In this section, you will find objective questions related to AC Fundamentals  which helps to analyze the networks consisting of various alternating current and voltage sources, resistances and inductive, capacitive reactances.

  1. The form factor in reference to alternating current wave form represents the ratio of the R.M.S. value to the average value.
  2. If the current and voltage are out of phase by 90Β°, thee power is Zero.
  3. The form factor of a 220 V, 50 Hz A.C. waveform is 1.11.
  4. If E1 = A sin πœ”π‘‘ and E2= A sin (πœ”π‘‘βˆ’πœƒ) then E1 leads E2 by𝜽.
  5. For the waveform shown in figure, average value is 1.5 A.
  6. For the above wave the rms value will be 1.528 A.
  7. Two sinusoidal quantities are said to be phase quadrature, when their phase difference is πŸ—πŸŽΒ°.
  8. The equation for 25 cycles current sine wave having rms value of 30 amperes, will be 42.4 sin50Ο€t.
  9. R.M.S value of a current given by 𝑖=10+5 cos (628𝑑+90Β°) is 10.6A
  10. What is the rms value of rectangular voltage wave with an amplitude of 10 V? 10V.
  11. The mean value of a.c related to peak value of a.c by the equation is π‘°π’Ž= πŸπ‘°π’π….
  12. In above case , the value of direct current which produces in the same conductor the same amount of heat in the same time will be 141 V.
  13. The current in a circuit follows the law 𝑖=100sinπœ”π‘‘, If the frequency is 25 Hz how long will it take for the current to rise to 50 amperes? 𝟏𝟏𝟎𝟎𝟎 sec.
  14. The voltage 𝑣=90cos(πœ”π‘‘βˆ’161.5Β°) may be represented as sine function by πŸ—πŸŽπ¬π’π§(πŽπ’•βˆ’πŸ•πŸ.πŸ“Β°).
  15. The equation of emf is given by =π‘°π’Žβˆš(π‘πŸ+πŸ’π›šπŸπ‹πŸ) .
  16. In above problem the frequency in hertz is πŽπ….
  17. The negative maximum of a cosine waveform occurs at 180Β°.
  18. The RMS value of sinusoidal 200 V peak to peak wave is 𝟐𝟎𝟎√𝟐 v. ???
  19. The positive maximum of a sine wave occurs at πŸ—πŸŽΒ°
  1. In figure 4 the combined impedance of the parallel circuits equals 120 V.
  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 𝑒=100sin628𝑑 then maximum value of voltage and frequency are 100V 100Hz.
  1. If 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 𝑖=42.42sin628𝑑. the average value of current is 27A.
  5. For the voltage waveform shown in figure 5 the rms value of voltage will be βˆšπŸ“πŸ‘ π’š π’Žπ’‚π’™.
  6. The average value of the above voltage is πŸπŸ‘ π’š π’Žπ’‚π’™.
  7. The peak factor for the above voltage is πŸ‘βˆšπŸ“ .
  1. The form factor for the above voltage isβˆšπŸ“π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 𝑒 = 200 𝑠𝑖𝑛 314𝑑 is applied to a device which offers an ohmic
    resistance of 200Ξ© 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 𝑰𝒐 = √𝟐 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, 𝑦 =π‘¦π‘šπ‘Žπ‘₯ π‘ π‘–π‘›πœ”π‘‘? π…πŸ’.
  1. What is the angle in NE quadrant corresponding to the average value of the quantity,
    𝑦 = π‘¦π‘šπ‘Žπ‘₯π‘ π‘–π‘›πœ”π‘‘? 0.69 rad.
  2. Two currents represented by 𝑖1 = 50 sin πœ”π‘‘ , 𝑖12 = 100 sin(πœ”π‘‘ + 45 Β°) are fed into
    a common conductor. The rms value of the current will be between 50 A and 100 A.