![\"Samacheer](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/08\/Samacheer-Kalvi.png\")
<\/p>\nQuestion 1.
\nA wire which is in North-South direction is dropped freely, will any emf be induced across its ends?
\nAnswer:
\nNo, because both horizontal and vertical component of Earth is magnetic field are not intercepted by the falling wire.<\/p>\n
Question 2.
\nDoes the change in magnetic flux induce emf or current?
\nAnswer:
\nemf is always induced.<\/p>\n
Question 3.
\nDuring the change of magnetic flux when will current be induced?
\nAnswer:
\nCurrent will be induced only when the circuit is complete.<\/p>\n
<\/p>\n
Question 4.
\nA train is moving with uniform speed from north to south. Across which the ends of its axle, will the emf appear?
\nAnswer:
\nYes, emf will be appear as the train is { intercepting vertical component of Earth’s magnetic field.<\/p>\n
Question 5.
\nWhat is meant by induced emf and induced current?
\nAnswer:
\nWhenever there is a change in the magnetic flux linked with a closed circuit an emf is produced. This emf is known as the induced emf and the current that flows in the closed circuit is called induced current.<\/p>\n
Question 6.
\nState Faraday’s laws of electromagnetic induction.
\nAnswer:
\nFaraday’s laws of electromagnetic induction are:
\nFirst law:
\nWhenever the amount of magnetic flux linked with a closed circuits changes, an emf is induced in the circuit. The induced emf lasts so long as the change in magnetic flux continues.<\/p>\n
Second law:
\nThe magnitude of emf induced in a closed circuits is directly proportional to the rate of change of magnetic flux linked with the circuit.<\/p>\n
<\/p>\n
Question 7.
\nDefine self inductance. Give its unit.
\nAnswer:
\nThe coefficient of self induction of a coil or self-inductance of a coil is numerically equal to the opposing em\/induced in the coil when the rate of change of current through the coil is unity. The unit of self inductance is henry (H).<\/p>\n
Question 8.
\nWhat are the applications of eddy currents?
\nAnswer:
\nEddy currents are used in<\/p>\n
Question 9.
\nDefine the unit of seif-inductance.
\nAnswer:
\nOne henry is defined as the self-inductance of a coil in which a change in current of one ampere per second produces an opposing emf of one volt.<\/p>\n
Question 10.
\nDefine mutual inductance in terms of magnetic flux.
\nAnswer:
\nMutual inductance is defined as the flux linkage of the second coil when a current of 1 A flows through the first coil.<\/p>\n
<\/p>\n
Question 11.
\nWhat is the unit of mutual inductance?
\nAnswer:
\nThe unit of mutual inductance is henry. Mutual inductance between two coils is defined as one herny if a current of 1A in one coil produces unit flux linkage in second coil.<\/p>\n
Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. In the step-up transformer the primary voltage is less than the secondary voltage. To maintain the power as constant the primary current in a step-up transformer is more than the secondary current.<\/p>\n Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Question 25. Question 26. Question 27. Question 28. Question 29. Question 30. Question 31. Question 32. where v is the frequency of the AC supply. Thus a pure capacitor offers infinite resistance to DC. But in an AC circuit, the reactance of the coil decreases with increase in frequency.<\/p>\n Question 33. Question 34. Question 35. Question 36. Question 37. = \\(\\frac{\\mathrm{E}_{s} \\mathrm{I}_{s}}{\\mathrm{E}_{p} \\mathrm{I}_{p}}\\)<\/p>\n Question 38. Question 39. Question 40. Question 41. Question 42. Question 43. In radio receivers the resonant. frequency of the circuit is tuned on the frequency of the signal desired to be detected. This is usually done by varying the capacitance of a capacitor.<\/p>\n Question 44. Question 45. Question 46. Question 47. Question 48. Question 49. Question 50. Question 51. Question 52. Question 53. Question 54. Question 55. Question 56. Question 57. Question 58. Question 59. Question 60. Question 61. Question 62. Question 63. Question 64. Question 65. Question 66. Question 67. Question 68. After that, when the electric current reaches a certain steady value, no deflection is observed in the galvanometer. Likewise if the primary circuit is broken, the electric current starts decreasing and there is again a sudden deflection but in the opposite direction.<\/p>\n When the electric current becomes zero, the galvanometer shows no deflection. From the above observations, it is concluded that whenever the electric current in the primary circuit changes, the galvanometer shows a deflection.<\/p>\n Question 69. To reduce these losses, the core of the transformer is made up of thin laminas insulated from one another in while for electric motor the winding is made up of a group of wires insulated from one another. The insulation used does not allow huge eddy currents to flow and hence losses are minimized.<\/p>\n Question 70. At the same time, when they recede away from one another, the magnetic flux linked with the coil decreases. The decrease in magnetic flux again induces an emf in opposite direction so an electric current flows in opposite direction.<\/p>\n Hence there is deflection in the galvanometer when there is a relative motion between the coil and the magnet. In the second experiment, when the primary coil P carries an electric current, a magnetic field is established around it. The magnetic lines of this field pass through itself and the neighbouring secondary coil S.<\/p>\n When the primary circuit is open, no electric current flows in it. Hence the magnetic flux linked with the secondary coil is zero.<\/p>\n However, when the primary circuit is closed, the increasing current builds up a magnetic field around the primary coil. Hence, the magnetic flux linked with the secondary coil increases. This increasing flux linked induces a transient electric current in the secondary coil.<\/p>\n When the electric current in the primary coil reaches a steady value, the magnetic flux linked with the secondary coil does not change and the electric current in the secondary coil will disappear.<\/p>\n Similarly, when the primary circuit is broken, the decreasing primary current induces an electric current in the secondary coil, but in the opposite direction So there is deflection in the galvanometer whenever there is a change in the primary current.<\/p>\n Question 71. Let us consider a pendulum that can be made to oscillate between the poles of a powerful electromagnet (fig (a)).<\/p>\n First the electromagnet is switched off, the pendulum is slightly displaced and released. It begins to oscillate and it executes a large number of oscillations before stopping. The air friction is the only damping force. When the electromagnet is switched on and the disc of the pendulum is made to oscillate, eddy currents are produced in it which will oppose the oscillation.<\/p>\n A heavy damping force of eddy currents will bring the pendulum to rest within a few oscillations (figure (b)). However if some slots are cut in the disc as shown in the (figure (c) eddy currents are reduced. The pendulum now will execute several oscillations before coming to rest. This clearly demonstrates the production of eddy current in the disc of the pendulum.<\/p>\n Question 72. The turn density of these solenoids are n1<\/sub> and n2<\/sub> respectively. As the field lines of \\(\\vec{B}\\)1<\/sub> are passing through the area bounded by solenoid 2, the magnetic flux is linked with each turn of solenoid 2 due to solenoid 1 and is given by<\/p>\n\\(\\Phi_{21}=\\int_{A_{2}} \\overrightarrow{\\mathrm{B}}_{1} \\cdot d \\overrightarrow{\\mathrm{A}}\\)\n = B1<\/sub> A2<\/sub> since \u03b8 = 0\u00b0 = (\u00b50<\/sub> n1<\/sub> i1<\/sub>)A2<\/sub><\/p>\n The flux linkage of solenoid 2 with total turns N2<\/sub> is N2<\/sub> \u03a621<\/sub> = (\u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l) i1<\/sub> ………..(1)<\/p>\n From equation. This is the expression for mutual inductance M21<\/sub> of the solenoid 2 with respect to solenoid 1. In the same way, we can find mutual inductance M12<\/sub> of solenoid 1 with respect to solenoid 2 as given below.<\/p>\n The magnetic field produced by the solenoid 2 when carrying a current i2<\/sub> is This magnetic field B2<\/sub> is uniform inside the solenoid 2 but outside the solenoid 2, it is almost zero. Therefore for solenoid 1, the area A2<\/sub> is the effective area over which the magnetic field B2<\/sub> is present; not area A1<\/sub>, Then the magnetic flux \u03a612<\/sub> linked with each turn of solenoid 1 due to solenoid 2 is<\/p>\n\\(\\Phi_{12}=\\int_{\\mathrm{A}_{2}} \\overrightarrow{\\mathrm{B}}_{2} \\cdot d \\overrightarrow{\\mathrm{A}}=\\mathrm{B}_{2} \\mathrm{~A}_{2}=\\left(\\mu_{0} n_{2} i_{2}\\right) \\mathrm{A}_{2}\\)\n The flux linkage of solenoid 1 with total turns N1<\/sub> is N1<\/sub> \u03a612<\/sub> = (\u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l) i2<\/sub> since N1<\/sub> \u03a612<\/sub> = M12<\/sub> i2<\/sub><\/p>\n M12<\/sub> i2<\/sub> = (\u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l) i2<\/sub><\/p>\n Therefore, we get From equations (3) and (4), we can write Question 73. To receive the signal of a particular station, tuning is done. The tuning is commonly achieved by varying capacitance of a parallel plate variable capacitor, thereby changing the resonant frequency of the circuit. When resonant frequency is nearly equal to the frequency of the signal of the particular station, the amplitude of the current in the circuit is maximum. Thus the signal of that station alone is received.<\/p>\n Question 74. refer figure P. No: 182 – (ii)<\/p>\n Consider a circuit containing a pure resistor of resistance R connected across an alternating voltage source. The instantaneous value of the alternating voltage is given by where \\(\\frac{\\mathrm{V}_{m}}{\\mathrm{R}}\\), the peak value of alternating current in the circuit. From equations (1) and (3), it is clear that the applied voltage and the current are in phase with each other in a resistive circuit. It means that they reach their maxima and minima simultaneously. This is indicated in the phasor diagram. The wave diagram also depicts that current is in phase with the applied voltage.<\/p>\n Question 75. Let Question be the instantaneous charge on the capacitor. The emf across the capacitor at that instant is \\(\\frac{q}{\\mathrm{C}}\\). According to Kirchoff\u2019s loop rule, By the definition of current,<\/p>\n From equation (1) and (2) The peak value of current Im<\/sub> is given by Im<\/sub> = \\(\\frac{\\mathrm{V}_{m}}{1 \/ \\mathrm{C\\omega}}\\).<\/p>\n Let us compare this equation Im<\/sub> = \\(\\frac{V_{m}}{R}\\) with from resistive circuit. The quantity The capacitive reactance (XC<\/sub>) varies inversely as the frequency. For a steady current, f = 0.<\/p>\n XC<\/sub> = \\(\\frac{1}{\\omega C}=\\frac{1}{2 \\pi f C}=\\frac{1}{0}=\\infty\\)<\/p>\n Thus a capacitive circuit offers infinite resistance to the steady current. So that steady current cannot flow through the capacitor.<\/p>\n Question 76. At resonance, the impedence is Therefore, the current in the circuit is Im<\/sub> = \\(\\frac{\\mathrm{V}_{m}}{\\mathrm{R}}\\)<\/p>\n The maximum current at series resonance is limited by the resistance of the circuit. For smaller resistance, larger current with sharper curve is obtained and vice versa.<\/p>\n Question 77. we obtain Question 78. The magnetic energy is If the two energies are plotted with an assumption of \u03a6 = 0, we obtain<\/p>\n From the graph, it can be noted that Question 79. The capacitor now begins to discharge through the inductor that establishes current i in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by When the charges in the capacitor are exhausted, its energy becomes zero i.e., UE<\/sub> = 0. The energy is fully transferred to the magnetic field of the inductor and its energy is maximum. This maximum energy is given Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transferred from the inductor back to the capacitor.<\/p>\n The total energy is the sum of the electrical and magnetic energies (figure (d)). When the current in the circuit reduces to zero, the capacitor becomes fully charged in the opposite direction. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical (figure (e)).<\/p>\n The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then starts to discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies (figure (f)).<\/p>\n As already explained, the processes are repeated in opposite direction (figure (g) and (h)). Finally, the circuit returns to the initial state (figure (a)). Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated.<\/p>\n These are known as LC oscillations. In the ideal LC circuit, there is no loss of energy. Therefore, the oscillations will continue indefinitely. Such oscillations are called undamped oscillations.<\/p>\n Question 80. Question 81. Question 82. Question 83. Self inductance of a solenoid L = \\(\\frac{\\mu_{0} \\mathrm{~N}^{2} \\mathrm{~A}}{l}\\)<\/p>\n Self inductance of air solenoid L = \\(\\frac{\\mu_{0} \\mathrm{~N}^{2} \\mathrm{~A}}{l}\\)<\/p>\n Area of a solenoid = \u03c0r\u00b2 = \\(\\frac{\\pi d^{2}}{4}\\)<\/p>\n Self inductance of the solenoid = \\(\\frac{4 \\times 3.14 \\times 10^{-7} \\times 16 \\times 10^{6} \\times 4 \\times 3.14 \\times 10^{-4}}{60 \\times 10^{-2}}\\)<\/p>\n = 42.067 \u00d7 10-7+6-4+2<\/sup>
\nDefine henry in terms of change of current.
\nAnswer:
\nMutual inductance between two coils is one henry if a current changing at the rate of 1 As-1<\/sup> in the first coil induces an opposing emf of 1 V in another coil.<\/p>\n
\nHow are eddy currents used in the application of brake to electrical trains?
\nAnswer:
\nA metallic drum is coupled to the wheels of a train. The drum rotates along with the wheel when the train is in motion. When the brake is applied, a strong magnetic field is developed f and hence, eddy currents are produced in the drum which oppose the motion of the drum. Hence, the train comes to rest.<\/p>\n<\/p>\n
\nDifferentiate between self-inductance and mutual inductance.
\nAnswer:<\/p>\n\n\n
\n Self inductance<\/td>\n \u00a0Mutual inductance<\/td>\n<\/tr>\n \n Self inductance is numerically equal to the opposing emf induced in the coil when the rate of change of current through the coil is unity.<\/td>\n \u00a0Mutual inductance is numerically equal to the emf induced in one coil, when the rate of change of current through the other coil is unity.<\/td>\n<\/tr>\n \n When a current of 1 ampere flows through coil in one second, an emf of e is induced. The self inductance is given by L = – e.<\/td>\n \u00a0When a current of 1 ampere flows through a primary coil in one second and emf induced in the secondary coil is es<\/sub> then the mutual inductance is given bv M = – es.<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nExplain the operation of motor used in fans.
\nAnswer:
\nEddy currents are produced in a metallic cylinder called rotor, when it is placed in a rotating magnetic field. The eddy current initially tries to decrease the relative motion between the cylinder and the rotating magnetic field. As the magnetic field continues to rotate, the metallic cylinder is set into a rotation. These motors are used in fans.<\/p>\n
\nMention the advantage of three phase alternation.
\nAnswer:<\/p>\n\n
<\/p>\n
\nIn a step-up transformer, the primary current is more than the secondary current. Explain why?
\nAnswer:
\nIn any transformer the power in the primary = power in the secondary.
\ni.e. V1<\/sub>I1<\/sub> = V2<\/sub>I2<\/sub><\/p>\n
\nCan a transformer be used for stepping up a DC? Support your answer with explanation.
\nAnswer:
\nThe transformer cannot be used for stepping up a DC. When the primary of the transformer is connected to DC there will be a momentary current in the secondary coil as the current in the primary builds up to its final value. Afterwards the primary current will be constant and there is no change of magnetic flux. Thus there will be no current in the secondary of the transformer.<\/p>\n
\nWhat are the energy losses in a transformer?
\nAnswer:
\nThe energy losses in a transformer are<\/p>\n\n
\nWhat is known as grid?
\nAnswer:
\nThe main transmission lines from power station to part of a common system called the grid. A large region of the country power from all the power stations in a particular region is fed into the grid. It forms a common pool from which power can be drawn where needed.<\/p>\n<\/p>\n
\nWhat are pylons?
\nAnswer:
\nThe cables used for transmitting power over long distances are suspended by the large porcelain insulators form large steel structures called pylons.<\/p>\n
\nWhat do you mean by mutual Induction?
\nAnswer:
\nThe phenomenon of producing an induced emf in a coil due to the change in current in the other coil is known as mutual induction.<\/p>\n
\nState the factors on which, the coefficient of mutual induction depends.
\nAnswer:
\nThe coefficient of mutual induction between a pair of coils depends on the following factors:
\n(i) Size and shape of the coils, number of turns and permeability of material on which the coils are wound.
\n(ii) Proximity of the coils.<\/p>\n
\nWhat is A.C. generator?
\nAnswer:
\nThe ac generator is a device used for converting mechanical energy into electrical energy.<\/p>\n<\/p>\n
\nState and explain the principle of A.C generator.
\nAnswer:
\nA.C. generator is based on the principle of electromagnetic inducion, according to which an emf is induced in a coil when it is rotated in a uniform magnetic field.<\/p>\n
\nMention the difference between a step up and step down transformer.
\nAnswer:<\/p>\n\n\n
\n Step up transformer<\/td>\n \u00a0Step down transformer<\/td>\n<\/tr>\n \n The current in the primary coil is more than that in the secondary coil.<\/td>\n \u00a0The current in the primary coil is less than that in the secondary coil.<\/td>\n<\/tr>\n \n The number of turns in the primary coil is less than that in the secondary coil.<\/td>\n \u00a0The number of turns in the primary coil is more than that in the secondary coil.<\/td>\n<\/tr>\n \n It is used to convert a low A.C. voltage into a high A.C. voltage.<\/td>\n \u00a0It is used to convert a high A.C. voltage into a low A.C. voltage.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nWhat is a poly phase AC generator?
\nAnswer:
\nIf a number of armature windings are used in the alternator it is known as polyphase alternator. It produces voltage waves equal to the number of windings or phases. Thus a polyphase system consists of a numerous windings which are placed on the same axis but displaced from one another by equal angle which depends on the number of phases.<\/p>\n
\nWhy a DC ammeter cannot read AC?
\nAnswer:
\nFor direct current XL<\/sub> = 0 and XC<\/sub> = \u221e. Alternating current varies in magnitude and direction periodically. But direct current is a unidirectional current and does not vary periodically. Hence a DC ammeter cannot read AC.<\/p>\n<\/p>\n
\nWhat is meant by RMS value of AC?
\nAnswer:
\nThe Root Mean Square value of an alternating current is defined as the square root of the mean squares of all currents over one cycle.<\/p>\n
\nWhat is phasor diagram?
\nAnswer:
\nThe diagram that indicates various phasors and their phase relations is called phasor diagram.<\/p>\n
\nDefine alternating current and give its expression.
\nAnswer:
\nAlternating current is defined as an electric current which is induced when a coil is rotated in a uniform magnetic field. Since the induced current varies in magnitude and direction periodically, it is called alternating current.
\nAlternating current is given by I = I0<\/sub> sin \u03c9t
\nIrms<\/sub> = \\(\\frac{\\mathrm{I}_{0}}{\\sqrt{2}}\\)<\/p>\n<\/p>\n
\nWhat is capacitive reactance?
\nAnswer:
\nCapacitive reactance is given by XC<\/sub> = \\(\\frac{1}{\\mathrm{C}_{\\omega}}\\) = \\(\\frac{1}{2 \\pi v C}\\)<\/p>\n
\nFor DC v = 0; XC<\/sub> = \u221e<\/p>\n
\nWhat is inductive reactance?
\nAnswer:
\nInductive reactance is given by,
\nXL<\/sub> = \u03c9L = 2\u03c0\u03c5L, where \u03c5 is the frequency of the AC supply.
\nFor DC \u03c5 = 0;
\n\u2234 XL<\/sub> = 0
\nThus a pure inductor offers zero resistance to DC. But in an AC circuit the reactance of the coil increases with increase in frequency.<\/p>\n
\nWhat is resonant frequency in LCR circuit?
\nAnswer:
\nThe particular frequency \u03c5o<\/sub> at which the impedance of the circuit becomes minimum and therefore the current becomes maximum is called resonant frequency of the circuit.<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-1.png\")
<\/p>\n<\/p>\n
\nDefine power factor.
\nAnswer:
\nThe average power dissipated in an AC circuit depends not only on the voltage and . current but also on the cosine of the phase angle \u03a6 between them.
\nPav<\/sub> = I2<\/sup>rms<\/sub> Z cos (\u03a6)
\nwhere Irms<\/sub> – RMS value of current
\nZ – impedance; cos \u03a6 – power factor.
\nHence, power factor is defined as the ratio of average power (Pav<\/sub>) dissipated in an AC circuit to the square of rms value of current (I2<\/sup>rms<\/sub>).<\/p>\n
\nWhat happens to the value of current in RLC series circuit, if frequency of the source is increased?
\nAnswer:
\nWhen the frequency of the source is increased, the value of current in RLC frequency, the current will be maximum since the impedance of the circuit is nearly zero.<\/p>\n
\nDefine efficiency of a transformer.
\nAnswer:
\nEfficiency of a transformer is defined as the ratio of output power to the input power, output power
\ninput power
\n\u03b7 = \\(\\frac{\\text { output power }}{\\text { input power }}\\)<\/p>\n<\/p>\n
\nWhat is efficiency of an ideal transformer? Why?
\nAnswer:
\nThe efficiency of an ideal transformer is 1. i.e., 100% where there is no power loss. Because, when there is no power loss, the output power is equal to input power.<\/p>\n
\nWhy the efficiency of transformer is less than one?
\nAnswer:
\nPractically there are numerous factors leading to energy loss in a transformer and hence the efficiency is always less than one.<\/p>\n
\nWhat is meant by alternating emf?
\nAnswer:
\nA rotating coil in a magnetic field, induces an alternating emf and hence an alternating current. Since, the emf induced in the coil varies in magnitude and direction periodically, it is called an alternating emf.<\/p>\n
\nWhat is the significance of alternating emf?
\nAnswer:
\nThe significance of an alternating emf is that it can be changed to lower or higher voltages conveniently and efficiently using a transformer. Also the frequency of the induced emf can be altered by changing the speed of the coil. This enables us to utilise the whole range of electromagnetic spectrum for one purpose or the other.<\/p>\n<\/p>\n
\nHow alternating current can be measured? Give reason.
\nAnswer:
\nSince alternating current varies continously with time, its average value over one complete cycle is zero. Hence its effect is measured by rms value of AC.<\/p>\n
\nWhat is acceptor circuit?
\nAnswer:
\nThe series resonant circuit is often called an ‘acceptor’ circuit. By offering minimum impedance to current at the resonant frequency it is able to select or accept most readily this particular frequency among many frequencies.<\/p>\n
\nWhat is alternating current?
\nAnswer:
\nWhen the electric current produced by a generator changes its direction of flow continuously and periodically in a circuit several times in a second, current is known as an alternating current.<\/p>\n
\nCan a transformer be used for stepping up a DC? Support your answer with explanation.
\nAnswer:
\nThe transformer cannot be used for stepping up a DC. When the primary of the transformer is connected to DC there will be a momentary current in the secondary coil as the current in the primary builds up to its final value. Afterwards the primary current will be constant and there is no change of magnetic flux. Thus there will be no current in the secondary of the transformer.<\/p>\n<\/p>\n
\nWhat is inductive reactance?
\nAnswer:
\nInductive reactance plays the same role as the resistance. Hence, the inductance impedes
\nthe flow of current in the circuit. It is given by XL<\/sub> = L\u03c9 = 2 \u03c0 \u03c5L.<\/p>\n
\nWhat is capacitive reactance?
\nAnswer:
\nThe capacitive reactance is the opposition offered by a capacitor to the flow of current through it. It is given by
\nXC<\/sub> = \\(\\frac{1}{\\mathrm{C}_{\\omega}}=\\frac{1}{2 \\pi v \\mathrm{C}}\\) .<\/p>\n
\nWhat is resonance in LCR circuit?
\nAnswer:
\nThe LCR circuit is said to be in resonance when its inductance reactance is equal to its capacitive reactance i.e., XL<\/sub> = XC<\/sub>. The current in the circuit is maximum. The voltage and current are in phase when LCR circuit is in resonance.<\/p>\n
\nWhat is the application of resonance in radio sets?
\nAnswer:
\nIn radio sets, the antenna circuit contains LCR circuit. Its capacitance is varied until the resonant frequency of the LCR circuit is the same as the carrier frequency of any particular radio station. At this state, the current in the LCR circuit is maximum and the receiver responds to the incoming signal.<\/p>\n<\/p>\n
\nWhat is meant by ‘wattful’ current?
\nAnswer:
\nThe component of current Irms<\/sub> cos (\u03a6) which is in phase with the voltage is called active component. The power consumed by this current = Vrms<\/sub> Irms<\/sub> cos (\u03a6). So that it is also known as \u2018Wattfui\u2019 current.<\/p>\n
\nMention various formulae for power factor.
\nAnswer:
\n(i) Power factor = cos \u03a6 = cosine of the angle of lead or lag.
\n(ii) Power factor = \\(\\frac{\\mathrm{R}}{\\mathrm{Z}}=\\frac{\\text { Resistance }}{\\text { Impedance }}\\)
\n(iii) Power factor = \\(\\frac{\\mathrm{VI} \\cos \\phi}{\\mathrm{VI}}=\\frac{\\text { True power }}{\\text { Apparent power }}\\)<\/p>\n
\nWhat are the examples of power factors?
\nAnswer:
\nSome examples for power factors:
\n(i) Power factor = cos 0\u00b0 = 1 for a pure resistive circuit because the phase angle <|) between voltage and current is zero. (ii) Power factor = cos (\u00b1 \\(\\frac{\\pi}{2}\\)) = 0 for a purely inductive or capacitive circuit because the phase angle c|> between voltage and current is \u00b1 \\(\\frac{\\pi}{2}\\)
\n(iii) Power factor lies between 0 and 1 for a circuit having R, L and C in varying proportions.<\/p>\n
\nState the advantages of AC over DC.
\nAnswer:
\n(i) The generation of AC is cheaper than that of DC.
\n(ii) When AC is supplied at higher voltages, the transmission losses are small compared to DC transmission.
\n(iii) AC can easily be converted into DC with the help of rectifiers.<\/p>\n<\/p>\n
\nWhat are the disadvantages of AC over DC?
\nAnswer:
\n(i) Alternating voltages cannot be used for certain applications Eg: charging of batteries, electroplating, electric traction etc.
\n(ii) At high voltages, it is more dangerous to work with AC than DC.<\/p>\n
\nDraw a diagram to illustrate the construction of a transformer.
\nAnswer:
\nrefer figure P. No: 206 – Q. No: 18<\/p>\n
\nWhat are copper losses? How are they minimised?
\nAnswer:
\nTransformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss that is minimized by using wires of larger diameter.<\/p>\n
\nWhat is meant by alternating voltage?
\nAnswer:
\nAn alternating voltage is the voltage which changes polarity at regular intervals of time and the direction of the resulting alternating current also changes accordingly.<\/p>\n<\/p>\n
\nRepresent squared wave form of alternating current diagramatically.
\nAnswer:<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-2.png\")
<\/p>\n
\nWhat is meant by sinusoidal alternating voltage?
\nAnswer:
\nIf the waveform of alternating voltage is a sine wave, then it is known as sinusoidal alternating voltage which is given by the relation.
\n\u03c5 = Vm<\/sub> sin \u03c9t.<\/p>\n
\nDraw curves explaining the variations em\/in a three phase generator.
\nAnswer:
\nrefer figure P. No: 205 – Q. No: 17 – (ii)<\/p>\n
\nWhat is a transformer?
\nAnswer:
\nTransformer is a stationary device used to transform electrical power from one circuit to another without changing its frequency. The applied alternating voltage is either increased or decreased with corresponding decrease or increase of current in the circuit.<\/p>\n<\/p>\n
\nDraw the phasor and wave diagrams for AC circuit with R.
\nAnswer:
\nrefer figure P. No: 182 – (ii)<\/p>\n
\nDraw the phasor and wave diagrams for A.C. current with L.
\nAnswer:
\nrefer figure P. No: 182 – (iii)<\/p>\n
\nDraw the phasor and wave diagrams for A.C. circuit with C.
\nAnswer:
\nrefer figure P. No: 182 – (iv)<\/p>\n
\nDraw a diagram to indicate the variation of (i) electrical energy and (ii) magnetic energy as a function of time.
\nAnswer:
\nrefer figure P. No: 188<\/p>\n<\/p>\n
\nAs soon as current is switched on in a high voltage wire, the bird sitting on the wire flies away: Give reason.
\nAnswer:
\nWhen current is switched on, current is induced in the body of the bird. Its wings experience mutual repulsion, due to the flow of opposite currents in them. So the bird flies away.<\/p>\n
\nDraw the phasor diagram its represent the variation of current with voltage.
\nAnswer:<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-3.png\")
<\/p>\n
\nProve that whenever the electric current changes in one circuit changes then the galvanometer shows a deflection using an experiment.
\nAnswer:
\nConsider two closed circuits. The circuit consisting of a coil P, a battery B and a key K is called as primary circuit while the circuit with a coil S and a galvanometer G is known as secondary circuit. The coils P and S are kept at rest in close proximity with respect to one another. If the primary circuit is closed, then electric current starts flowing in the primary circuit. Due to this, the galvanometer gives a momentary deflection.<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-4.png\")
<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-5.png\")
<\/p>\n<\/p>\n
\nDiscuss the drawbacks of eddy currrents.
\nAnswer:
\nWhen eddy currents flow in the conductor, a large amount of energy is dissipated in the form of heat. The energy loss due to the flow of eddy current is inevitable but it can be reduced to a greater extent with suitable measures. The design of transformer core and electric motor armature is crucial in order to minimise the eddy current loss.<\/p>\n<\/p>\n
\nExplain Faraday’s experiments that lead to Faraday’s laws of electromagnetic induction.
\nAnswer:
\nIn the first experiment, when a bar magnet is placed close to a coil, some of the magnetic lines of force of the bar magnet pass through the coil. It infers that the magnetic flux is linked with the coil. When the bar magnet and the coil approach each other, the magnetic flux linked with the coil increases. So this increase in magnetic flux induces an emf
\nHence a transient electric current flows in the circuit in one direction.<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-6.png\")
<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-7.png\")
<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-8.png\")
<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-9.png\")
<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-10.png\")
<\/p>\n<\/p>\n
\nExplain a simple demonstration for the production of eddy current.
\nAnswer:<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-11.png\")
<\/p>\n
\nDrive expression for mutual inductance between two long co-axial solenoids.
\nAnswer:
\nConsider two long co-axial solenoids of same length l. The length of these solenoids is large when compared to their radii so that the magnetic field produced inside the solenoids is uniform and the fringing effect at the ends may be ignored. Let A1<\/sub> and A2<\/sub> be the area of cross section of the solenoids with A1<\/sub> being greater A2<\/sub>.<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-12.png\")
<\/p>\n
\nLet i1<\/sub> be the current flowing through solenoid 1, then the magnetic field produced inside it is B1<\/sub> = \u00b50<\/sub> n1<\/sub> i1<\/sub><\/p>\n
\nN2<\/sub> \u03a621<\/sub> = (n2<\/sub> l) (\u00b50<\/sub> n1<\/sub> i1<\/sub>) A2<\/sub> since N2<\/sub> = n2<\/sub> l<\/p>\n
\nN2<\/sub> \u03a621<\/sub> = M21<\/sub> i1<\/sub> ………….(2)
\nComparing the equation (1) and (2)
\nM21<\/sub> = \u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l ……………..(3)<\/p>\n
\nB2<\/sub> = \u00b50<\/sub>n2<\/sub>i2<\/sub><\/p>\n
\nN1<\/sub> \u03a612<\/sub> = (n1<\/sub> l) (\u00b50<\/sub> n2<\/sub> i2<\/sub>) A2<\/sub> since N1<\/sub> = n1<\/sub> l<\/p>\n
\n\u2234 M12<\/sub> = \u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l …………(4)<\/p>\n
\nM12<\/sub> = M21<\/sub> = M ………..(5)
\nIn general, the mutual inductance between two long co-axial solenoids is given by
\nM = \u00b50<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l …….(6)
\nIf a dielectric medium of relative permeability \u00b5r<\/sub> is present inside the solenoids, then
\nM = \u00b5 n1<\/sub> n2<\/sub> A2<\/sub> l
\n(or) M = \u00b50<\/sub> \u00b5r<\/sub> n1<\/sub> n2<\/sub> A2<\/sub> l<\/p>\n<\/p>\n
\nExplain the applications of series RLC resonant circuit.
\nAnswer:
\nRLC circuits have many applications like filter circuits, oscillators, voltage multipliers etc.. An important use of series RLC resonant circuits is in the tuning circuits of radio and TV systems. The signals from many broadcasting stations at different frequencies are available in the air.<\/p>\n
\nProve that in a pure resistive circuit the current is in phase with the applied voltage.
\nAnswer:<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-13.png\")
<\/p>\n
\n\u03c5 = Vm<\/sub> sin \u03c9t ………..(1)
\nAn alternating current i flowing in the circuit due to this voltage, develops a potential drop across R and is given by
\nVR<\/sub> = iR …………(2)
\nKirchhoff \u2019s loop rule states that the algebraic sum of potential differences in a closed circuit is zero. For this resistive circuit,
\n\u03c5 – VR<\/sub> = 0
\nFrom equation (1) and (2),
\nVm<\/sub> sin \u03c9t = iR
\ni = \\(\\frac{\\mathrm{V}_{m}}{\\mathrm{R}}\\) sin \u03c9t
\ni = Im<\/sub> sin \u03c9t ……………….(3)<\/p>\n<\/p>\n
\nDetermine the phase relationship between voltage and current in a pure capacitive circuit.
\nAnswer:
\nLet us consider a circuit containing a capacitor of capacitance C connected across an alternating voltage source. The alternating voltage is given by
\n\u03c5 = Vm<\/sub> sin \u03c9t ………….(1)<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-14.png\")
<\/p>\n
\n\u03c5 = \\(\\frac{q}{\\mathrm{C}}\\) = 0
\nQuestion = CVm<\/sub> sin \u03c9t<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-15.png\")
<\/p>\n
\nIt is clear that current leads the applied voltage by \\(\\frac{\\pi}{2}\\) in a capacitive circuit. This is shown pictorially in figure. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by 90\u00b0.
\nCapacitive reactance XC<\/sub><\/p>\n
\n\\(\\frac{1}{\\mathrm{C} \\omega}\\) plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance (XC<\/sub>). It measured in ohm.
\nXC<\/sub> = \\(\\frac{1}{\\mathrm{C} \\omega}\\) …….(3)
\nIts unit is ohm.<\/p>\n<\/p>\n
\nDiscuss the effects of series resonance.
\nAnswer:
\nWhen series resonance occurs, the impedance of the circuit is minimum and is equal to the resistance of the circuit. As a result of this, the current in the circuit becomes maximum. This is shown in the -resonance curve drawn between current and frequency.<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-16.png\")
<\/p>\n
\nZ = \\(\\sqrt{\\mathrm{R}^{2}+\\left(\\mathrm{X}_{\\mathrm{L}}-\\mathrm{X}_{\\mathrm{C}}\\right)^{2}}\\) = R
\nsince XL<\/sub> = XC<\/sub><\/p>\n
\nIm<\/sub> = \\(\\frac{V_{m}}{\\sqrt{R^{2}+\\left(X_{L}-X_{C}\\right)^{2}}}\\)<\/p>\n
\nDerive an expression for angular frequency of LC oscillations.
\nAnswer:
\nAngular frequency of LC oscillations
\nBy differentiating equation
\nq(t) = Qm<\/sub> cos(\u03c9t + \u03a6) …………(1)
\ntwice, we get
\n\\(\\frac{d^{2} q}{d t}\\) = – Qm<\/sub> \u03c92<\/sup> cos(\u03c9t + \u03a6) ………….(2)
\nSubstituting equations
\n(1) and (2) in equation,
\nL\\(\\frac{d^{2} q}{d t^{2}}\\) + \\(\\frac{1}{\\mathrm{C}}\\) Question = 0<\/p>\n
\nL[-Qm<\/sub> \u03c92<\/sup> cos(\u03c9t + \u03a6)] + \\(\\frac{1}{\\mathrm{C}}\\) Qm<\/sub>
\ncos (\u03c9t + \u03a6) = 0
\nRearranging the terms, the angular frequency of LC oscillations is given by
\n\u03c9 = \\(\\frac{1}{\\sqrt{\\mathrm{LC}}}\\) ………..(4)
\nThis equation is the same as that obtained from qualitative analogy.<\/p>\n
\nDiscuss the oscillations of electrical and magnetic energy.
\nAnswer:
\nThe electrical energy of the LC oscillator is
\nUE<\/sub> = \\(\\frac{q^{2}}{2 \\mathrm{C}}=\\frac{\\mathrm{Q}_{m}^{2}}{2 \\mathrm{C}} \\cos ^{2}(\\omega t+\\phi)\\)<\/p>\n
\nUB<\/sub> = \\(\\frac{1}{2} L i^{2}=\\frac{Q_{m}^{2}}{2 C} \\sin ^{2}(\\omega t+\\phi)\\)<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-17.png\")
<\/p>\n
\n(i) At any instant UE<\/sub> + UB<\/sub> = \\(\\frac{\\mathrm{Q}_{m}^{2}}{2 \\mathrm{C}}\\) = constant
\n(ii) The maximum values of UE<\/sub> and UB<\/sub> are both \\(\\frac{\\mathrm{Q}_{m}^{2}}{2 \\mathrm{C}}\\)
\n(iii) When UE<\/sub> is Maximum, UB<\/sub> is zero and vice versa.<\/p>\n<\/p>\n
\nExplain the generation of LC oscillations.
\nAnswer:
\nLet us assume that the capacitor is fully charged with maximum charge Qm at the initial stage. So that the energy stored in the capacitor is maximum and is given by
\nUE<\/sub> = \\(\\frac{\\mathrm{Q}_{m}^{2}}{2 \\mathrm{C}}\\)
\nAs there is no current in the inductor, the energy stored in it is zero i.e., UB<\/sub> = 0. Therefore, the total energy is wholly electrical (fig (a)).<\/p>\n
\nUB<\/sub> = \\(\\frac{\\mathrm{L} i^{2}}{2}\\). As the charge in the capacitor decreases, the energy stored in it also decreases and is given by UE<\/sub> = \\(\\frac{q^{2}}{2 C}\\). Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies (figure (,b)).<\/p>\n
\nUB<\/sub> = \\(\\frac{\\mathrm{LI}_{m}^{2}}{2}\\)
\nwhere Im<\/sub> that is maximum current flowing in the circuit. The total energy is wholly magnetic (figure (c)).<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-18.png\")
<\/p>\n
\nCompare the analogies between electrical and mechanical quantities.
\nAnswer:<\/p>\n![\"TN](\"https:\/\/samacheer-kalvi.com\/wp-content\/uploads\/2021\/09\/TN-State-Board-12th-Physics-Important-Questions-Chapter-4-Electromagnetic-Induction-and-Alternating-Current-19.png\")
<\/p>\n<\/p>\n
\nThe number of turns of a long solenoid is 1000 turns. When a current of 4A flows through it the magnetic flux linked with each , turn of the solenoid \u00a1s 4 x iO Wb. Calculate the self induct\u00e1nce of the solenoid.
\nAnswer:
\nNumber of turns N = 1000
\nCurrent I = 4A
\nMagnetic flux = 4 \u00d7 10-3<\/sup>
\n\u2234 self inductance of the so\u00ccenoid L = \\(\\frac{\\mathrm{N} \\phi}{\\mathrm{I}}\\)
\nL = \\(\\frac{1000 \\times 4 \\times 10^{-3}}{4}\\) = 1H<\/p>\n
\nA coil of area of cross section 0.5 m2<\/sup> with 10 turns is in a plane perpendicular to a uniform magnetic field of 0.2 Wb\/m2<\/sup>. Calculate the flux through the coil.
\nAnswer:
\nArea of a coil A = 0.5 M2<\/sup>
\nNumber of turns N = 10
\nMagnetic field B= 0.2 Wb \/m2<\/sup>
\nMagnetic flux (\u03a6) = ?
\n\u03a6 = NBA cos \u03b8 = NBA
\n\u2235 \u03b8 = 0\u00b0
\n= 10 \u00d7 0.2 \u00d7 0.5 = 1 Wb<\/p>\n
\nAir core solenoid having a diameter of 4 cm and length 60 cm is wound with 4000 turns. If a current of 5 A flows in the solenoid, calculate the energy stored in the solenoid.
\nAnswer:
\nDiameter of a solenoid d = 4 \u00d7 10-2<\/sup> m
\nLength of the solenoid \/ = 60 \u00d7 10-2<\/sup> m
\nNumber of turns N = 4000;
\nCurrent I = 5 A
\nEnergy stored of a solenoid L = \\(\\frac{1}{2}\\) LI0<\/sub>2<\/sup><\/p>\n
\nL = \\(\\frac{4 \\pi \\times 10^{-7} \\times(4000)^{2} \\times 4 \\pi \\times 10^{-4}}{60 \\times 10^{-2}}\\)<\/p>\n
\n= 42.067 \u00d7 10-3<\/sup> henry
\n= 42.067 \u00d7 10-3<\/sup>
\nEnergy stored E = \\(\\frac{1}{2}\\) LI0<\/sub>