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 =

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?