Network Elements and the Concept of Circuit

Network Elements and the Concept of Circuit

Easy Questions

Network Theory

Network Elements and the Concept of Circuit
Network theory miscellaneous
  1. Consider the circuit shown below—

    The voltage V1 and V2 are related as—










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    V1 = 16 V
    on applying KVL in loop

    V1 – 6 ia – 8 – V2 = 0
    V1 = V2 + 8 + 6 ia

    Correct Option: B

    V1 = 16 V
    on applying KVL in loop

    V1 – 6 ia – 8 – V2 = 0
    V1 = V2 + 8 + 6 ia

  1. If a resistor of 10 Ω is placed in parallel with voltage source in the circuit shown below, the current i will be—










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    If we place any circuit element (like resistor, capacitor or conductor) in parallel with the voltage source. It means this is extra redundant element present in the circuit, therefore, the current will remain constant in the circuit.

    Correct Option: C

    If we place any circuit element (like resistor, capacitor or conductor) in parallel with the voltage source. It means this is extra redundant element present in the circuit, therefore, the current will remain constant in the circuit.

  1. Two elements are connected in series as shown below element-1 supplies 36 W of power. Element-2—










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    Current through element 1 is =
    36
    		 = 6 A
    6

    This current will pass through element-2. Since the polarities of element 1 and 2 are opposite to each other it means element-1 supply power so direction of current will be form negative to positive, i.e. element-2 will absorb power,
    P = V, i = 4 × 6 = 24 W

    Correct Option: B

    Current through element 1 is =
    36
    		 = 6 A
    6

    This current will pass through element-2. Since the polarities of element 1 and 2 are opposite to each other it means element-1 supply power so direction of current will be form negative to positive, i.e. element-2 will absorb power,
    P = V, i = 4 × 6 = 24 W

  1. Let i(t) = 3 te– 100 t A and V (t) = 0.6 (0.01 – t) e– 100 t V for the network of fig. shown below. The power being absorbed by the network element at t = 5 ms is—










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    Power absorbed
    P = Vi = 0.6 (0.01 – t) e–100t .3.t.e– 100t
    Power at t = 5 ms = 5 × 10– 3
    P = 0.6 (0.01 – 5 × 10– 3) e– 100 × 5 × 10– 3 3.5 × 10– 3 e– 100 × 5 × 10– 3
    = 16.6 µW.
    Hence alternative (C) is the correct choice.

    Correct Option: C

    Power absorbed
    P = Vi = 0.6 (0.01 – t) e–100t .3.t.e– 100t
    Power at t = 5 ms = 5 × 10– 3
    P = 0.6 (0.01 – 5 × 10– 3) e– 100 × 5 × 10– 3 3.5 × 10– 3 e– 100 × 5 × 10– 3
    = 16.6 µW.
    Hence alternative (C) is the correct choice.

  1. Calculate the voltage between terminal ab for figure shown below—










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    Apply KCL at node C

    2 + i1 = 8
    or i1 = 8 – 2 = 6A
    Vba = 0.2i1 × 2 = .4 × 6
    = 2.4 × V
    or Vab = – 2.4 V

    Correct Option: D

    Apply KCL at node C

    2 + i1 = 8
    or i1 = 8 – 2 = 6A
    Vba = 0.2i1 × 2 = .4 × 6
    = 2.4 × V
    or Vab = – 2.4 V