4.7 THE SOLENOID We shall discuss a long solenoid. By long solenoid we mean that the solenoid’s length is large compared to its radius. It consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced. So each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enamelled wires are used for winding so that turns are insulated from each other. FIGURE 4.15 (a) The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the exterior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. (b) The magnetic field of a finite solenoid. Figure 4.15 displays the magnetic field lines for a finite solenoid. We show a section of this solenoid in an enlarged manner in Fig. 4.15(a). Figure 4.15(b) shows the entire finite solenoid with its magnetic field. In Fig. 4.15(a), it is clear from the circular loops that the field between two neighbouring turns vanishes. In Fig. 4.15(b), we see that the field at the interior mid-point P is uniform, strong and along the axis of the solenoid. The field at the exterior mid-point Q is weak and moreover is along the axis of the solenoid with no perpendicular or normal component. As the FIGURE 4.16 The magnetic field of a very long solenoid. We consider a 121 rectangular Amperian loop abcd to determine the field. Reprint 2025-26 Physics solenoid is made longer it appears like a long cylindrical metal sheet. Figure 4.16 represents this idealised picture. The field outside the solenoid approaches zero. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis. Consider a rectangular Amperian loop abcd. Along cd the field is zero as argued above. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h. Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law [Eq. 4.13 (b)] BL = µ0Ie, B h = µ0I (n h) B = µ0 n I (4.16) The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field. We shall see in the next chapter that a large field is possible by inserting a soft iron core inside the solenoid. Example 4.8 A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? Solution The number of turns per unit length is, 500 n = = 1000 turns/m 4.8 0.5 The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a. Hence, we can use the long solenoid formula, namely, Eq. (4.20) B = µ0n I = 4π × 10–7 × 103 × 5 EXAMPLE = 6.28 × 10–3 T 4.8 FORCE BETWEEN TWO PARALLEL CURRENTS, THE AMPERE We have learnt that there exists a magnetic field due to a conductor carrying a current which obeys the Biot-Savart law. Further, we have learnt that an external magnetic field will exert a force on a current-carrying conductor. This follows from the Lorentz force formula. Thus, it is logical to expect that two current-carrying conductors placed near each other will exert (magnetic) forces on each other. In the period 1820-25, Ampere studied the nature of this FIGURE 4.17 Two long straight magnetic force and its dependence on the magnitude of parallel conductors carrying steady the current, on the shape and size of the conductors, as currents Ia and Ib and separated by a well as, the distances between the conductors. In this distance d. Ba is the magnetic field section, we shall take the simple example of two parallelset up by conductor ‘a’ at conductor ‘b’. current- carrying conductors, which will perhaps help us 122 to appreciate Ampere’s painstaking work. Reprint 2025-26 Moving Charges and Magnetism Figure 4.17 shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents Ia and Ib, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). Its magnitude is given by Eq. [4.15(a)] or from Ampere’s circuital law, µ0 I a Ba = 2 π d The conductor ‘b’ carrying a current Ib will experience a sideways force due to the field Ba. The direction of this force is towards the conductor ‘a’ (Verify this). We label this force as Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by Eq. (4.4), Fba = I b LB a µ0 I a I b = L (4.17) 2πd It is of course possible to compute the force on ‘a’ due to ‘b’. From considerations similar to above we can find the force Fab, on a segment of length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to Fba, and directed towards ‘b’. Thus, Fba = –Fab (4.18) Note that this is consistent with Newton’s third Law. Thus, at least for parallel conductors and steady currents, we have shown that the Biot-Savart law and the Lorentz force yield results in accordance with Newton’s third Law*. We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other. Thus, Parallel currents attract, and antiparallel currents repel. This rule is the opposite of what we find in electrostatics. Like (same sign) charges repel each other, but like (parallel) currents attract each other. Let fba represent the magnitude of the force Fba per unit length. Then, from Eq. (4.17), µ0 I a I b f ba = 2 π d (4.19) The above expression is used to define the ampere (A), which is one of the seven SI base units. * It turns out that when we have time-dependent currents and/or charges in motion, Newton’s third law may not hold for forces between charges and/or conductors. An essential consequence of the Newton’s third law in mechanics is conservation of momentum of an isolated system. This, however, holds even for the case of time-dependent situations with electromagnetic fields, provided 123 the momentum carried by fields is also taken into account. Reprint 2025-26 Physics The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. This definition of the ampere was adopted in 1946. It is a theoretical definition. In practice, one must eliminate the effect of the earth’s magnetic field and substitute very long wires by multiturn coils of appropriate geometries. An instrument called the current balance is used to measure this mechanical force. The SI unit of charge, namely, the coulomb, can now be defined in terms of the ampere. When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C). Example 4.9 The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Solution F = Il × B F = IlB sinθ The force per unit length is f = F/l = I B sinθ (a) When the current is flowing from east to west, θ = 90° Hence, f = I B = 1 × 3 × 10–5 = 3 × 10–5 N m–1 This is larger than the value 2×10–7 Nm–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere. 4.9 The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors. (b) When the current is flowing from south to north, θ = 0o f = 0 EXAMPLE Hence there is no force on the conductor. 4.9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE 4.9.1 Torque on a rectangular current loop in a uniform magnetic field We now show that a rectangular loop carrying a steady current I and placed in a uniform magnetic field experiences a torque. It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.11).124 Reprint 2025-26 Moving Charges and Magnetism We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in Fig. 4.18(a). The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop. Its magnitude is, F1 = I b B Similarly, it exerts a force F2 on the arm CD and F2 is directed out of the plane of the paper. F2 = I b B = F1 Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces F1 and F2. Figure 4.18(b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anticlockwise. This torque is (in magnitude), a a τ = F1 + F2 2 2 a a = IbB + IbB = I (ab ) B 2 2 FIGURE 4.18 (a) A rectangular = I A B (4.20) current-carrying coil in uniform where A = ab is the area of the rectangle. magnetic field. The magnetic moment We next consider the case when the plane of the loop, m points downwards. The torque τ is is not along the magnetic field, but makes an angle with along the axis and tends to rotate the it. We take the angle between the field and the normal to coil anticlockwise. (b) The couple acting on the coil.the coil to be angle θ (The previous case corresponds to θ = π/2). Figure 4.19 illustrates this general case. The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are F1 and F2. They too are equal and opposite, with magnitude, F1 = F2 = I b B But they are not collinear! This results in a couple as before. The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.19(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is, a a τ = F1 sin θ+ F2 sin θ 2 2 = I ab B sin θ = I A B sin θ (4.21) 125 Reprint 2025-26 Physics As θ à 0, the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. The torques in Eqs. (4.20) and (4.21) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as, m = I A (4.22) where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.18. Then as the angle between m and B is θ , Eqs. (4.20) and (4.21) can be expressed by one expression (4.23) This is analogous to the electrostatic case (Electric dipole of dipole moment pe in an electric field E). τ = p e × E As is clear from Eq. (4.22), the dimensions of the magnetic moment are [A][L2] and its unit is Am2. FIGURE 4.19 (a) The area vector of the loop From Eq. (4.23), we see that the torque τ ABCD makes an arbitrary angle θ with vanishes when m is either parallel or antiparallel the magnetic field. (b) Top view of to the magnetic field B. This indicates a state of the loop. The forces F1 and F2 acting equilibrium as there is no torque on the coil (this on the arms AB and CD also applies to any object with a magnetic moment are indicated. m). When m and B are parallel the equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position. When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. If the loop has N closely wound turns, the expression for torque, Eq. (4.23), still holds, with m = N I A (4.24) Example 4.10 A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis 4.10 of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated EXAMPLE 126 by 90°? The moment of inertia of the coil is 0.1 kg m2. Reprint 2025-26 Moving Charges and Magnetism Solution (a) From Eq. (4.12) µ0 NI B = 2R Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence, 4 × 10 −5 × 10 = − 1 (using π × 3.2 = 10) 2 × 10 = 2 × 10–3 T The direction is given by the right-hand thumb rule. (b) The magnetic moment is given by Eq. (4.24), m = N I A = N I π r2 = 100 × 3.2 × 3.14 × 10–2 = 10 A m2 The direction is once again given by the right-hand thumb rule. (c) τ = m × B [from Eq. (4.23)] = m B sin θ Initially, θ = 0. Thus, initial torque τi = 0. Finally, θ = π/2 (or 90º). Thus, final torque τf = m B = 10 × 2 = 20 N m. (d) From Newton’s second law, I where I is the moment of inertia of the coil. From chain rule, d ω d ω d θ d ω = = ω d t d θ d t d θ Using this, I ωd ω = m B sin θ d θ Integrating from θ = 0 to θ = π/2, EXAMPLE 4.10 Example 4.11 (a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis). EXAMPLE(b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of 4.11 127 Reprint 2025-26 Physics the total field (external field + field produced by the loop) is maximum. (c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape? Solution (a) No, because that would require τ to be in the vertical direction. But τ = I A × B, and since A of the horizontal loop is in the vertical direction, τ would be in the plane of the loop for any B. (b) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this 4.11 orientation,direction as theexternalmagneticfield,fieldbothproducednormal byto thethe loopplaneis inof thethe sameloop, thus giving rise to maximum flux of the total field. (c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater EXAMPLE area than any other shape. 4.9.2 Circular current loop as a magnetic dipole In this section, we shall consider the elementary magnetic element: the current loop. We shall show that the magnetic field (at large distances) due to current in a circular current loop is very similar in behaviour to the electric field of an electric dipole. In Section 4.5, we have evaluated the magnetic field on the axis of a circular loop, of a radius R, carrying a steady current I. The magnitude of this field is [(Eq. (4.11)], µ0 I R 2 B = 2 2 3/2 2 x + R ( ) and its direction is along the axis and given by the right-hand thumb rule (Fig. 4.10). Here, x is the distance along the axis from the centre of the loop. For x >> R, we may drop the R2 term in the denominator. Thus, µ0 IR 2 B = 3 2x Note that the area of the loop A = πR2. Thus, µ0 IA B = 3 2 πx As earlier, we define the magnetic moment m to have a magnitude IA, m = I A. Hence, B ≃µ0 m3 2 πx µ0 2 m = 3 [4.25(a)] 4 π x The expression of Eq. [4.25(a)] is very similar to an expression obtained earlier for the electric field of a dipole. The similarity may be seen if we substitute,128 µ0 → 1/ε0 Reprint 2025-26 Moving Charges and Magnetism m → p e (electrostatic dipole) B → E (electrostatic field) We then obtain, 2 pe E = 3 4 π ε0 x which is precisely the field for an electric dipole at a point on its axis. considered in Chapter 1, Section 1.9 [Eq. (1.20)]. It can be shown that the above analogy can be carried further. We had found in Chapter 1 that the electric field on the perpendicular bisector of the dipole is given by [See Eq.(1.21)], pe E ≃ 4 πε0 x 3 where x is the distance from the dipole. If we replace p à m and µ0 → 1/ ε0 in the above expression, we obtain the result for B for a point in the plane of the loop at a distance x from the centre. For x >>R, m x >> R B ≃µ0 3 ; [4.25(b)] 4π x The results given by Eqs. [4.25(a)] and [4.25(b)] become exact for a point magnetic dipole. The results obtained above can be shown to apply to any planar loop: a planar current loop is equivalent to a magnetic dipole of dipole moment m = I A, which is the analogue of electric dipole moment p. Note, however, a fundamental difference: an electric dipole is built up of two elementary units — the charges (or electric monopoles). In magnetism, a magnetic dipole (or a current loop) is the most elementary element. The equivalent of electric charges, i.e., magnetic monopoles, are not known to exist. We have shown that a current loop (i) produces a magnetic field (see Fig. 4.10) and behaves like a magnetic dipole at large distances, and (ii) is subject to torque like a magnetic needle. This led Ampere to suggest that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment, not accounted by circulating currents.

4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current