10), we have d i v v m = = sin ωt (7.11) d t L L Equation (7.11) 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/L. To obtain the current, we integrate di/dt with respect to circuits: time: di v m series d t = sin(ωt )d t ∫ d t L ∫ RLC and get, and v m i = − cos( ω t ) + constant C ω L L, The integration constant has the dimension of current and is time- R, 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. containing, Using  π  circuits − cos(ωt ) = sin ω t − , we have  2  ac of  π  i = i m sin  ωt − 2  diagrams v m (7.12) where i m = is the amplitude of the current. The quantity w L is ω L analogous to the resistance and is called inductive reactance, denoted Phasor by XL: on XL = w L (7.13) The amplitude of the current is, then v m animation i m = (7.14) X L The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (W). The inductive reactance limits the current in a Interactive http://www.animations.physics.unsw.edu.au//jw/AC.html purely 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. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by p/2 or one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t1. The current phasor I is p/2 behind the voltage phasor V. When rotated with frequency w counter- clockwise, they generate the voltage and current given by Eqs. (7.1) and 182 (7.12), respectively and as shown in Fig. 7.6(b). Reprint 2025-26 Alternating Current FIGURE 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5. (b) Graph of v and i versus wt. We see that the current reaches its maximum value later than the  T π/ 2 voltage by one-fourth of a period = . You have seen that an  4 ω  inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is  π  p L = i v = i m sin ω t − ×v m sin (ωt )  2  = −i m vm cos (ωt ) sin (ωt ) i m v m = − sin ( 2ωt ) 2 So, the average power over a complete cycle is i m v m PL = − sin ( 2ωt ) 2 i m v m = − sin ( 2ωt ) = 0, 2 since the average of sin (2wt) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Example 7.2 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 L = 2 πνL = 2 × 3 .14 × 50 × 25 × 10 –3 Ω = 7.85W The rms current in the circuit is EXAMPLE V 220 V I = = = 28 A 7.2 7.85 Ω X L 183 Reprint 2025-26 Physics 7.5 AC VOLTAGE APPLIED TO A CAPACITOR Figure 7.7 shows an ac source e generating ac voltage v = vm sin wt 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 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. 7.7, it limits or regulates the current, but FIGURE 7.7 An ac source does not completely prevent the flow of charge. The connected to a capacitor. 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 is q v = (7.15) C From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal, q v m sin ωt = C d q To find the current, we use the relation i = dt d i = (vm C sin ωt ) = ωC v m cos(ωt ) d t  π  Using the relation, cos(ωt ) = sin  ω t + 2  , we have  π  (7.16) i = i m sin  ωt + 2  where the amplitude of the oscillating current is im = wCvm. We can rewrite it as vm i m = (1/ωC ) Comparing it to im= vm/R for a purely resistive circuit, we find that (1/wC) plays the role of resistance. It is called capacitive reactance and is denoted by Xc, Xc= 1/wC (7.17) so that the amplitude of the current is vm i m = (7.18)184 X C Reprint 2025-26 Alternating Current The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. A comparison of Eq. (7.16) with the FIGURE 7.8 (a) A Phasor diagram for the circuit equation of source voltage, Eq. (7.1) shows that in Fig. 7.7. (b) Graph of v and i versus ωt. the current is π/2 ahead of voltage. Figure 7.8(a) shows the phasor diagram at an instant t1. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.8(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 = i v = im cos(ωt)vm sin(ωt) = imvm cos(ωt) sin(ωt) i m vm = sin(2ωt ) (7.19) 2 So, as in the case of an inductor, the average power i m v m i m v m PC = sin(2ωt ) = sin(2ωt ) = 0 2 2 since <sin (2ωt)> = 0 over a complete cycle. Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2. Example 7.3 A lamp 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 lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. EXAMPLE Reducing C will increase reactance and the lamp will shine less brightly than before. 7.3 Example 7.4 A 15.0 µF 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 1 1 = = 212 Ω X C = EXAMPLE 2 π (50Hz)(15.0 × 10 −6 F ) 2 π νC The rms current is 7.4 185 Reprint 2025-26 Physics V 220 V I = = = 1.04 A X C 212 Ω The peak current is m = 2 I = (1.41)(1.04 A ) = 1.47 A 7.4 i This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by p/2. If the frequency is doubled, the capacitive reactance is halved and EXAMPLE consequently, the current is doubled. Example 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9. FIGURE 7.9 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. 7.5 SolutionmagnetizesAsthetheironiron increasingrod is inserted,the magneticthe magneticfieldfieldinsideinsideit. theHence,coil 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 EXAMPLE across the bulb. Therefore, the glow of the light bulb decreases. 7.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Figure 7.10 shows a series LCR circuit connected to an ac source e. As usual, we take the voltage of the source to be v = vm sin wt. If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule: d i q L + i R + = v (7.20) d t C 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 FIGURE 7.10 A series LCR circuit phasors and in the second method, we solve connected to an ac source. Eq. (7.20) analytically to obtain the time– 186 dependence of i. Reprint 2025-26 Alternating Current 7.6.1 Phasor-diagram solution From the circuit shown in Fig. 7.10, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be i = im sin(wt+f) (7.21) where fis 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 diagram for the present case. Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that VR is parallel to I, VC is p/2 behind I and VL is p/2 ahead of I. VL, VR, VC and I are shown in Fig. 7.11(a) with apppropriate phase- relations. The length of these phasors or the amplitude of VR, VC and VL are: vRm = im R, vCm = im XC, vLm = im XL (7.22) The voltage Equation (7.20) for the circuit can be written as vL + vR + vC = v (7.23) The phasor relation whose vertical component gives the above equation is FIGURE 7.11 (a) Relation between the phasors VL, VR, VC, and I, (b) Relation VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC) This relation is represented in Fig. 7.11(b). Since for the circuit in Fig. 7.10. VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude ½vCm – vLm½. Since V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC + VL), the pythagorean theorem gives: 2 2 2 vm = v Rm + (v Cm − v Lm ) Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above equation, we have v m2 = (i m R )2 + (i m X C − i m X L )2 = i m2  R 2 + ( X C − X L )2  v m or, i m = 2 2 [7.25(a)] R + ( X C − X L ) By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit: v m i m = [7.25(b)] Z where Z = R 2 + ( X C − X L )2 (7.26) 187 Reprint 2025-26 Physics Since phasor I is always parallel to phasor VR, the phase angle f is the angle between VR and V and can be determined from Fig. 7.12: vCm − v Lm tan φ = v Rm Using Eq. (7.22), we have X C − X L tan φ = (7.27) R Equations (7.26) and (7.27) are graphically shown in Fig. (7.12). FIGURE 7.12 Impedance This is called Impedance diagram which is a right-triangle with diagram. Z as its hypotenuse. Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified. If XC > XL, f is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC < XL, f is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage. Figure 7.13 shows the phasor diagram and variation of v and i with wt 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 analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t1, 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 FIGURE 7.13 (a) Phasor diagram of V and I. solution. Additionally, we do have a (b) Graphs of v and i versus w t for a series LCR transient solution which exists even for circuit where XC > XL. v = 0. The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. 7.6.2 Resonance An interesting characteristic of the series RLC 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 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 swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the Reprint 2025-26 Alternating Current 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 w, we found that the current amplitude is given by v m v m i m = = 2 2 Z R + ( X C − X L ) with Xc = 1/wC and XL = wL. So if w is varied, then at a particular frequency Z = R 2 + 0 2 = R . Thisw0, Xc = XL, and the impedance is minimum ( ) frequency is called the resonant frequency: 1 X c = X L or = ω0 L ω0 C 1 or ω0 = (7.28) LC At resonant frequency, the current amplitude is maximum; im = vm/R. Figure 7.16 shows the variation of im with w in a RLC series circuit with L = 1.00 mH, C =