1. For the same peak value of voltage, which waveform will have the least rms value? Triangular wave.
  2. For the same peak value of voltage, which waveform will have the highest rms value? Square wave.
  3. Which wave has the highest value of form factor? Half wave rectified sine wave.
  4. Which wave has the least value of form factor? Square wave.
  5. Which of the following waves has form factor of 1.0? Square wave.
  6. If one cycle of ac waveform occurs every milli-second, the frequency will be 1000Hz.
  7. Which of the following frequencies has the longest period? 10kHz.
  8. RMS value and the mean value is the same in case of Square wave.
  9. If emf in a circuit is given by ๐‘’=100sin628๐‘ก then maximum value of voltage and frequency are 100V 100Hz.
  1. For a wave form more peaky than a sine wave, the form factor will be more than 1.11.
  2. For a triangular wave, the form factor is 1.15.
  3. For a square wave, the form factor is 1.0.
  4. The equation of alternating current is ๐‘–=42.42sin628๐‘ก. the average value of current is 27A.
  5. For the voltage waveform shown in figure 5 the rms value of voltage will be โˆš๐Ÿ“๐Ÿ‘ ๐’š ๐’Ž๐’‚๐’™.
  6. The average value of the above voltage is ๐Ÿ๐Ÿ‘ ๐’š ๐’Ž๐’‚๐’™.
  7. The peak factor for the above voltage is ๐Ÿ‘โˆš๐Ÿ“ .
  1. The form factor for the above voltage isโˆš๐Ÿ“๐Ÿ.
    Questions 38 to 40 refer to the following data:
    A current has following steady values in amperes for equal intervals of time changing
    simultaneously from one to the next 0, 10, 20, 30, 20, 10, 0, -10, -20, -30, -20, -10, -0.
  2. The rms value of the current will be 17.8 A.
  3. The form factor will be 1.187.
  4. The peak value of a wave isMaximum valuer.m.s value
  1. In the above case the value of which parameter for full sine wave will be more?
    Average value.
  2. An alternating voltage ๐‘’ = 200 ๐‘ ๐‘–๐‘› 314๐‘ก is applied to a device which offers an ohmic
    resistance of 200ฮฉ to the flow of current in one direction while entirely preventing the
    flow in the opposite direction. The value of current will be 5 A. ??
  3. In the above case, the average value of the current will be 3.18 A.
  4. When the sole purpose of the alternating current is to produce heat, the selection of
    conductor is based on rms value of current.
  5. The r.m.s value of a.c. is related to peak value of a.c. by the equation ๐‘ฐ๐’ = โˆš๐Ÿ I r.m.s.
  6. The form factor of dc supply voltage is always Unity.
  7. What is the angle in the NE quadrant corresponding to a rms value of the quantity, ๐‘ฆ =๐‘ฆ๐‘š๐‘Ž๐‘ฅ ๐‘ ๐‘–๐‘›๐œ”๐‘ก?
  1. What is the angle in NE quadrant corresponding to the average value of the quantity,
    ๐‘ฆ = ๐‘ฆ๐‘š๐‘Ž๐‘ฅ๐‘ ๐‘–๐‘›๐œ”๐‘ก? 0.69 rad.
  2. Two currents represented by ๐‘–1 = 50 sin ๐œ”๐‘ก , ๐‘–12 = 100 sin(๐œ”๐‘ก + 45 ยฐ) are fed into
    a common conductor. The rms value of the current will be between 50 A and 100 A.

Questions 51 to 54 refer to the fig7.

  1. A 1000 W heater is rated to operate at a direct current of 10A. If the heater is supplied
    alternating current, for producing the same quantity of heat, the value of current should
    be ๐‘ฐ๐’“๐’Ž๐’” = ๐Ÿ๐ŸŽ๐‘จ.
  2. The rms value of the voltage shown in Fig. 7 is๐’š๐’Ž๐’‚๐’™โˆš๐Ÿ‘
  1. The peak factor for the above voltage is 1.41. ??
  2. The average value for above voltage is๐’š๐’Ž๐’‚๐’™๐Ÿ.
  3. The form factor for the above voltage is๐Ÿโˆš๐Ÿ‘.
  4. With reference to Fig.8 which statement is correct? Current ๐’Š๐Ÿ is leading current ๐’Š๐Ÿ.

Questions 57 to 60 refer to the following data:
An alternating quantity increases uniformly from 0 to 0ยฐ at ๐น๐‘š at ๐›ผ; remains constant
from ๐›ผ to (๐œ‹ โˆ’ ๐›ผ) and decreases uniformly from ๐น๐‘š at (๐œ‹ โˆ’ ๐›ผ) to 0 at ๐œ‹.

  1. The rms value of the wave for one half cycle wil be๐‘ญ๐’Ž(๐…โˆ’๐œถ)๐….
  2. The average value of the wave for one half cycle will be๐Ÿ๐‘ญ๐’Ž๐….
  3. In above case when ๐Ÿ‘๐‘ญ๐’Ž๐Ÿ“.
  4. In the above case when ๐‘ญ๐’Ž๐Ÿ

Questions 61 to 63 refer to the fig9.

  1. For the the waveform shown in figure, the equation for voltage may be written as ๐’— =๐‘ฝ๐’Ž ๐ฌ๐ข๐ง ๐Ž๐’•.
  1. In the figure shown the equation for the current may be written as ๐’Š = ๐‘ฐ๐’Ž ๐ฌ๐ข๐ง(๐Ž๐’• โˆ’ โˆ…).
  2. For the waveform the vectorical reperesentation is correctly shown in which of thefollowing ig.10Figure D
  3. The period of voltage 12 sin(800๐œ‹๐‘ก + 0.125๐œ‹)V is 1.33 milli-second.
  4. The period of voltage 3 cos 2500๐‘ก + 4 cos(2500๐‘ก +๐œ‹2) is 2.5 milli-second.
  5. The period of voltage 2 cos 4500๐œ‹๐‘ก + 7 sin 7500 ๐œ‹๐‘ก is 2.51 milli-second.
  6. The average value of the square of the wave from 3 cos 2๐‘ก + 4 cos(2๐‘ก +๐œ‹4) is nearly20.5.
  7. The average value of the square of the waveform (4 cos 2๐‘ก + 5 cos 3๐‘ก) is nearly 21.
  8. A load draws 10W power from a 10V source drawing 10A current. The power factor is0.1.
  9. The vector can be represented in rectangular form as ๐’‘ + ๐’‹๐’’.
  10. The vector ๐›ผ can be represented in polar as โˆš๐’‘๐Ÿ + ๐’’๐Ÿ < ๐œƒ.
  1. The vector ๐›ผ can be represented in exponential form as โˆš๐’‘๐Ÿ + ๐’’๐Ÿ๐’†๐’† .
  2. The vector ๐›ผ can be represented in trigonometrical form asโˆš๐’‘๐Ÿ + ๐’’๐Ÿ)(๐’„๐’๐’”๐œฝ ๐’‹ ๐’”๐’Š๐’๐œฝ).
  3. The rms value of the waveform shown in Fig.12 is 66.7.
  4. The multiplication of the vector (๐‘ + ๐‘—๐‘ž) and (๐‘Ÿ + ๐‘—๐‘ ) will be (๐’‘๐’“ โˆ’ ๐’’๐’”) +๐’‹(๐’’๐’“ + ๐’‘๐’”).
  5. ๐‘Ž < โˆ… . ๐‘ < ๐œƒ = ๐’‚๐’ƒ/๐‹ + ๐œฝ.
  6. Two sinusoidal quantities are said to be in phase quadrature, when their phasedifference is ๐Ÿ—๐ŸŽยฐ.
  7. Which of the following relation is incorrect ?๐‘ช๐’๐’๐’…๐’–๐’„๐’•๐’‚๐’๐’„๐’†๐‘บ๐’–๐’”๐’„๐’†๐’‘๐’•๐’‚๐’๐’„๐’†.
  8. The capacitor for power factor correction are rated in terms of KVAR.
  9. Poor power factor results in all of the following EXCEPT: reduction in power loss.
  10. Power factor of an inductive circuit can be improved by connecting a capacitor to it inparallel.
  11. For the same load, if the power factor of load is reduced, it will draw more current.
  12. For the saw tooth current waveform, the rms value is 5.77A.
  13. The effects due to electric current are I. Magnetic effect, II. Heating effect, III.
    Luminous effect. Appliance working on which effect can be used on ac as well as dc
    supply? II. and III. Only.
  14. The rms value of the waveform shown in Fig.14 will be 57.7.
  15. The effects due to electric current are I. Thermal effect, II. Luminous effect, III.
    Chemical effect, IIII. Magnetic effect. Which two effects are significant when current
    flows through transmission lines I and IV only.
  16. A current in ac circuit measures 4 A. Then , the maximum instantaneous magnitude of
    this current will be ๐Ÿ’ ร— โˆš๐Ÿ A.
  17. The equation of an a.c voltage is ๐‘‰ = 200 sin 50๐œ‹๐‘ก. Then, the r.m.s value of voltage is๐Ÿ๐ŸŽ๐ŸŽโˆš๐Ÿ V. ??
  18. When a.c. current flows through a resistance, then current and e.m.f. are in phase.
  19. The power factor of incandescent bulb is unity.
  20. A incandescent bulb can work on both ac as well as dc.
  21. It is know that a given capacitor will fail if the terminal voltage exceeds 180 V. If the
    circuit is operating on sinusoidal steady voltage, the maximum rms voltage that may
    applied to the capacitor will be 127 V.
  22. For an alternating voltage form factor is๐‘๐Œ๐’ ๐•๐š๐ฅ๐ฎ๐ž๐Œ๐ž๐š๐ง ๐ฏ๐š๐ฅ๐ฎ๐ž.
  23. In and a.c. circuit, the current any of the above, depending upon the elements (L,Cor R) of the Circuit.
  24. Power factor of the magnetizing component of a transformer is zero.
  25. Which of the following statement is not necessarily valid for ac currents: Alternating
    currents is suitable of charging batteries.
    Questions 97 and 98 refer to the data given below:

    A current is given by ๐‘– = 45.24 sin 377๐‘ก.
  26. The maximum value of the current is 45.24 A.
  27. The frequency is๐Ÿ‘๐Ÿ•๐Ÿ•๐Ÿ๐…Hz.
  28. A circuit has impedance of (3 + ๐‘—4). If a voltage (100 + ๐‘—50) is applied, the power inthe circuit will be 500 W.
  29. Which will draw least current 40W lamp.
    Questions 101 and 102 refer to the given data below:

Since wave A has a frequency of 100 Hz and an rms applied of 100 mA. Since wave B has the same frequency, has an rms amplitude of 200 mAand lags sine wave A by 60ยฐ.

  1. The instantaneous value of current ๐‘–๐ด at ๐‘ก=0.35 ๐‘Ÿ๐‘š๐‘  will be 20.5 mA.
  2. The instantaneous value of current ๐‘–๐ต at ๐‘ก=0.35 ๐‘Ÿ๐‘š๐‘  will be –20.6 mA.
  3. What value of dc voltage will produce the same average power as 150V peak sine wave? 106 V.
  4. What current will an ammeter show when connected in series with a 32 ohm load that is issipating APeak power of 288 W? 2.12 A.
  5. A 160 W soldering iron is operated from the 60 Hz power line. The resistance of the soldering iron is 75.6 ohms.
  6. A certain 1-Khz sine wave reaches -70 V at ๐‘ก=0.6 ๐‘š๐‘ , The peak value of this sine wave will be 118 V.
  7. A certain sine wave is expressed as ๐‘’=sin (4000๐‘ก). The frequency of this sine wave will be 637 Hz.
  8. A sine wave of voltage produces a peak current of 12 mA in a 3 Kฮฉ resistor. The average power dissipated in the resistor will be 216 W.
  9. In a.c. circuits, the a.c. meters measures r.m.s values.
  10. Radio frequency choke is air cored to keep inductive reactance low.
  11. Hot wire ammeters are used for measuring only a.c.
  12. When ๐‘ฃ1=50sin๐œƒ,๐‘ฃ2=30sin(๐œƒ+25ยฐ) and v2=25sin(๐œƒโˆ’90ยฐ),the resultant of (๐‘ฃ1 +๐‘ฃ2โˆ’๐‘ฃ3) is given by 8๐Ÿ”๐ฌ๐ข๐ง(๐œฝ+๐Ÿ๐Ÿ”ยฐ).
  13. 12<30ยฐ in rectangular coordinates can be represented as ๐Ÿ๐ŸŽ.๐Ÿ’+๐’‹๐Ÿ”.
  14. 270/1.7 ฯ€ can be represented in rectangular coordinates as ๐Ÿ๐Ÿ“๐Ÿ—โˆ’๐’‹ ๐Ÿ๐Ÿ๐Ÿ–.
  15. A choke coil is used for controlling current in an a.c.circuit only.
  16. Ohmโ€™s law (๐ธ=๐ผ๐‘…) can be applied to a.c. but after replacing R by Z (impedance).
  17. What frequency is one octave above 110 Hz? 220 Hz.
  18. What frequency is the fourth harmonic of 4 Mhz? 16 Mhz.
    Questions 119 to 122 refer to data given below:

A 60 Hz power-line voltage of 120 V is applied across a resistance of 10 ohms.

  1. The rms current in the circuit is 12 A.
  2. The frequency of the current will be 60 Hz.
  3. The phase angle between the current and the voltage will be ๐ŸŽยฐ.
  1. The dc applied voltage necessary for the same heating affect in the resistance will be
    120 V.
    Questions 123 to 124 refer to data given below:

    An a.c wave form with a frequency of 1.5 kHz has a peak value of 3.3 V.
  2. The instantaneous value of voltage at 0.65 micro-second will be 20.2 mV.
  3. The instantaneous value at 1.2 milli-seconds will be -3.1 V.
    Questions 125 to 128 refer to data given below:
    The circuit shown in figure below has a 60 Hz supply voltage with a maximum value
    of 160 V.
  4. The instantaneous current at ๐œ‹
    โ„4 radians will be 11.3 A.

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