Physical electronics devices and ics miscellaneous

Physical electronics devices and ics miscellaneous

Easy Questions

Physical Electronics Devices and ICs

Physical electronics devices and ics miscellaneous
  1. Determine the value of he load resistor—










  1. View Hint View Answer Discuss in Forum

    From the given information, we can make the equivalent circuit as shown below:
    From below(i.e., fig-2) circuit applying KVL 12–0.3–07–2m A RL = 0 or

    RL =
    11V
    		2 mA
    5.5kΩ


    Correct Option: B

    From the given information, we can make the equivalent circuit as shown below:
    From below(i.e., fig-2) circuit applying KVL 12–0.3–07–2m A RL = 0 or

    RL =
    11V
    		2 mA
    5.5kΩ


  1. Determine the peak value of the current through the load resistor—








  1. View Hint View Answer Discuss in Forum

    Considering the given circuit for first half cycle in order to determine the peak value of the current through the load resistor.
    From above figure

    I =
    V
    Req

    Where, Req = 2 || (2 + 2) =
    4
    		kΩ.
    3

    Now, I =
    (10 – 0.7) V
    		=
    9.3 x 3
    		mA
    4 / 3 kΩ4

    = 6.975mA
    and IL(peak) = I ×
    2
    		= 6.975 x
    2
    		= 2.325 mA.
    2 + 46


    Correct Option: A

    Considering the given circuit for first half cycle in order to determine the peak value of the current through the load resistor.
    From above figure

    I =
    V
    Req

    Where, Req = 2 || (2 + 2) =
    4
    		kΩ.
    3

    Now, I =
    (10 – 0.7) V
    		=
    9.3 x 3
    		mA
    4 / 3 kΩ4

    = 6.975mA
    and IL(peak) = I ×
    2
    		= 6.975 x
    2
    		= 2.325 mA.
    2 + 46


  1. Determine the average value of the current through the load resistor—










  1. View Hint View Answer Discuss in Forum

    Upto Irms or I(peak) follow the solution of above question. We know average value of a full wave rectifies is

    Iav =
    Irms
    		=
    Ipeak
    		=
    2.325 mA
    		= 1.67 mA
    221.414

    Correct Option: C

    Upto Irms or I(peak) follow the solution of above question. We know average value of a full wave rectifies is

    Iav =
    Irms
    		=
    Ipeak
    		=
    2.325 mA
    		= 1.67 mA
    221.414

  1. What best describes the circuit?










  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: A

    NA

  1. Determine the peak value of the output waveform—










  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: B

    NA