Power electronics and drives miscellaneous

Power electronics and drives miscellaneous

Easy Questions
Moderate Questions

Power Electronics and Drives

Power electronics and drives miscellaneous
  1. The average power delivered to an impedance (4 – j3) Ω by a current 5 cos (100 πt + 100) A is








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    Z = 4 – j3 = RL – jXC; RL = 4;
    I = 5 cos (100πt + 100) = Im cos2 (ωt + α)

    P =
    1
    		Im2 .RL =
    1
    		 × 52 × 4 = 50 W
    22

    Correct Option: B

    Z = 4 – j3 = RL – jXC; RL = 4;
    I = 5 cos (100πt + 100) = Im cos2 (ωt + α)

    P =
    1
    		Im2 .RL =
    1
    		 × 52 × 4 = 50 W
    22

  1. Thyristor T in the figure below is initially off and is triggered with a single pulse of width 10 µs.
    It is given that L =
    100
    		 H and C =
    100
    		 F.
    ππ

    Assuming latching and holding currents of the thyristor are both zero and the initial charge on C is zero, T conducts for









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    When thyristor ON


    Is
    		 = L[bI / s] +
    1
    I(s)
    scs

    Correct Option: C

    When thyristor ON


    Is
    		 = L[bI / s] +
    1
    I(s)
    scs

  1. A voltage 1000 sin ωt Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts is









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    ‘D’ 0 for all +

    Note :
    ⇒ All diode conducts only done negative half.
    ⇒ XW is at symmetrical point so voltage across XW is zero for all time.

    Correct Option: D

    ‘D’ 0 for all +

    Note :
    ⇒ All diode conducts only done negative half.
    ⇒ XW is at symmetrical point so voltage across XW is zero for all time.

  1. In the circuit shown below, the knee current of the ideal Zener diode is 10 mA. To maintain 5 V across RL, the minimum value of RL in Ω and the minimum power rating of the Zener diode in mW respectively are









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    Is = Iz + IL
    Is – Iz = IL

    Two extreme condition :
    If Iz (min),then IL (max)
    If Iz (max) then IL (min) = 0

    Iz (max) = Is =
    10 - 5
    		 = 50 mA
    10

    Iz (min) = Is – IL (max)
    IL (max) = Is – Iz (min) = Is – Iz = (50 – 10) = 40mA
    RL (min) =
    V
    		 =
    5
    		K = 125 Ω
    IL (max)40

    Pz = Vz × Iz (max) = 5 × 50 mA = 250 mw

    Correct Option: B

    Is = Iz + IL
    Is – Iz = IL

    Two extreme condition :
    If Iz (min),then IL (max)
    If Iz (max) then IL (min) = 0

    Iz (max) = Is =
    10 - 5
    		 = 50 mA
    10

    Iz (min) = Is – IL (max)
    IL (max) = Is – Iz (min) = Is – Iz = (50 – 10) = 40mA
    RL (min) =
    V
    		 =
    5
    		K = 125 Ω
    IL (max)40

    Pz = Vz × Iz (max) = 5 × 50 mA = 250 mw

  1. In the feedback network shown below, if the feedback factor k is increased, then the









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    Input Independence of a voltage-voltage feedback circuit = Zi (1 + A0 k)
    Zi = initial input impedance (without feed output
    Impedance of a voltage-voltage feedback circuit = Z0 / (1 + A0 k)
    Z0 = initial output impedance (without feedback)
    Hence, As K is increased, the input impedance will increase and output impedance will decrease.

    Correct Option: A

    Input Independence of a voltage-voltage feedback circuit = Zi (1 + A0 k)
    Zi = initial input impedance (without feed output
    Impedance of a voltage-voltage feedback circuit = Z0 / (1 + A0 k)
    Z0 = initial output impedance (without feedback)
    Hence, As K is increased, the input impedance will increase and output impedance will decrease.