Network Elements and the Concept of Circuit


Network Elements and the Concept of Circuit

  1. In the circuit shown in figure, the switch S is closed at time t = 0. The voltage across the inductance at t = 0+, is—











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    The given circuit
    at t = 0+ capacitor is replaced by short circuit and inductor is replaced by open circuit. The given circuit can be redrawn as

    Now, VL (0+) =
    2 × 10
    = 4V
    2 + 3

    Hence alternative (B) is the correct choice.


    Correct Option: B

    The given circuit
    at t = 0+ capacitor is replaced by short circuit and inductor is replaced by open circuit. The given circuit can be redrawn as

    Now, VL (0+) =
    2 × 10
    = 4V
    2 + 3

    Hence alternative (B) is the correct choice.



  1. The initial current in the circuit shown below when the switch is opened for t > 0—











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    So in steady state when switch closed current 0 Amp.
    ∵ at initially capacitor is uncharged and there is no current flow due to open switch.


    Correct Option: B

    So in steady state when switch closed current 0 Amp.
    ∵ at initially capacitor is uncharged and there is no current flow due to open switch.




  1. The initial current in the circuit shown below when the switch is opened for t > 0 is—











  1. View Hint View Answer Discuss in Forum

    Applying KVL in the loop
    20 – 10I – 2I = 0
    at t < 0 20 = 12I
    or I = 1·67 A


    Correct Option: A

    Applying KVL in the loop
    20 – 10I – 2I = 0
    at t < 0 20 = 12I
    or I = 1·67 A



  1. The circuit shown below is under steady-state condition with the switch closed. The switch is opened at t = 0. What is the time constant of the circuit—











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    As we know that τ =
    L
    =
    2
    Req10

    or τ = 0·2 S


    Correct Option: B

    As we know that τ =
    L
    =
    2
    Req10

    or τ = 0·2 S




  1. For the network shown below, the switch S is closed at t = 0 with the capacitor uncharged. The value of di(t) / dt at t = 0+ will be—











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    When capacitor opened
    at t = 0+

    IC(0) =
    V
    =
    .1000
    R100

    = 0.1 Amp
    at t = o+
    ⇒ 100 = 1000 IC + 1 / C ∫ICdt
    After differentiating, we get
    0 = 1000
    dIC
    +
    1
    lc
    dtC

    Capacitor in short circuited
    -1000
    dIC
    =
    Ic
    =
    .1
    dtC1 × 10 -6

    dI(+)
    =
    -
    .1 × 10 +6
    dt103

    = – 100 Ampere.
    Hence alternative (B) is the correct choice.


    Correct Option: B

    When capacitor opened
    at t = 0+

    IC(0) =
    V
    =
    .1000
    R100

    = 0.1 Amp
    at t = o+
    ⇒ 100 = 1000 IC + 1 / C ∫ICdt
    After differentiating, we get
    0 = 1000
    dIC
    +
    1
    lc
    dtC

    Capacitor in short circuited
    -1000
    dIC
    =
    Ic
    =
    .1
    dtC1 × 10 -6

    dI(+)
    =
    -
    .1 × 10 +6
    dt103

    = – 100 Ampere.
    Hence alternative (B) is the correct choice.