8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

2.15 ENERGY STORED IN A CAPACITOR A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by 73 Reprint 2025-26 Physics bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Fig 2.30 ). In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges Q¢ and FIGURE 2.30 (a) Work done in a small –Q¢ respectively. At this stage, the potential difference step of building charge on conductor 1 V¢ between conductors 1 to 2 is Q¢/C, where C is thefrom Q¢ to Q¢ + d Q¢. (b) Total work done in charging the capacitor may be capacitance of the system. Next imagine that a small viewed as stored in the energy of charge d Q¢ is transferred from conductor 2 to 1. Work electric field between the plates. done in this step (d W), resulting in charge Q¢ on conductor 1 increasing to Q¢+ d Q¢, is given by Q ′ δW = V ′δQ ′ = δQ ′ (2.68) C Integrating eq. (2.68) Q 2 Q 2 Q ′ 1 Q ′ Q δ Q ’ = = W = ∫ C 2 2C 0 C 0 We can write the final result, in different ways Q 2 1 2 1 W = = CV = QV (2.69) 2C 2 2 Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (2.69)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area A (of each plate) and separation d between the plates]. Energy stored in the capacitor 1 Q 2 ( Aσ)2 d = = × (2.70) 2 C 2 ε0 A The surface charge density s is related to the electric field E between the plates, σ E = (2.71) ε0 From Eqs. (2.70) and (2.71) , we get Energy stored in the capacitor 2 × A d (2.72) U = (1/2 ) ε0 E 74 Reprint 2025-26 Electrostatic Potential and Capacitance Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.72) shows that Energy density of electric field, u =(1/2)e0E 2 (2.73) Though we derived Eq. (2.73) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges. Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system? FIGURE 2.31 Solution (a) The charge on the capacitor is Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C The energy stored by the capacitor is = (1/2) CV 2 = (1/2) QV = (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J (b) In the steady situation, the two capacitors have their positive EXAMPLE plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V¢. The 2.10 75 Reprint 2025-26 Physics charge on each capacitor is then Q¢ = CV¢. By charge conservation, Q¢ = Q/2. This implies V¢ = V/2. The total energy of the system is 1 1 − 6 = 2 × Q ' V ' = QV = 2.25 × 10 J 2 4 Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone? 2.10 There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in EXAMPLE the form of heat and electromagnetic radiation. SUMMARY 1. Electrostatic force is a conservative force. Work done by an external force (equal and opposite to the electrostatic force) in bringing a charge q from a point R to a point P is q(VP–VR), which is the difference in potential energy of charge q between the final and initial points. 2. Potential at a point is the work done per unit charge (by an external agency) in bringing a charge from infinity to that point. Potential at a point is arbitrary to within an additive constant, since it is the potential difference between two points which is physically significant. If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by 1 Q V ( r ) = 4 πεo r 3. The electrostatic potential at a point with position vector r due to a point dipole of dipole moment p placed at the origin is 1 p.rˆ V ( r ) = 2 4 πεo r The result is true also for a dipole (with charges –q and q separated by 2a) for r >> a. 4. For a charge configuration q1, q2, ..., qn with position vectors r1, r2, ... rn, the potential at a point P is given by the superposition principle 1 q1 q 2 qn V = ( + + ... + ) 4 πε0 r1P r2P rnP where r1P is the distance between q1 and P, as and so on. 5. An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential. Reprint 2025-26 Electrostatic Potential and Capacitance 6. Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by 1 q1 q 2 U = 4 πε0 r12 where r12 is distance between q1 and q2. 7. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E. 8. Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by σ E = nˆ where ˆn is the unit vector along the outward normal to the ε0 surface and s is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero. 9. A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V –1. For a parallel plate capacitor (with vacuum between the plates), A C = ε0 d where A is the area of each plate and d the separation between them. 10. If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), C = KC0 where K is the dielectric constant of the insulating substance. 11. For capacitors in the series combination, the total capacitance C is given by 1 1 1 1 = + + + ... C C1 C 2 C 3 In the parallel combination, the total capacitance C is: C = C1 + C2 + C3 + ... where C1, C2, C3... are individual capacitances. 77 Reprint 2025-26 Physics 12. The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is 1 1 2 1 Q 2 U = QV = CV = 2 2 2 C The electric energy density (energy per unit volume) in a region with electric field is (1/2)e0E2. Physical quantity Symbol Dimensions Unit Remark Potential or V [M1 L2 T–3 A–1] V Potential difference is physically significant Capacitance C [M–1 L–2 T–4 A2] F Polarisation P [L–2 AT] C m-2 Dipole moment per unit volume Dielectric constant K [Dimensionless] POINTS TO PONDER 1. Electrostatics deals with forces between charges at rest. But if there is a force on a charge, how can it be at rest? Thus, when we are talking of electrostatic force between charges, it should be understood that each charge is being kept at rest by some unspecified force that opposes the net Coulomb force on the charge. 2. A capacitor is so configured that it confines the electric field lines within a small region of space. Thus, even though field may have considerable strength, the potential difference between the two conductors of a capacitor is small. 3. Electric field is discontinuous across the surface of a spherical charged ˆn outside. Electric potential is, however shell. It is zero inside and σε0 continuous across the surface, equal to q/4pe0R at the surface. 4. The torque p × E on a dipole causes it to oscillate about E. Only if there is a dissipative mechanism, the oscillations are damped and the dipole eventually aligns with E. 5. Potential due to a charge q at its own location is not defined – it is infinite. 6. In the expression qV (r) for potential energy of a charge q, V (r) is the potential due to external charges and not the potential due to q. As seen in point 5, this expression will be ill-defined if V (r) includes potential 78 due to a charge q itself. Reprint 2025-26 Electrostatic Potential and Capacitance 7. A cavity inside a conductor is shielded from outside electrical influences. It is worth noting that electrostatic shielding does not work the other way round; that is, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. EXERCISES

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 =