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<h1><br>Basic Electrical Engineering MCQs<br>Unit 1<br><br><br>1. If the length of a wire of resistance R is uniformly stretched to n times its original value, its<br>new resistance is,<span id="cke_bm_186S" style="display: none;"> </span><span id="cke_bm_183S" style="display: none;"> </span><span id="cke_bm_180S" style="display: none;"> </span><span id="cke_bm_177S" style="display: none;"> </span><span id="cke_bm_174S" style="display: none;"> </span><br><span id="cke_bm_174E" style="display: none;"> </span><span id="cke_bm_175S" style="display: none;"> </span><span id="cke_bm_177E" style="display: none;"> </span><span id="cke_bm_178S" style="display: none;"> </span><span id="cke_bm_180E" style="display: none;"> </span><span id="cke_bm_181S" style="display: none;"> </span><span id="cke_bm_183E" style="display: none;"> </span><span id="cke_bm_184S" style="display: none;"> </span><span id="cke_bm_186E" style="display: none;"> </span><span id="cke_bm_187S" style="display: none;"> </span>(a) nR (b) R/n (c) n^2R (d) R/n^2<span id="cke_bm_187E" style="display: none;"> </span><span id="cke_bm_188C" style="display: none;"> </span><span id="cke_bm_184E" style="display: none;"> </span><span id="cke_bm_185C" style="display: none;"> </span><span id="cke_bm_181E" style="display: none;"> </span><span id="cke_bm_182C" style="display: none;"> </span><span id="cke_bm_178E" style="display: none;"> </span><span id="cke_bm_179C" style="display: none;"> </span><span id="cke_bm_175E" style="display: none;"> </span></h1><h1> </h1><h1>2. Two wires A and B of the same material and length L and 2L have radius L and 2L<br>respectively. The ratio of their specific resistance will be,<br>(a) 1:1 (b) 1:2 (c) 1:4 (d) 1:8</h1><h1> </h1><h1>3. The current through an electrical conductor is 1 ampere when the temperature of the<br>conductor is 0oc and 0.7 ampere when the temperature is 100oc. the current when the temperature<br>of conductor is 1200oc must be (a) 0.08 amp (b) 0.16 amp (c) 0.32 amp (d) 0.64 amp</h1><h1> </h1><h1>4. A length of the wire having resistance of 1 .<br>is cut into four equal parts and these four<br>parts are bundled together side by side to form the wire. The new resistance in Ω will be,<br>(a) 1/4Ω.<br>(b) 1/16Ω.<br>(c) 4Ω.<br>(d) 16Ω.</h1>
<h1> </h1><h1>5. The hot resistance of the filament of a bulb is higher than the cold resistance because the<br>temperature coefficient of the filament is (a) negative (b) infinite (c) zero (d) positive</h1><h1> </h1><h1>6. The insulation resistance of the insulating material should be,<br>(a) high (b) low (c) zero (d) none of these</h1><h1> </h1><h1>7. The resistance of a wire is R ohms; it is stretched to double its length. The new resistance of<br>the wire in ohms is,<br>(a) R/2Ω.<br>(b) 2RΩ.<br>(c) 4RΩ.<br>(d) R/4Ω.</h1><h1> </h1><h1>8. In which of the following substances, the resistance decreases with the increase in<br>temperature.<br>(a) carbon (b) constantan (c) copper (d) silver<br> </h1><h1>9. The resistance of the wire of uniform diameter d and length l is R. The resistance of another<br>wire of same material but diameter 2d and length 4l will be,<br>(a) 2R (b) R (c) R/2 (d) R/4</h1><h1> </h1><h1>10. The temperature coefficient of the resistance of the wire is 0.00125oC. At 300oK, its<br>resistance is 1 ohm. The resistance of the wire will be 2 ohm at,<br>(a) 1154˚ k (b) 1100˚ k (c) 1400˚ k (d) 1127˚ k</h1><h1> </h1><h1>11. The resistance of the 20 cm long wire is 5 ohm. The wire is stretched to a uniform wire of 40<br>cm length. The resistance (ohm) now will be<br>(a) 5 (b) 10 (c) 20 (d) 200</h1><h1> </h1><h1>12. The current of 4.8 amps is flowing in a conductor. The number of electrons flowing per<br>second through the conductor will be,<br>(a)3×10^19 (b)76.8×10^20 (c)7.68×10^20 (d)3×10^20</h1><h1> </h1><h1>13. A carbon resistor has colored strips as brown, green, orange and silver respectively. The<br>resistance is,<br>(a) 15k±10% (b) 10k±10% (c) 15k±5% (d) 10k±5%</h1><h1> </h1><h1>14. A wire has the resistance of 10. It is stretched by one tenth of its original length. Then its<br>resistance will be,<br>(a) 10.<br>(b) 12.1.<br>(c) 9.<br>
(d) 11.</h1><h1> </h1><h1>15. A 10 m long wire of resistance 20 ohm is connected in series with a battery of EMF 3 volts<br>(with negligible internal resistance) and a resistance of 10 ohms. The potential gradient along the<br>wire in volt per meter is,<br>(a) 0.02 (b) 0.1 (c) 0.2 (d) 1.2</h1><h1> </h1><h1>16. The diameter of an atom is about,<br>(a)10^-10m (b)10^-8m (c)10m (d)10m</h1><h1> </h1><h1>17. 1 cm3 of copper at room temperature has about……….. no of free electrons.<br><br>(a) 200 (b) 20×10^10(c) 8.5×10^22(d) 3×10^5</h1><h1> </h1><h1>18. The electric current is due to flow of …………..charges.<br>(a) positive (b) negative (c) both (d) neutral.<br> </h1><h1>19. The quantity of charge that will be transferred by a current flow of 10 amps over 1 hour<br>period is,<br><br>(a) 10C (c)2.4x10^4C<br>(b) 3.6×10<span style="font-size: 11px;"><span id="cke_bm_150C" style="display: none;"> </span><span id="cke_bm_152C" style="display: none;"> </span><span id="cke_bm_154C" style="display: none;"> </span><span id="cke_bm_156C" style="display: none;"> </span><span style="font-size: 24px;">^</span></span>4C <span style="font-size: 11px;"><span style="font-size: 24px;">(d) 1.6×10</span><span style="font-size: 22px;">^</span><span style="font-size: 24px;">2C</span></span></h1><h1> </h1><h1>20. The drift velocity of the Electrons is of the order of,<br><br>(a) 1 m/s (b) 10^-3m/s (c) 10^6m/s (d) 3×10^8m/s</h1><h1> </h1><h1>21. Insulators have…………..temperature coefficient of resistance.<br>(a) zero<br>(b) positive (c) negative (d) none of these</h1><h1> </h1><h1>22. Eureka has almost………….temperature coefficient of resistance.<br>(a) almost zero (b) positive (c) negative (d) none of these</h1><h1> </h1><h1>23. Constantan wire is used to make standard resistance because it has,<br>(a) low resistivity<br>(b) high resistivity<br>(c) negligibly small RTC (d) high melting point</h1><h1> </h1><h1>24. Two resistors A and B have resistances RA and RB respectively with RA < RB. The<br>resistivities of the materials are A and B, then<br>(a) A> B (b) A= B (c) A< B (d) unable to define</h1><h1> </h1>
<h1>25. in case of liquids, ohms law is………..<br>(a) fully obeyed<br>(b) partially obeyed<br>(c) no relation between current and p.d. (d) None of the above</h1><h1> </h1><h1>26. Three 9 ohm resistances are connected to form<br>triangle. What is the resistance between any<br>Two corner<br>(a)2 (b) 4 (c) 6 (d) 8</h1><h1> 27.The thickness of insulation provided on the conductor depends on<br><br>(a) The magnitude of current flow through it.<br>(b) The magnitude of the voltage of conductor<br>(c) both a and b (d) none of above<br> </h1><h1> </h1><h1>28. Two heaters A and B are connected in parallel across a suitable supply. The heater A<br>produces heat 250 Kcal in 10 minutes while the heater B produces 500 kcal in 5 minutes. If the<br>resistance of heater A is 10 ohm, then the resistance of the heater B is<br>( a) 1.5 ohm (b) 2.5 ohm (c) 3.5 ohm (d) 4.5 ohm</h1><h1> </h1><h1>29. The electrical energy required to raised the temperature of a given amount of water is 700<br>Kwh . If heat losses are 30 % the total heat energy required is<br>( a) 900 kwh (b) 1000 kwh (c) 1100kwh (d) 1200 kwh</h1><h1> </h1><h1>30. Five 100 W, 110 V heaters are connected in series across a 400v supply. The total power<br>consumed by heater under this condition will be<br>( a) 650 W (b) 540 W (c) 430 W (d) 320W</h1><h1> </h1><h1>31. If a 230 v heater is used on 115 v supply , heat produce will be<br>(a) four times the normal<br>(b) two times the normal<br>(c) one half of the normal<br>(d) one fourth of the normal</h1><h1> </h1><h1>32. For a given line voltage, four heating coil will produce maximum heat when connected<br>(a) all in series<br>(b) all in parallel<br>(c ) with two parallel pairs in series<br>(d) one pair in parallel with other two in series<br> </h1><h1>Basic electrical Engineering<br>Unit 1 MCQ Answers<br><br>QUESTION ANSWERS<br>1 -----------------c<br>2 -----------------b<br>3 -----------------b<br>4 -----------------b<br>5 -----------------d<br>6 -----------------a<br>
7 -----------------c<br>8 -----------------a<br>9 -----------------b<br>10-----------------b<br>11-----------------c<br>12-----------------a<br>13-----------------a<br>14-----------------b<br>15-----------------c<br>16----------------a<br>17----------------c<br>18----------------c<br>19----------------b<br>20----------------b<br>21----------------c<br>22----------------a<br>23----------------c<br>24----------------d<br>25----------------a<br>26----------------C<br>27----------------b<br>28----------------b<br>29----------------b.<br>30----------------d.<br>31----------------d.<br>32----------------b.<br><br><br>BASIC ELECTRICAL ENGINEERING<br>Unit 2 objective questions<br><br>1.<br>When a current carrying conductor is brought into magnetic field, the force that moves the<br>Conductor depends upon<br>a.Direction of current<br>b.Length of conductor<br>c.Value of current<br>d.None of the these</h1><h1> </h1><h1>2.Two current carrying conductors laying parallel to each other are exerting a force of attraction<br>on each other. The current are<br>a.Very high<br>b.In opposite direction<br>c.Low<br>d.In same direction</h1><h1> </h1><h1>3.Two conductors are laying parallel and close to each other. They are carrying current in opposite<br>direction . The force between them is<br>a.Repulsive<br>b.Attractive<br>c.Zero<br>d.None of these</h1><h1> </h1><h1>4.Permeance is analogous to<br>
a.Resistance<br>b.Reluctance<br>c.Conductance<br>d.None of these</h1><h1> </h1><h1>5.When a coil consisting of single turn rotates at a uniform speed in magnetic field . the induced<br>emf is<br>a.Steady<br>b.Alternating<br>c.Changing<br>d.Reversing</h1><h1> </h1><h1>6.The emf induced in a conductor of length 1 meter moving at right angles to a uniform magnetic<br>field of flux density 1.5 Wb/m square with velocity 50 m/s is<br>a.0v<br>b.1.5v<br>c.75v<br>d.100v</h1><h1> </h1><h1>7.Which of the following statement is incorrect?<br>a.Whenever the flux linking with the coil or circuit change , an emf is induced.<br>b.The direction of dynamically induced emf can be determined by Fleming’s right hand rule.<br>c.The coefficient of self inductance is proportional to the square of number of turns on it.<br>d.Coefficient of coupling for tightly coupled coil is zero.</h1><h1> </h1><h1>8.The coefficient of self inductance of a coil is defined as<br>a.f/NI<br>b.NI/f<br>c.N/I<br>d.FI/N</h1><h1> </h1><h1>9.The induced emf in a coil of 0.08 mH carrying 2 A current reversed in 0.4 s is<br>a.0.4v<br>b.0.008 v<br>c.0.16 v<br>d.0.064 v</h1><h1> </h1><h1>10. An iron cored coil of 10 turns has reluctance of 100 AT/wb. The inductance of the coil is<br>a.0.1H<br>b.1H<br>c.10H<br>d.2H</h1><h1> </h1><h1>11.A coil of radius R has 400 turns and a self inductance of 32 mH. The inductance of similar coil of<br>300 turns will be<br>a.4.8A<br>b.0.1A<br>c.1.5A<br>d.2.2A</h1>
<h1> </h1><h1>12.The self inductance of two coils are 3 H and 27 H. They are so wound that 50 % of the flux of one<br>coil links the other. The mutual inductance is<br>a.4.5H<br>b.24H<br>c.10H<br>d.30H</h1><h1> </h1><h1>3.The two coils have self inductance of 0.09 H and 0.01 H and a mutual inductance of 0.015 H. The<br>coefficient of coupling between coil is<br>a.0.5<br>b.1.0<br>c.0.05<br>d.0.75</h1><h1> </h1><h1>14.Two coils X and Y are placed in a circuit such that a current changes by 2 A in coil X and the<br>magnetic flux change of 0.4 Wb occurs in coil Y. The mutual inductance of coil is<br>a.0.8H<br>b.1.6H<br>c.0.2H<br>d.1H</h1><h1> </h1><h1>15.Mutual inductance between two coils is 4 H . If current in one coil changes at a rate of 2 A/s,<br>then Emf induced in the other coil is<br>a.8v<br>b.1v<br>c.2v<br>d.0.5v</h1><h1> </h1><h1>16. Relative permeability of vacuum is<br>a.4. x 10 -7 H/m<br>b.1H/m<br>c.1<br>d.1/4.</h1><h1> </h1><h1>17. Unit of magnetic flux is Weber<br>a.Weber<br>b.Ampere turn<br>c.Tesla<br>d.Coulomb</h1><h1> </h1><h1>18.Point out the wrong statement-<br>The magnetizing force at the centre of a circular coil varies<br>a.Directly as the number of its turns<br>b.Directly as the current<br>c. Directly as radius<br>d. D inversely as its radius.</h1><h1> </h1><h1>19. When a magnet is heated<br>a. Its gain magnetism<br>b. Its loss magnetism<br>c. Its neither gain nor loss the magnetism<br>
d. None of the above</h1><h1> </h1><h1>20. The magnetic material used in permanent magnet is<br>a. Iron<br>b. Soft steel<br>c. Nickel<br>d. Hard steel</h1><h1> </h1><h1>21. The magnetic material used in temporary magnet is<br>a. Hard steel<br>b. Cobalt steel<br>c. Soft iron<br>d. Tungsten steel</h1><h1> </h1><h1>22. Magnetic flux density is a<br>a. Vector quantity<br>b. Scalar quantity<br>c. Phasor<br>d. None of the above</h1><h1> </h1><h1>23. The relative permeability of a ferromagnetic material is 1000 its absolute permeability will be<br>a. 106 H/m<br>b.4πx 10^-3 H/m<br>c.4πx 10^-11H/m<br>d.None of the above</h1><h1> </h1><h1>24. The main advantages of temporary magnet is that we can<br>a.change the magnetic flux<br>b.use any magnetic material<br>c.decrease the hysteresis loss<br>d.None of the above</h1><h1> </h1><h1>25. One weber is equal to<br>a.10^6 b.10^12 c.10^7<br>d.10^8</h1><h1> </h1><h1>26. Magnetic field intensity is a<br>a.Scalar quantity<br>b.Vector quantity<br>c.Phasor<br>d.None of the above</h1><h1> </h1><h1>27. When the relative permeability of material is slightly less than 1, it is called a<br>a.Diamagnetic material<br>b.Paramagnetic material<br>c.Ferromagnetic material<br>d. None of the above</h1><h1> </h1><h1>28. When the relative permeability of material is much greater than 1, it is called a<br>a. Diamagnetic material<br>b. Paramagnetic material<br>
c. Ferromagnetic material<br>d. None of the above</h1><h1> </h1><h1>29. The source of magnetic field is<br>a. An isolated magnetic pole<br>b. Static electric charge<br>c. Magnetic substances<br>d. Current loop</h1><h1> </h1><h1>30. A magnetic needle is kept in a uniform magnetic field. It experience<br>a. A force and a torque<br>b. A force but not a torque<br>c. A torque but not force<br>d. Neither a torque nor a force</h1><h1> </h1><h1>31. The unit of pole strength is<br>a. A/m^2<br>b. Am<br>c. Am^2<br>d. Wb/ m^2</h1><h1> </h1><h1>32. AT/m is the unit of<br>a. Mmf<br>b. Reluctance<br>c. Magnetizing force<br>d. Magnetic flux density<br> </h1><h1> </h1><h1>33. A magnetic needle is kept in a non uniform magnetic field. It experiences<br>a.A force and a torque<br>b.A force but not a torque<br>c.A torque but not force<br>d.Neither a torque nor a force</h1><h1> </h1><h1>34. Magnetic flux passes more readily through<br>a.Air<br>b.Wood<br>c.Vacuum<br>d.Iron</h1><h1> </h1><h1>35. Demagnetizing of magnet can be done by<br>a.Rough handling<br>b.Heating<br>c.Magnetizing in opposite direction<br>d.All the above</h1><h1> </h1><h1>36.magnetic circuit has mmf of 400 AT and a reluctance of 2X10^5 AT/wb. The magnetic flux in the<br>magnetic circuit is<br>a.3X10^-5 Wb<br>b.2X10^-3 Wb<br>c.1.5 x 10^-2 Wb<br>d.2.5 X 10^-4 Wb</h1><h1> </h1>
<h1>37. A 2 Cm long coil has 10 turns and caries a current of 750 m A . The magnetizing force of the coil<br>is<br>a. 225 AT/m<br>b. 675AT/m<br>c. 450 AT/m<br>d. 375 AT/m</h1><h1> </h1><h1>38.. The reluctance of a magnetic circuit is ……….. Relative permeability of the material comprising<br>the circuit..<br>a. Directly proportional to<br>b. Inversely proportional to<br>c. Independent of<br>d. None of the above</h1><h1> </h1><h1>39. MMF in a magnettic circuit corresponding to …………….. In an electric circuit.<br>a. Voltage drop<br>b. Potential difference<br>c. Electric intensity<br>d. Emf</h1><h1> </h1><h1>40. The magnitude of AT required for aiir gap iis much greatter than that required for iron part of<br>magnetic circuit because......<br>a. Air is gas<br>b. Air has the lowest relative permeability<br>c. Air is a conductor of magnetic flux<br>d. None of the above</h1><h1> </h1><h1>41. The reluctance of9 the magnetic circuit depends upon<br>a.Current in the coil<br>b.Number of turns of coil<br>c.Flux density in the circuit<br>d.None of the above</h1><h1> </h1><h1>42. The B-H curves Curve of ……… will not be a straight line<br>a.Air<br>b.Soft iron<br>c.Harden steel<br>d.Silicon steel<br> </h1><h1> </h1><h1>43. Whatever may be the flux density in ……….. the material will never saturate.<br>a. Soft iron<br>b. Cobalt steel<br>c. air<br>d. Silicon steel</h1><h1> </h1><h1>44. The B-H curve of …… will not be a straight line.<br>a. air<br>b. copper<br>c. wood<br>d. soft iron</h1><h1> </h1>
<h1>45. The B-H curve is used to find the mmf of ……. In a magnetic circuit.<br>a. Air gap<br>b. Iron path<br>c. Both air gap and iron part<br>d. None of the above</h1><h1> </h1><h1>46. The material used for the core of a good relay should have …….. hysteresis loop<br>a. Large<br>b. Very large<br>c. Narrow<br>d. None of the above</h1><h1> </h1><h1>47. The magnetic material used for ………. Should have a large hysteresis loop<br>a. Transformer<br>b. Dc generator<br>c. Ac motors<br>d. Permanent magnet</h1><h1> </h1><h1>48. The basic requirement for inducing emf in a coil is that<br>a. Flux should link the coil<br>b. There should be change in flux linking the coil.<br>c. Coil should form a closed loop<br>d. None of the above</h1><h1> </h1><h1>49. The emf induced in the coil of N turns is given by ……..<br>a. dφ/dt<br>b. N dφ/dt<br>c. - N dφ/dt<br>d. N dt/dφ</h1><h1> </h1><h1>50. The direction of the induced emf in a conductor (or coil ) can be determine by ……….<br>a. Work law<br>b. Ampere’s law<br>c. Fleming,s right hand rule<br>d. Flemin,s left hand rule</h1><h1> </h1><h1>51. The emf induced in a ………. Is the statically induced emf. a. Dc generator<br>b. Transformer<br>c. Dc motor<br>d. None of the above</h1><h1> </h1><h1>52. The emf induced in a ……………. Is dynamically induced emf.<br>a. Alternator<br>b. Transformer<br>c. Dc generator<br>d. None of the above</h1><h1> </h1><h1>53. Inductance apposes ……. In the current in a circuit.<br>a. Only increases<br>b.Only decreases<br>c.Change<br>d.None of the above</h1>
<h1> </h1><h1>54.If the relative permeability of the material surrounding the coil is increased, the inductance of<br>the coil…………….<br>a.Is increased<br>b.Is decreased<br>c.Remains unchanged<br>d.None of the above</h1><h1> </h1><h1>55. Inductance in the circuit…<br>a.Prevents the current from changing<br>b.Delays the change in current<br>c.Causes power loss<br>d.Causes the current to lead the voltage</h1><h1> </h1><h1>56. The inductance of a coil is ….. the reluctance of magnetic path.<br>a.Independent<br>b.Directly proportional<br>c.Inversely proportional<br>d.None of the above</h1><h1> </h1><h1>57. If the number of turns of a coil is increased two times, its inductance is ……………..<br>a.Increased two times<br>b.Decreased two times<br>c.Decreased four times<br>d.Increased four times<br> </h1><h1>Answers Basic Electrical Engineering unit 2<br><br><br>1. C 2. D 3. A 4. C 5. B<br>6. c 7. D 8. C 9. B 10. B<br>11. B 12. A 13. A 14. C 15. A<br>16. C 17 A 18 A 19 B 20 D<br>21. C 22. A 23. B 24. A 25. D<br>26. B 27. A 28. C 29. D 30. C<br>31. B 32. C 33. A 34. D 35. D<br>36. B 37. D 38. B 39. D 40. B<br>41. C 42 A 43. C 44. D 45. B<br>46. C 47. D 48. B 49. C 50. D<br>51. B 52. C 53. C 54. A 55. B<br>56. C 57. D<br><br><br> </h1><div id="cke_pastebin" style="overflow: hidden; position: absolute; top: 2243.8px; width: 1px; height: 1px; left: -1000px;">
<p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;"> <em><strong><span style="font-size: 26px;"><span style="background-color: rgb(255, 240, 245);">1) Basic Circuit Elements</span></span></strong></em></span></span></h1><h1><span style="font-size: 20px;">Some others books</span></h1><h1><span style="font-size: 16px;"><a href="http://files.sai34.webnode.com/200000008-45ef346e87/DC%20CIRCUIT.pdf"><span style="color: rgb(178, 34, 34);">DC CIRCUIT.pdf (669839)</span></a></span></h1><h1><span style="font-size: 18px;"><a href="http://files.sai34.webnode.com/200000009-b809cb9039/EEE%201.pdf"><span style="color: rgb(165, 42, 42);">EEE 1.pdf (565046)</span></a></span></h1><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;"><u><strong>1) Resistance</strong></u></span></span></h1><p><span style="font-size: 16px;">When voltage is applied to a piece of metal wire, as shown in figure 1.2 (a), the current I flowing through the wire is proportional to the voltage V across two points in the wire. This property is known as Ohm's law, which reads<img alt="" height="50" src="http://files.sai34.webnode.com/200000010-3ba973ca36/img10.gif" width="189">where <em>R</em> is called resistance, and <em>G</em> is called conductance. The resistance <em>R</em> and the conductance <em>G</em> of the same piece of wire is related by <em>R</em> = 1/<em>G</em>. Resistance is measured in <em>ohms</em> (Ω) and conductance in <em>siemens</em> (S or Ʊ ).</span></p><p><img alt="" height="132" src="http://files.sai34.webnode.com/200000012-4337c4431d/img13.gif" width="405"></p><p><span style="font-size: 14px;"><strong>Figure 1.2:</strong> Ohm's law. (a) Metal wire; (b) circuit symbol</span></p><p><span style="font-size: 16px;">Any apparatus/device that has this property is called a <em>resistor.</em> Study of the physics of resistance shows that it is proportional to the length of the metal wire, <em>l</em>, and inversely proportional to the cross-sectional area, <em>A</em>, i.e., <img alt="" height="54" src="http://files.sai34.webnode.com/200000013-9e7c89f771/img15.gif" width="85">where the proportionality constant (ρ) is known as the <em>resistivity</em> of the metal.</span></p><p> </p><p><span style="font-size: 16px;">We may calculate the power required to pass current <em>I</em> through a resistor of resistance <em>R</em> using the previously derived formula, i.e.,</span><br><img alt="" height="11" src="http://files.sai34.webnode.com/200000014-9048c9141d/img16.gif" width="274"><br><span style="font-size: 16px;">Using the Ohm's law equation, we get <img alt="" height="55" src="http://files.sai34.webnode.com/200000015-7dac27ea6a/img17.gif" width="301">The last inequality defines a property called <em>passivity.</em></span></p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;"><strong>2) Independent and Dependent Sources</strong></span></span></h1><p> </p><p><span style="font-size: 16px;">There are two principal types of source, namely <em>voltage source</em> and <em>current source</em>.
Sources can be either independent or dependent upon some other quantities.</span></p><p><span style="font-size: 16px;">An <em>independent voltage source</em> maintains a voltage (fixed or varying with time) which is not affected by any other quantity. Similarly an <em>independent current source</em> maintains a current (fixed or time-varying) which is unaffected by any other quantity. The usual symbols are shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node6.html#ch0source">1.3</a>.</span></p><p><img alt="" height="149" src="http://files.sai34.webnode.com/200000016-d0cfcd1ca1/img18.gif" width="374"></p><p><span style="font-size: 14px;"><strong>Figure 1.3:</strong> Symbols for independent sources </span></p><p><span style="font-size: 16px;">Some voltage (current) sources have their voltage (current) values varying with some other variables. They are called <em>dependent</em> voltage (current) sources or <em>controlled</em> voltage (current) sources , and their usual symbols are shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node6.html#ch0dsource">1.4</a>.</span></p><p><span style="font-size: 16px;"><em>Remarks</em> -- It is not possible to force an independent voltage source to take up a voltage which is different from its defined value. Likewise, it is not possible to force an independent current source to take up a current which is different from its defined value. Two particular examples are short-circuiting an independent voltage source and open-circuiting an independent current source. Both are not permitted.</span></p><p> </p><p><img alt="" height="150" src="http://files.sai34.webnode.com/200000017-c39acc4933/img19.gif" width="390"></p><p> </p><p><span style="font-size: 14px;"><strong>Figure 1.4:</strong> Symbols for dependent sources. Variables in brackets are the controlling variables whose values affect the value of the source. </span></p><p> </p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 16px;"><span style="font-size: 22px;"><strong>3) Circuit</strong></span></span></span></h1><p><span style="font-size: 16px;">A collection of devices such as resistors and sources in which terminals are connected together by connecting wires is called an <em>electric circuit.</em> These wires converge in <em>nodes</em>, and the devices are called <em>branches</em> of the circuit, as shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node7.html#ch0circuit">1.5</a>.</span></p><p><span style="font-size: 16px;">The general circuit problem is to find all currents and voltages in the branches of the circuit when the intensities of the sources are known. Such a problem is usually referred to as <em>circuit analysis.</em></span></p><p><span style="font-size: 16px;"><em>Remarks</em> -- While the current in a resistor has a fixed relationship with the voltage across it, the current flowing in a voltage source, or the voltage across a current source, is theoretically unrestricted and can assume whatever value governed by the external circuit.</span></p>
<p><img alt="" height="163" src="http://files.sai34.webnode.com/200000018-430c044052/img20.gif" width="399"></p><p><span style="font-size: 14px;"><strong>Figure 1.5:</strong> Part of an electric circuit </span></p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;"><strong>4) Kirchhoff's Current Law (KCL)</strong></span></span></h1><p><span style="font-size: 16px;">As a direct consequence of the conservation of charge, namely charge can neither be created nor destroyed, the node, being of negligible physical size, holds no charge. For instance, referring to figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node9.html#ch0kcl">1.6</a>, the sum of I1, I2 and I3 must equal zero.</span></p><p><img alt="" height="96" src="http://files.sai34.webnode.com/200000019-6d46a6e40f/img21.gif" width="254"></p><p><span style="font-size: 14px;"><strong>Figure 1.6:</strong> Kirchhoff's current law </span></p><p><span style="font-size: 16px;">Formally, KCL states that <em>the algebraic sum of the currents in all the branches that converge in a common node is equal to zero.</em> In mathematical form, for <em>n</em> branches converging into a node, KCL states that <img alt="" height="50" src="http://files.sai34.webnode.com/200000020-9de929ee35/img22.gif" width="173">where (I k)is the current flowing in the <em>k</em>th branch and its direction is assumed to be pointing towards the node.</span></p><p> </p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;">5) <strong>Kirchhoff's Voltage Law (KVL)</strong></span></span></h1><p><span style="font-size: 16px;">When a charged particle is moved from a point to another, the work done is , where <em>V</em>.</span>δq<span style="font-size: 16px;"> is the voltage across the two points and </span>(δq )<span style="font-size: 16px;">is the amount of charge on the particle. Consider a particular case where the two points are actually the same point in the circuit. In this case, the work done is zero. By the same argument, if a unit charge is moved around a closed path such as the square path shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node10.html#ch0kvl">1.7</a>, the work done is zero, i.e.,</span></p><p><img alt="" height="13" src="http://files.sai34.webnode.com/200000021-572965823a/img23.gif" width="317"></p><p><img alt="" height="180" src="http://files.sai34.webnode.com/200000022-5bed85ce7f/img24.gif" width="384"></p><p><span style="font-size: 14px;"><strong>Figure 1.7:</strong> Kirchhoff's voltage law </span></p><p> </p><p><span style="font-size: 16px;">Formally, KVL states that <em>the algebraic sum of the voltages between successive nodes in a closed path in a circuit is equal to zero.</em> In mathematical form, for a closed path with successive nodes 1,2,.....π, KVL states that</span></p><p><img alt="" height="50" src="http://files.sai34.webnode.com/200000023-1dbb21eb43/img25.gif" width="253"></p><p><span style="font-size: 16px;">where V<span style="font-size: 10px;">j1k </span>is the voltage between nodes <em>j</em> and <em>k</em>.</span></p>
<p><span style="font-size: 16px;"><em> </em></span></p><p><span style="font-size: 16px;"><em> Remarks</em> -- Any circuit that has a solution must satisfy Kirchhoff's laws. From the properties of independent sources, we can immediately conclude that a circuit cannot be solved if there exists a loop that is formed exclusively of independent voltage sources. Thus, short-circuiting an independent voltage source, as remarked earlier, is a particular case where KVL is violated. Similarly, a circuit cannot be solved if there exists a node to which only independent current sources are connected. Also, open-circuiting an independent current source is a particular case where KCL is violated.</span></p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;">6) Series Circuit</span></span></h1><p><span style="font-size: 16px;">When devices are ``chained'' up such that each of the nodes is incident to just two devices, the resulting circuit is said to be a series circuit.</span></p><p><img alt="" height="111" src="http://files.sai34.webnode.com/200000024-a37b6a4743/img26.gif" width="485"></p><p><span style="font-size: 14px;"><strong>Figure 1.8:</strong> Series connection </span></p><p><img alt="" height="105" src="http://files.sai34.webnode.com/200000025-dfeb3e0e5b/img27.gif" width="248"></p><p><span style="font-size: 16px;"><strong>Figure 1.9:</strong> Equivalent resistance for <em>n</em> resistors in series </span></p><p><span style="font-size: 16px;">A series circuit consisting of <em>n</em> resistors is shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node12.html#ch0series">1.8</a>. Clearly, KVL gives V<span style="font-size: 10px;">1(n1+1)</span>= V<span style="font-size: 10px;">12</span>+V<span style="font-size: 10px;">23</span>+......+V<span style="font-size: 10px;">n(n+1)</span>. Also, from Ohm's Law,V<span style="font-size: 10px;">1(n+1)</span>=(R<span style="font-size: 10px;">1</span>+R<span style="font-size: 10px;">2</span>+.....R<span style="font-size: 10px;">n</span>)I . Therefore, the equivalent circuit seen from nodes 1and <em>n</em>+1 is a resistor of resistance ,R<span style="font-size: 10px;">1</span>+R<span style="font-size: 10px;">2</span>+....R<span style="font-size: 10px;">n</span> as shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node12.html#ch0serres">1.9</a>.</span></p><p> </p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;">6) Parallel Circuit</span></span></h1><p><span style="font-size: 16px;">When devices are connected such that one terminal of each device is connected to a node of the circuit while the other terminals of the elements are connected to another node of the circuit, the resulting circuit is said to be a parallel circuit.</span></p><p><img alt="" height="114" src="http://files.sai34.webnode.com/200000026-a253aa34e3/img28.gif" width="363"></p><p> </p><p><span style="font-size: 14px;"><strong>Figure 1.10:</strong> Parallel connection </span></p>
<p><span style="color: rgb(255, 255, 255);"><img alt="" height="105" src="http://files.sai34.webnode.com/200000027-444b845428/img29.gif" width="250"></span></p><p><span style="font-size: 14px;"><strong>Figure 1.11:</strong> Equivalent conductance for <em>n</em> conductors in parallel </span></p><p> </p><p><span style="font-size: 16px;">A parallel circuit consisting of <em>n</em> resistors is shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node13.html#ch0parallel">1.10</a>. Note that we use conductance instead of resistance in this case. Clearly, KCL gives </span><img alt="" height="20" src="http://files.sai34.webnode.com/200000028-5e3355f2c5/img30.gif" width="131"><span style="font-size: 16px;">Since all voltages across the resistors are equal to <em>V</em>, we have <img alt="" height="23" src="http://files.sai34.webnode.com/200000029-080be0905f/img31.gif" width="169">and the equivalent conductance as seen from the left end is</span><img alt="" height="20" src="http://files.sai34.webnode.com/200000030-520765300c/img32.gif" width="120"> <span style="font-size: 16px;">, as shown in figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node13.html#ch0parres">1.11</a>. Note that the expression</span><img alt="" height="20" src="http://files.sai34.webnode.com/200000031-e63f3e738b/img33.gif" width="150"> <span style="font-size: 16px;">can be written as</span></p><p><img alt="" height="31" src="http://files.sai34.webnode.com/200000032-9603f96fd9/img34.gif" width="424"></p><p><span style="font-size: 16px;">For example, if there are only two resistors, the equivalent resistance of the parallel circuit is</span></p><p><img alt="" height="31" src="http://files.sai34.webnode.com/200000033-34367352f9/img35.gif" width="291"></p><p><img alt="" src="file:///C:/Users/SAI/AppData/Local/Temp/moz-screenshot-14.jpg"></p><p><span style="font-size: 18px;"><strong>Example 1.1: Illustration of series/parallel reduction</strong> -- It is possible to reduce an assembly of resistors whose configuration is based on series and parallel connections. Referring to the circuit of figure <a href="http://www.eie.polyu.edu.hk/%7Ecktse/linear_circuits/main/node13.html#ch0serp">1.12</a>, we can use an equivalent resistance to replace the circuit such that the input current and voltage are unaffected.</span></p><p><img alt="" height="220" src="http://files.sai34.webnode.com/200000034-79dbd7bcf1/img36.gif" width="309"></p><p><span style="font-size: 14px;"><strong>Figure 1.12:</strong> Series/parallel reduction process </span></p><p> </p><p><span style="font-size: 16px;">First of all, we observe that R<span style="font-size: 10px;">4</span> and R<span style="font-size: 10px;">5</span> are in parallel and can be replaced by an equivalent resistance <em>R</em>' which is given by</span></p><p> </p><p><img alt="" height="31" src="http://files.sai34.webnode.com/200000035-77527784e3/img37.gif" width="322"></p><p><span style="font-size: 16px;">This <em>R</em>' is connected in series with R<span style="font-size: 10px;">2</span> , and the resulting sub-circuit is connected in parallel with R<span style="font-size: 10px;">3</span> . Thus,
using the series formula, followed by the parallel formula, we get an equivalent resistance <em>R</em>'' which </span></p><p><span style="font-size: 16px;">represents the part of the circuit covering R</span>2,<span style="font-size: 16px;">R</span>3,<span style="font-size: 16px;">R</span>4 <span style="font-size: 16px;">and R</span>5 i.e.,</p><p><img alt="" height="16" src="http://files.sai34.webnode.com/200000036-63e2c64ded/img38.gif" width="310"></p><p><span style="font-size: 16px;">Finally, adding R<span style="font-size: 10px;">1</span> to <em>R</em>'' yields the equivalent resistance R<span style="font-size: 10px;">eq</span> as required</span></p><p><img alt="" height="66" src="http://files.sai34.webnode.com/200000037-c2304c32ae/img39.gif" width="358"></p><h1><span style="color: rgb(255, 0, 0);"><span style="font-size: 22px;"><span style="background-color: rgb(255, 255, 0);">ACCORDING TO TEXTBOOK</span></span></span></h1><p><span style="color: rgb(0, 128, 0);"><span style="font-size: 22px;">FORMULAE:</span></span></p><h4><u><strong><span style="color: rgb(255, 215, 0);"><span style="font-size: 18px;"><span style="background-color: rgb(255, 0, 0);">1) Resistance:-</span></span></span></strong></u></h4><p><u><strong><span style="color: rgb(255, 215, 0);"><span style="font-size: 18px;"> </span></span></strong></u><span style="font-size: 18px;">(1) R=V/I</span></p><p><span style="font-size: 18px;"> (2) p=I*I*R</span></p><p><span style="font-size: 18px;"> (3) E=I*I*R*t</span></p><p><span style="font-size: 18px;"> (4) R= ϱ<span style="font-size: 16px;">l</span>/A</span></p><p><span style="font-size: 18px;"> (5) ϱ= RA/l</span></p><p><u><span style="color: rgb(255, 215, 0);"><span style="font-size: 18px;"><strong><span style="background-color: rgb(255, 0, 0);">2)Effect Of Temperature of R :-</span></strong></span></span></u></p><p><span style="font-size: 18px;"> (1) ΔR α R<span style="font-size: 12px;">0</span>*t<span style="font-size: 12px;">1</span></span></p><p><span style="font-size: 18px;"> (2) α<span style="font-size: 12px;">o</span>= ΔR/R<span style="font-size: 12px;">o</span>t<span style="font-size: 12px;">1</span></span></p><p><span style="font-size: 18px;">  
; <span style="color: rgb(0, 255, 0);">TIME CONSTANT λ=CR</span></span></p><h1><span style="color: rgb(0, 0, 205);"><span style="font-size: 24px;"> <span style="background-color: rgb(0, 255, 0);">PROBLEMS</span></span></span></h1><p> <span style="font-size: 20px;"><span style="font-size: 16px;"><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;">EX 1:- </span></span> A coil has a resistance of 100 Ω when the mean temperature is 20˚c and 110 Ωwhen the mean</span></span><span style="font-size: 20px;"><span style="font-size: 16px;"> temperature</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">is </span></span><span style="font-size: 20px;"><span style="font-size: 16px;">45˚c. Find it's mean temperature rise when its resistance is 130 Ω and surrounding</span></span><span style="font-size: 20px;"><span style="font-size: 16px;"> temperature </span></span><span style="font-size: 16px;"> </span><span style="font-size: 20px;"><span style="font-size: 16px;">is15˚</span></span><span style="font-size: 16px;">c.</span><span style="font-size: 16px;"> </span></p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 18px;"><strong><em>(ANS:- 80˚c)</em></strong></span></span></p><p> </p><p><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> EX 2:-</span></span><span style="font-size: 18px;"> <span style="font-size: 16px;">A potential difference of 250V is applied to copper field coil at a temperature of 15˚c and the</span></span><span style="font-size: 18px;"><span style="font-size: 16px;"> current is 5A. what will be the mean temperature of the coil when the current has fallen to 4A, the</span></span><span style="font-size: 18px;"><span style="font-size: 16px;"> applied voltage being the sameas before ? take αo=1/234.5˚c</span></span> </p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 18px;"><strong><em>(ANS:-77.375˚c)</em></strong></span></span></p><p> </p><p><span style="color: rgb(255, 0, 0);"><span style="font-size: 18px;"><span style="font-size: 20px;"> EX 3:-</span></span></span> <span style="font-size: 16px;">A temperature coefficient of copper material at 20</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c is 1</span>/<span style="font-size: 16px;">234.5 per</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c .Find the temperature coefficient </span><span style="font-size: 16px;">of copper at 40</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c and 60</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c<span style="font-style: italic;"><span style="font-weight: bold;"> </span></span></span><strong><em><span style="font-size: 16px;"> </span></em></strong></p><p><span style="color: rgb(0, 0, 255);"><strong><em><span style="font-size: 16px;"><span style="font-size: 18px;">(ANS:- </span></span><span style="font-size: 18px;">0.00364298˚c, 0.0033955)</span></em></strong></span></p><p> </p>
<p><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> EX 4:- </span></span><span style="font-size: 16px;">Two types of resistance wire are available :</span><span style="font-size: 16px;">Type 1: Resistance=1.6</span><span style="font-size: 20px;"><span style="font-size: 16px;">Ω/mt,RTC=0.006/</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c.</span></p><p><span style="font-size: 16px;"> Type 2: Resistance=0.8</span><span style="font-size: 20px;"><span style="font-size: 16px;">Ω/mt,RTC=0.003</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">/</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c</span></p><p><span style="font-size: 16px;">A resistance of 260 </span><span style="font-size: 20px;"><span style="font-size: 16px;">Ω is to be constructed by connecting suitable lengths of the two types in series so as</span></span><span style="font-size: 20px;"><span style="font-size: 16px;"> to have RTC of combination as 0.0042</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">/</span></span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c.Calculate the required lengths of the types. All the given</span><span style="font-size: 16px;"> figures belong to the same temperature.</span><span style="font-size: 18px;"><strong> </strong></span></p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 18px;"><strong>(<em>ANS :- 65 m,195m)</em></strong></span></span></p><p> </p><p><span style="font-size: 20px;"><span style="color: rgb(255, 0, 0);"> EX 5:-</span></span> <span style="font-size: 16px;">Find the current flowing at the instant of switching 25 watts lamp on 230 V circuit, gives the</span><span style="font-size: 16px;"> incandescent filaments temperature is 2000</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c and RTC at 15</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c is 0.005</span>/<span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c</span> </p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 18px;"><em><strong>(ANS :- 1.2505 A)</strong></em></span></span></p><p> </p><p><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> EX 6:-</span></span><span style="font-size: 16px;">Two resistance A and B are connected in series .At˚c resistance of B is four times the resistance of A.</span><span style="font-size: 16px;"> At 0˚c temperature coefficent of resistance of A is 0.1 and that of series combination is 0.08. Find
the</span><span style="font-size: 16px;"> temperature coefficent of resistance of B at 0</span><span style="font-size: 20px;"><span style="font-size: 16px;">˚</span></span><span style="font-size: 16px;">c </span></p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 16px;"><strong><span style="font-size: 18px;">(ANS :- 0.075/</span></strong></span><strong><span style="font-size: 18px;">˚c)</span></strong></span></p><p> </p><p><span style="font-size: 20px;"><span style="color: rgb(255, 0, 0);"> EX 7:-</span> </span><span style="font-size: 16px;">A copper conductor has its specific resistance of 1.6*10^-6 Ω-cm at 0˚c and a resistaence </span><span style="font-size: 16px;">temperature coefficent of 1/254.5/˚c at 20˚c.Find resistance temperature coefficent and specific</span><span style="font-size: 16px;"> resistance at 60˚c</span><span style="font-size: 16px;"> </span></p><p><span style="color: rgb(0, 0, 255);"><span style="font-size: 16px;"><span style="font-size: 18px;"><strong>(ANS :- 0.0033955</strong></span></span><span style="font-size: 18px;"><strong>˚c, 2.009344Ω-cm )</strong></span></span></p><p> </p><p><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> EX 8:-</span></span><span style="font-size: 16px;"><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> </span></span>The current flowing at the instant of switching on a 40 W lamp to a 240 V supply is 2 A. Resistance</span><span style="font-size: 16px;"> temperature coefficent of the filament material is 0.0055 at the room temperature of 20˚c. Calculate the</span><span style="font-size: 16px;"> working temperature of the filaments and current taken during normal working.</span> </p><p><span style="color: rgb(0, 0, 255);"><strong><span style="font-size: 18px;">(ANS :- 2020˚c , 0.16667 Amp)</span></strong></span></p><p> </p><p><span style="font-size: 20px;"><span style="color: rgb(255, 0, 0);"> EX 9:- </span></span><span style="font-size: 16px;">Singal core cable has conductor diameter of 1.5 cm and insulation thickness of 2.5 cm. the resistivite </span><span style="font-size: 16px;">of insulating material is 9*10^12 Ω- meter.Determine the insulation resistance per meter length of</span><span style="font-size: 16px;"> cable.if working voltage is leakage current per km length of cable <span style="font-weight: bold;">? </span></span></p><p><span style="color: rgb(0, 0, 255);"><strong><span style="font-size: 18px;">(ANS :- 2100.373MΩ , 0.4761m Amp)</span></strong></span></p><p> </p><p><span style="color: rgb(255, 0, 0);"><span style="font-size: 20px;"> EX 10:- </span></span><span style="font-size: 16px;">A singal core cable has conductor dimeter of 1.2 cm and insulation thickness of 1.4 cm. The</span><span style="font-size: 16px;"> </span><span style="font-size: 16px;">insulation resistance per km is 6550 M Ω. What would be insulation thickness of insulation resistance</span><span style="font-size: 16px;"> per km is 669.85 M-Ω.</span> </p><p><span style="color: rgb(0, 0, 255);"><strong><span style="font-size: 18px;">(ANS :- 1.999952 Cm)</span></strong></span></p><p> </p>
CIRCUIT ELEMENTS</a></li><li><a href="/eee/a2-magnetic-circuit/">2) MAGNETIC CIRCUIT</a></li><li class="open last selected activeSelected"><a href="/eee/a3-ac-
fundamental/">3) AC FUNDAMENTAL</a><ul class="level1">
<h1>3) AC FUNDAMENTAL</h1><p><strong><u>Ac Fundamentals MCQ's</u></strong></p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>1) EMF generated in coil depends upon -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) Strength of magnetic field b) Number of turns in coil</strong></span></p><p><span style="font-size:16px;"><strong> c) Speed at which coil or magnetic field rotates</strong></span></p><p><span style="font-size:16px;"><strong> d) All of these </strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>2) Equation of alternating emf is given by -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) e = NCOSθ b) e = N_COSθ </strong></span></p><p><span style="font-size:16px;"><strong> c) e = N d) None of these</strong></span></p><p> </p><p> </p>
<p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>3) The induced alternating emf is maximum at an angle of -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) θ = 0</strong><strong><sup>0</sup></strong><strong> b) θ = 90</strong><strong><sup>0</sup></strong></span></p><p><span style="font-size:16px;"><strong> c) θ = 180</strong><strong><sup>0</sup></strong><strong> d) None of these</strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>4) Equation of alternating emf is zero at an angle of -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) θ = 0</strong><strong><sup>0</sup></strong><strong> b) θ = 90</strong><strong><sup>0</sup></strong></span></p><p><span style="font-size:16px;"><strong> c) θ = 180</strong><strong><sup>0</sup></strong> <strong> d) None of these</strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>5) RMS value is also known as - </strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) Effective value b) Virtual value of AC current or Voltage</strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) & b) d) None of these</strong></span></p><p> </p><p> </p><p><span style="font-family:comic sans ms,cursive;"><em><span style="font-size: 20px;"><strong>6) Mean square value of instantaneous current is equal to -</strong></span></em></span></p><p><span style="font-size:16px;"><strong> a) Imax / 1.414 b) I / 1.414 c) 0.707 Imax d) Both a) & c)</strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>7) Average or Mean value of current is equal to -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) 2 Imax / 3.14 b) 0.637 Imax </strong></span></p><p><span style="font-size:16px;"><strong> c) 2 I / 3.14 &n
bsp; d) Both a) & b)</strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>8) Form factor <u>kf</u></strong> <strong>is equal to -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) Irms / I avg b) Iavg / Irms</strong></span></p><p><span style="font-size:16px;"><strong> c) 1.11 d) Both a) & c)</strong></span></p><p> </p><p> </p><p><span style="font-size:20px;"><em><span style="font-family: comic sans ms,cursive;"><strong>9) Peak factor is also known as -</strong></span></em></span></p><p><span style="font-size:16px;"><strong> a) Crest factor b) Amplitude factor</strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) & b) d) None of these</strong></span></p><p> </p><p> </p><p><span style="font-size:20px;"><em><span style="font-family: comic sans ms,cursive;"><strong>10) In pure resistive circuit, current i is maximum when </strong></span></em></span></p><p><span style="font-size:16px;"><strong> a) wt = 3.14 / 2 b) Sin wt = 1</strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) and b) d) None of these</strong></span></p><p> </p><p> </p><p><em><span style="font-family:comic sans ms,cursive;"><span style="font-size: 20px;"><strong>11) In pure resistive circuit, applied voltage and current are -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) Out of phase b) In phase</strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) and b) d) None of these</strong></span></p><p> </p><p> </p><p><em><span style="font-size:20px;"><span style="font-family: comic sans ms,cursive;"><strong>12) For pure resistive circuit -</strong></span></span></em></p>
<p><span style="font-size:16px;"><strong> a) i = Imax Sin wt b) i =<u> Vmax Sin wt </u>/ R</strong></span></p><p><span style="font-size:16px;"><strong> c) i = I Sin wt d) Both a) and b)</strong></span></p><p> </p><p> </p><p><em><span style="font-size:20px;"><span style="font-family: comic sans ms,cursive;"><strong>13) Average power in pure resistive circuit is equal to - </strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) Vmax . Imax / 1.414 b) Irms . Vrms </strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) and b) d) None of these</strong></span></p><p> </p><p> </p><p><em><span style="font-size:20px;"><span style="font-family: comic sans ms,cursive;"><strong>14) In pure resistive circuit, power is -</strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) zero b) never zero c) Maximum d) Minimum</strong></span></p><p> </p><p> </p><p><em><span style="font-size:20px;"><span style="font-family: comic sans ms,cursive;"><strong>15) For pure inductive circuit, </strong></span></span></em></p><p><span style="font-size:16px;"><strong> a) i = Imax Sin wt b) i = Imax Sin (wt – π/2) </strong></span></p><p><span style="font-size:16px;"><strong> c) i = Imax Sin (wt+ π/2) d) i = v.r</strong></span></p><p><span style="font-size:16px;"><strong>16) Average power in pure inductive circuit - </strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) maximum c) never 0 d) None of these</strong></span></p><p><span style="font-size:16px;"><strong>17) For pure capacitive circuit</strong></span></p><p><span style="font-size:16px;"><strong> a) Applied voltage and current are in phase</strong></span></p><p><span style="font-size:16px;"><strong> b) Applied voltage and current are out of phase</strong></span></p><p><span style="font-size:16px;"><strong> c) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong>18) For pure inductive circuit</strong></span></p>
<p><span style="font-size:16px;"><strong> a) Applied voltage and current are in phase</strong></span></p><p><span style="font-size:16px;"><strong> b) Applied voltage and current are out of phase</strong></span></p><p><span style="font-size:16px;"><strong> c) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong>19) In R – L series circuit current is </strong></span></p><p><span style="font-size:16px;"><strong> a) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> b) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> c) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong>20) In R – L series circuit voltage is </strong></span></p><p><span style="font-size:16px;"><strong> a) voltage lags behind current by π/2</strong></span></p><p><span style="font-size:16px;"><strong> b) voltage leads current by π/2</strong></span></p><p><span style="font-size:16px;"><strong> c) voltage lags behind currentby π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) voltage leads applied current by π/2</strong></span></p><p><span style="font-size:16px;"><strong>21) In R – L series circuit the equation of current is given by </strong></span></p><p><span style="font-size:16px;"><strong> a) i = Imax Sinωt b) i = Imax Sin (ωt- π/2 ) </strong></span></p><p><span style="font-size:16px;"><strong> c) i = Imax Sin (ωt+π/2) d) i = v.r</strong></span></p><p><span style="font-size:16px;"><strong>22) In R – L series circuit the equation of voltage is given by</strong></span></p><p><span style="font-size:16px;"><strong> a) v = Vmax Cosωt b) v = Vmax Cos (ωt-π/2)</strong></span></p><p><span style="font-size:16px;"><strong> c) v = Vmax Cos (ωt+ π/2 ) d) v = Vmax Sin (ωt –π/2) </strong></span></p><p><span style="font-size:16px;"><strong>23) In R – L series circuit the power factor in terms of average power is given by - </strong></span></p><p><span style="font-size:16px;"><strong> a) Cos θ = Pavg . Vrms . Irms b) Cos θ = Vrms . Irms / Pavg </strong></span></p><p><span style="font-size:16px;"><strong> c) Cos θ = Pavg / Vrms . Irms d) Cos θ = 0</strong></span></p><p><span style="font-size:16px;"><strong>24) In R – L series circuit the rms current in terms of average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) I = Pavg . Vrms . Cos θ b) I = Cos θ / Pavg . Vrms</strong></span></p>
<p><span style="font-size:16px;"><strong> c) I = Pavg .Vrms / Cos θ d) I = Pavg / Vrms . Cos θ </strong></span></p><p><span style="font-size:16px;"><strong>25) In R – L series circuit the rms voltage in terms of average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) V = 1 / Pavg . Vrms . Irms b) V = Pavg / Vrms . Irms</strong></span></p><p><span style="font-size:16px;"><strong> c) V = Pavg / Irms . Cos θ d) V = Pavg.Cos θ / Irms</strong></span></p><p><span style="font-size:16px;"><strong>26) In R – L series circuit the average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) Pavg = 0 b) Pavg = infinity c) P = V.I.Cos θ d) V.I / Cos θ</strong></span></p><p><span style="font-size:16px;"><strong>27) In R – C circuit</strong></span></p><p><span style="font-size:16px;"><strong> a) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> b) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> c) Current lags behind applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) Current leads applied voltage by π/2</strong></span></p><p><span style="font-size:16px;"><strong>28) In R – C series circuit voltage is </strong></span></p><p><span style="font-size:16px;"><strong> a) voltage lags behind current by π/2</strong></span></p><p><span style="font-size:16px;"><strong> b) voltage leads current by π/2</strong></span></p><p><span style="font-size:16px;"><strong> c) voltage lags behind currentby π/2</strong></span></p><p><span style="font-size:16px;"><strong> d) voltage leads applied current by π/2</strong></span></p><p><span style="font-size:16px;"><strong>29) In R – C series circuit the equation of current is given by </strong></span></p><p><span style="font-size:16px;"><strong> a) i = Imax Sin ωt b) i = Imax Sin (ωt -π/2 ) </strong></span></p><p><span style="font-size:16px;"><strong> c) i = Imax Sin (ωt + π/2 ) d) i = v.r</strong></span></p><p><span style="font-size:16px;"><strong>30) In R – C series circuit the equation of voltage is given by</strong></span></p><p><span style="font-size:16px;"><strong> a) v = Vmax Cos ωt b) v = Vmax Sin ωt</strong></span></p><p><span style="font-size:16px;"><strong> c) v = Vmax Cos (ωt+π/2 ) d) v = Vmax Sin (ωt- π/2 ) </strong></span></p><p><span style="font-size:16px;"><strong>31) In R – C series circuit the power factor in terms of average power over a complete cycle is given by - </strong></span></p><p><span style="font-size:16px;"><strong> a) Pavg = Vrms . Irms Cos ф b) Pavg = 1</strong></span></p>
<p><span style="font-size:16px;"><strong> c) Pavg = Cos ф / Vrms . Irms d) Pavg = 0</strong></span></p><p><span style="font-size:16px;"><strong>32) In R – C series circuit the power factor in terms of average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) Cos ф = 1 b) Cos ф = Vrms . Irms / Pavg </strong></span></p><p><span style="font-size:16px;"><strong> c) Cos ф = Pavg / Vrms . Irms d) Cos ф = infinity </strong></span></p><p><span style="font-size:16px;"><strong>33) In R – C series circuit the rms voltage in terms of power factor is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) V = Pavg . Irms. Cos ф b) V = Cos ф / Pavg. Irms</strong></span></p><p><span style="font-size:16px;"><strong> c) V = Pavg / Irms . Cos ф d) V = Cos ф. Irms / Pavg</strong></span></p><p> </p><p> </p><p> </p><p><span style="font-size:16px;"><strong>37) In R – C series circuit current in terms of the average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) I = Vrms / Pavg.Cos ф b) I = Pavg / Vrms . Cos ф </strong></span></p><p><span style="font-size:16px;"><strong> c) I = Pavg.Vrms.Cos ф d) I = 0</strong></span></p><p><span style="font-size:16px;"><strong>38) In R – C series circuit the average power is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) Pavg = 0 b) Pavg = infinity c) P = V.I.Cos ф d) V.I / CoS ф</strong></span></p><p><span style="font-size:16px;"><strong>39) In R – C series circuit the power factor in terms of impedance is given by – </strong></span></p><p><span style="font-size:16px;"><strong> a) Cos ф = Z / R b) Cos ф = R / Z</strong></span></p><p><span style="font-size:16px;"><strong> c) Cos ф = R.Z d) Cos ф = 1 / R.Z</strong></span></p><p><span style="font-size:16px;"><strong>40) In R – C series circuit the current in terms of impedance is given by </strong></span></p><p><span style="font-size:16px;"><strong> a) I = X.L.L / V b) I = V / <u>XC</u></strong></span></p><p><span style="font-size:16px;"><strong> c) I = V.<u>XC</u> d) I = V / </strong></span></p><p><span style="font-size:16px;"><strong>41) In R – L series circuit the equation of current in terms of impedance is </strong></span></p><p><span style="font-size:16px;"><strong> a) I = V / <u>XL.XL</u> b) I = / V</strong></span></p><p><span style="font-size:16px;"><strong> c) I = V /
d) I = XC / V</strong></span></p><p><span style="font-size:16px;"><strong>42) In R – L series circuit the power factor in terms of impedance is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) infinity b) 0 c) R / Z d) Z / R</strong></span></p><p> </p><p><span style="font-size:16px;"><strong>43) In R-L-C series circuit the power factor is given by -</strong></span></p><p><span style="font-size:16px;"><strong> a) Cos ф = R.Z b) Cos ф = 1 /Z.R</strong></span></p><p><span style="font-size:16px;"><strong> c) Cos ф = Z / R d) Cos ф = R /Z</strong></span></p><p><span style="font-size:16px;"><strong>44) In R-L-C series circuit, current and voltage are –</strong></span></p><p><span style="font-size:16px;"><strong> a) In parallel with each other</strong></span></p><p><span style="font-size:16px;"><strong> b) Mutually perpendicular with each other</strong></span></p><p><span style="font-size:16px;"><strong> c) Out of phase</strong></span></p><p><span style="font-size:16px;"><strong> d) In phase</strong></span></p><p><span style="font-size:16px;"><strong>45) Power factor is defined as </strong></span></p><p><span style="font-size:16px;"><strong> a) Cosine of the angle between Impedance – Z & Current – I</strong></span></p><p><span style="font-size:16px;"><strong> b) Sine of the angle between Impedance – Z & Voltage – V </strong></span></p><p><span style="font-size:16px;"><strong> c)Sine of the angle between Reactive inductance & Reactive capacitance</strong></span></p><p><span style="font-size:16px;"><strong> d) Cosine of the angle between Voltage & Current</strong></span></p><p><span style="font-size:16px;"><strong>46) Power factor is nothing but –</strong></span></p><p><span style="font-size:16px;"><strong> a) Cosine of the angle of lead or lag</strong></span></p><p><span style="font-size:16px;"><strong> b) The numerical value of cosine of the phase angle between voltage and current</strong></span></p><p><span style="font-size:16px;"><strong> c) Both a) & b)</strong></span></p><p><span style="font-size:16px;"><strong> d) None of the above</strong></span></p><p><span style="font-size:16px;"><strong>47) In R – C series & parallel circuit the power factor is</strong></span></p><p><span style="font-size:16px;"><strong> a) Independent of each other b) Unity</strong></span></p><p><span style="font-size:16px;"><strong> c) Zero d) Infinity </strong></span></p><p><span style="font-size:16px;"><strong>48) In R – C series circuit the effective impedance is –</strong></span></p>
<p><span style="font-size:16px;"><strong> a) Z =</strong><strong><img height="35" src="file:///C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp\msohtmlclip1\01\clip_image002.jpg" width="137"></strong><strong> b) Z = R</strong></span></p><p><span style="font-size:16px;"><strong> c) Z = </strong><strong><img height="35" src="file:///C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp\msohtmlclip1\01\clip_image004.jpg" width="83"></strong> <strong>d) Z = V / I</strong></span></p><p><span style="font-size:16px;"><strong>49) In R – C parallel circuit the effective impedance is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Z =</strong><strong><img height="35" src="file:///C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp\msohtmlclip1\01\clip_image006.jpg" width="85"></strong><strong> b) Z =1/</strong><strong><img height="35" src="file:///C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp\msohtmlclip1\01\clip_image006.jpg" width="85"></strong></span></p><p><span style="font-size:16px;"><strong> c) Z = L / CR d) Z = CR / L</strong></span></p><p><span style="font-size:16px;"><strong>50) In R – C series circuit</strong></span></p><p><span style="font-size:16px;"><strong> a) Current magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong> b) Impedance magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong> c) Resonance takes place</strong></span></p><p><span style="font-size:16px;"><strong> d) Voltage magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong>51) In R – C paralle circuit</strong></span></p><p><span style="font-size:16px;"><strong> a) Current magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong> b) Impedance magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong> c) Resonance takes place</strong></span></p><p><span style="font-size:16px;"><strong> d) Voltage magnification takes place</strong></span></p><p><span style="font-size:16px;"><strong>52) In below figure, power factor is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) KVA / KW b) KVAR / KW</strong></span></p><p><span style="font-size:16px;"><strong> c) KW / KVAR d) KVAR / KVA</strong></span></p><p><span style="font-size:16px;"><strong>53) In series circuit, net reactance is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) Infinity b) Log 1 c) 1 / Z d) R</strong></span></p><p><span style="font-size:16px;"><strong>54) In pure resistive circuit, instantaneous current is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) i = Imax Sin wt b) i = Imax Cos wt</strong></span></p>
<p><span style="font-size:16px;"><strong> c) i = Imax Cos ( wt +π/2) d) i = V / R</strong></span></p><p><span style="font-size:16px;"><strong>55) In pure inductive circuit, instantaneous current is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) i = Imax Sin (wt - π/2) b) i = Imax Sin (wt + π/2)</strong></span></p><p><span style="font-size:16px;"><strong> c) i = Imax Cos (wt - π/2) d) i = V / R</strong></span></p><p><span style="font-size:16px;"><strong>56) In pure inductive circuit, instantaneous current is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) i = Imax Sin wt b) i = Imax Sin (wt + π/2)</strong></span></p><p><span style="font-size:16px;"><strong> c) i = Imax Cos (wt - π/2) d) i = V / R</strong></span></p><p><span style="font-size:16px;"><strong>57) The average power required to complete half cycle of R-L series circuit is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Zero b) V.I.Cos ф</strong></span></p><p><span style="font-size:16px;"><strong> c) Vmax.Imax.Cosф / 2 d) Both b) & c)</strong></span></p><p> </p><p> </p><p><span style="font-size:16px;"><strong>58) Reactive power is given by –</strong></span></p><p><span style="font-size:16px;"><strong> a) P = V.I.Cosф b) P = V.I.Sinф </strong></span></p><p><span style="font-size:16px;"><strong> c) P = V.I d) P = V.I.tanф</strong></span></p><p><span style="font-size:16px;"><strong>59) Power factor is the ratio of –</strong></span></p><p><span style="font-size:16px;"><strong> a) Real power to Apparent power</strong></span></p><p><span style="font-size:16px;"><strong> b) Apparent power to Real power</strong></span></p><p><span style="font-size:16px;"><strong> c) Real power to Reactive power</strong></span></p><p><span style="font-size:16px;"><strong> d) Reactive power to Apparent power</strong></span></p><p><span style="font-size:16px;"><strong>60) The power in purely resistive circuit is equal to product of –</strong></span></p><p><span style="font-size:16px;"><strong> a) Voltage and Current </strong></span></p><p><span style="font-size:16px;"><strong> b) RMS voltage and current</strong></span></p><p><span style="font-size:16px;"><strong> c) Voltage and RMS current</strong></span></p>
<p><span style="font-size:16px;"><strong> d) RMS Voltage and RMS current </strong></span></p><p><span style="font-size:16px;"><strong>61) In purely resistive circuit the power consumed is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) Infinity c) never 0 d) All of these</strong></span></p><p><span style="font-size:16px;"><strong>62) In purely inductive circuit the power consumed is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) Infinity c) never 0 d) All of these</strong></span></p><p><span style="font-size:16px;"><strong>63) In purely capacitive circuit the power consumed is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) Infinity c) never 0 d) All of these</strong></span></p><p><span style="font-size:16px;"><strong>64) In R-C parallel circuit, the impedance is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Z = </strong><strong>(R</strong><strong>^2+Xl^2)^1/2 Z = R / Cosф</strong></span></p><p><span style="font-size:16px;"><strong> c) Z = Infinity d) Z = Maximum</strong></span></p><p><span style="font-size:16px;"><strong>65) The value of an alternating quantity at a particular instant is a –</strong></span></p><p><span style="font-size:16px;"><strong> a) Constant value b) Instantaneous value</strong></span></p><p><span style="font-size:16px;"><strong> c) Average value d) None of the above</strong></span></p><p><span style="font-size:16px;"><strong>66) If the Vrms value is 100 then the Vmax is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 100*1.414 b) 100 / 1.414</strong></span></p><p><span style="font-size:16px;"><strong> c) 0.637 d) 0.707</strong></span></p><p><span style="font-size:16px;"><strong>67) The ratio of maximum value of an alternating quantity to its RMS value is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Peak factor b) Form factor</strong></span></p><p><span style="font-size:16px;"><strong> c) Average power d) Power factor</strong></span></p><p><span style="font-size:16px;"><strong>68) The ratio of RMS value to Average value is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Peak factor b) Form factor</strong></span></p><p><span style="font-size:16px;"><strong> c) Average power d) Power factor </strong></span></p>
<p><span style="font-size:16px;"><strong>69) Peak factor for sinusoidal (i.e. alternating) current or voltage is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 1.11 b) 1.414 c) 0.707 d) 0.637 </strong></span></p><p><span style="font-size:16px;"><strong>70) Form factor for sinusoidal (i.e. alternating) current or voltage is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 1.11 b) 1.414 c) 0.707 d) 0.637</strong></span></p><p><span style="font-size:16px;"><strong>71) The diagram in which different sinusoidal (i.e. alternating) quantities of the same frequency are represented is a –</strong></span></p><p><span style="font-size:16px;"><strong> a) Phasor diagram</strong></span></p><p><span style="font-size:16px;"><strong> b) Power triangle</strong></span></p><p><span style="font-size:16px;"><strong> c) Wave form</strong></span></p><p><span style="font-size:16px;"><strong> d) All of the above</strong></span></p><p><span style="font-size:16px;"><strong>72) The value of average current over a complete cycle is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) 1.11 c) 0.637 Imax d) 1.414</strong></span></p><p><span style="font-size:16px;"><strong>73) The power factor for pure resistive circuit is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 b) 1 c) - 1 d) 1.414</strong></span></p><p><span style="font-size:16px;"><strong>74) The power factor for pure capacitive circuit is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 ( leading ) b) 1 c) - 1 d) 0 ( lagging )</strong></span></p><p><span style="font-size:16px;"><strong>75) The power factor for pure inductive circuit is –</strong></span></p><p><span style="font-size:16px;"><strong> a) 0 ( leading ) b) 1 c) - 1 d) 0 ( lagging )</strong></span></p><p><span style="font-size:16px;"><strong>76) The power triangle which shows correct relation between G, B, Y –</strong></span></p><p><span style="font-size:16px;"><strong> a) b) c) d)</strong></span></p><p><span style="font-size:16px;"><strong>77) Above figure is called as -</strong></span></p><ol style="list-style-type:lower-alpha;">
<li><span style="font-size:16px;"><strong>Voltage triangle b) Power triangle</strong></span></li></ol><p style="margin-left:17.4pt;"><span style="font-size:16px;"><strong>c) Impedance triangle d) Admittance triangle</strong></span></p>
<p><span style="font-size:16px;"><strong>78) The triangle which shows relation between active & reactive power is –</strong></span></p><ol style="list-style-type:lower-alpha;">
<li><span style="font-size:16px;"><strong>Voltage triangle b) Power triangle</strong></span></li></ol><p style="margin-left:17.4pt;"><span style="font-size:16px;"><strong>c) Impedance triangle d) Admittance triangle</strong></span></p><p><span style="font-size:16px;"><strong>79) The product of RMS value of voltage and current is –</strong></span></p><p><span style="font-size:16px;"><strong> a) Apparent power b) Active power</strong></span></p><p><span style="font-size:16px;"><strong> c) Reactive power d) None</strong></span></p><p><span style="font-size:16px;"><strong>80) The product of voltage and current perpendicular to the voltage –</strong></span></p><p><span style="font-size:16px;"><strong> a) Active power b)</strong> <strong>Reactive power</strong></span></p><p><span style="font-size:16px;"><strong> c) None d) Apparent power </strong></span></p><p> </p><p style="margin-left:17.4pt;"> </p>
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