Electric circuits miscellaneous


Electric circuits miscellaneous

  1. In the circuit shown in the given figure, the values of I(0+) and I(∞), will be, respectively










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    I(s) =
    3
    (s + 1)1 +
    1
    2s

    =
    3
    (s + 0.5)(s + 1)

    =
    -3
    +
    6
    (s + 0.5)(s + 1)

    ⇒ i (t) = L– 1(I(s)) = 6e– t – 3e– 0.5t
    At t = 0, i (0+) = 3A,
    and at t = ∞, i (∞) = 0

    Correct Option: C

    I(s) =
    3
    (s + 1)1 +
    1
    2s

    =
    3
    (s + 0.5)(s + 1)

    =
    -3
    +
    6
    (s + 0.5)(s + 1)

    ⇒ i (t) = L– 1(I(s)) = 6e– t – 3e– 0.5t
    At t = 0, i (0+) = 3A,
    and at t = ∞, i (∞) = 0


  1. Consider the graph and tree (dotted) of the given figure. The fundamental loops include the set of lines










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    NA

    Correct Option: D

    NA



  1. A connected network of N > 2 nodes has at most one branch directly connecting any pair of nodes. The graph of the network









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    NA

    Correct Option: D

    NA


  1. The number of independent loops for a network with n nodes and b branch is









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    NA

    Correct Option: C

    NA



  1. In the circuit shown in the figure, X is an element which always absorbs power during a particular operation it sets up a current of 1 amp in the direction shown and absorbs a power Px. It is possible that X can absorb the same power Px for another current I. the value of this current is











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    As element absorbs power, let it be Rx.

    i = 1 =
    6
    Rx + 1

    ⇒ Rx = 5 Ω
    P =
    6
    2 × 5 = 5 W
    6

    For 5A, P = 52 × 5 = 125 W
    For 3 + √14, P = 227 W
    For 3 – √14, P = 2.75 W
    Hence none of the above is correct.

    Correct Option: D

    As element absorbs power, let it be Rx.

    i = 1 =
    6
    Rx + 1

    ⇒ Rx = 5 Ω
    P =
    6
    2 × 5 = 5 W
    6

    For 5A, P = 52 × 5 = 125 W
    For 3 + √14, P = 227 W
    For 3 – √14, P = 2.75 W
    Hence none of the above is correct.