Power electronics and drives miscellaneous

Power electronics and drives miscellaneous

Easy Questions
Moderate Questions

Power Electronics and Drives

Power electronics and drives miscellaneous
  1. A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure.

    If the load current is sinusoidal and is zero at 0, π, 2π...., the node voltage VAO has the waveform








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    NA

    Correct Option: D

    NA

  1. In the circuit shown below what is the output voltage (Vout) in Volts if a silicon transistor Q and an ideal op-amp are used?









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    I = I, e I = I0e VRE / VT , I = I0e -V0 / VT

    -V0 = VT ln
    I
    		 = VT ln
    I3ms
    I0Is

    V0 = - 0.7
    RL = √Rs² + Xs² = √4² + 3² = 5

    Correct Option: B


    I = I, e I = I0e VRE / VT , I = I0e -V0 / VT

    -V0 = VT ln
    I
    		 = VT ln
    I3ms
    I0Is

    V0 = - 0.7
    RL = √Rs² + Xs² = √4² + 3² = 5

  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.

  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. 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