Power electronics and drives miscellaneous


Power electronics and drives miscellaneous

Power Electronics and Drives

  1. A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.

    The proper configuration for realizing switches S1 to S6 is









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    NA

    Correct Option: C

    NA


  1. A lossy capacitor Cx, rated operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor Cp is parallel with a resistor Rp. The value of Cp is found to be 0.102 μF and the value of Rp = 1.25 MΩ. Then the power loss and tan δ of the lossy capacitor operating at the rated voltage, respectively, are









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    Power - loss =
    V2
    (5 KV)2
    = 20 W
    RP1.25 MΩ

    tan δ =
    1
    ωCP RP

    =
    1
    = 0.025
    (2π 50)(0.102 × 10-6)(1.25 × 106)


    Correct Option: C

    Power - loss =
    V2
    (5 KV)2
    = 20 W
    RP1.25 MΩ

    tan δ =
    1
    ωCP RP

    =
    1
    = 0.025
    (2π 50)(0.102 × 10-6)(1.25 × 106)




  1. Circuit turn-off time of an SCR is defined as the time
    (a)
    (b)
    (c)
    (d)









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    NA

    Correct Option: C

    NA


  1. A voltage commutated chopper circuit, operated at 500 Hz, is shown below.

    If the maximum value of load current is 10 A, then the maximum current through the main (M) and auxiliary (A) thyristors will be









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    iAmax = Io = 10 A

    Correct Option: A


    iAmax = Io = 10 A



  1. In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that ∆ic is 1.6 A peak-to-peak and I0 is 5 A dc, the peak current in S is










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    The size in inductor current during ON period.

    ∆IL =
    Vs - V0
    TON , TON =
    L∆L
    LVs - V0

    and during OFF - period
    -∆IL =
    - V0
    TOFF , TOFF =
    L∆L
    TV0

    Switching frequency,
    f =
    1
    =
    1
    .
    V0(Vs - V0)
    TON + TOFF L∆ILVs

    ⇒ ∆IL =
    VsK(1 - K)
    .K =
    V0
    fLVs

    The maximum source current is given by,
    Imax = Io +
    ∆L
    = 5 +
    1.6
    = 5.8 A
    22

    Correct Option: C



    The size in inductor current during ON period.

    ∆IL =
    Vs - V0
    TON , TON =
    L∆L
    LVs - V0

    and during OFF - period
    -∆IL =
    - V0
    TOFF , TOFF =
    L∆L
    TV0

    Switching frequency,
    f =
    1
    =
    1
    .
    V0(Vs - V0)
    TON + TOFF L∆ILVs

    ⇒ ∆IL =
    VsK(1 - K)
    .K =
    V0
    fLVs

    The maximum source current is given by,
    Imax = Io +
    ∆L
    = 5 +
    1.6
    = 5.8 A
    22