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. For the networks shown in the figures (a) and (b) to be duals, it is necessary that R′, L′ and C′ are respectively equal to—










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    In order to make the dual network. simply replace voltage by current, current by voltage, inductance by capacitance resistance by conductance and capacitance by inductance.
    From fig. (a)

    V = iR + L
    di
    		+
    1
    		∫ i dt ......(i)
    dtC

    from fig. (b)
    I = V′ G′ + C′
    dV
    		+
    1
    		∫ V´ dt ......(ii)
    dtL

    On comparing equation (i) and (ii), we get
    R = G′
    L′ = C
    C′ = L

    Correct Option: A

    In order to make the dual network. simply replace voltage by current, current by voltage, inductance by capacitance resistance by conductance and capacitance by inductance.
    From fig. (a)

    V = iR + L
    di
    		+
    1
    		∫ i dt ......(i)
    dtC

    from fig. (b)
    I = V′ G′ + C′
    dV
    		+
    1
    		∫ V´ dt ......(ii)
    dtL

    On comparing equation (i) and (ii), we get
    R = G′
    L′ = C
    C′ = L

  1. The time constant of the network shown in the figure is—










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    τ = R eq. C eq Ceq = C + C = 2C

    so, time constant Req =
    2R . 2R
    		= R
    2R + 2R

    τ = R. 2C = 2RC

    Correct Option: A

    τ = R eq. C eq Ceq = C + C = 2C

    so, time constant Req =
    2R . 2R
    		= R
    2R + 2R

    τ = R. 2C = 2RC

  1. The value of current I flowing in 5 Ω resistor in the figure, is—










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    Apply the superposition theorem in the given circuit
    Case I. When 10 V source is treated 5 A current source is replaced by their internal resistance (i.e. open circuited)

    Now, 10 = 5 I ⇒ 1 = 2 Amp.
    Case II. When current source (5 A) is treated and voltage source (10 V) S.C. the circuit becomes

    According to the current division rule I will be zero so, the net current I = 2 amp.

    Correct Option: B

    Apply the superposition theorem in the given circuit
    Case I. When 10 V source is treated 5 A current source is replaced by their internal resistance (i.e. open circuited)

    Now, 10 = 5 I ⇒ 1 = 2 Amp.
    Case II. When current source (5 A) is treated and voltage source (10 V) S.C. the circuit becomes

    According to the current division rule I will be zero so, the net current I = 2 amp.

  1. The Y11-parameters for the given network shown below are—










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    We given network

    Compare this network with standard network

    Ya =
    1
    		+
    1
    		=
    1
    		+
    3

    25 / 2 x (s)25s / 3

    Yb =
    1
    		+
    1
    		=
    1
    		+
    2

    35 / 2 x (s)35s / 3

    Yc =
    1
    		+
    1
    		=
    1
    		 + 6s
    51 / 6 s5

    Y11 = Ya + Yc =
    1
    		+
    3
    		+
    1
    		 + 6s
    25s5

    Correct Option: A

    We given network

    Compare this network with standard network

    Ya =
    1
    		+
    1
    		=
    1
    		+
    3

    25 / 2 x (s)25s / 3

    Yb =
    1
    		+
    1
    		=
    1
    		+
    2

    35 / 2 x (s)35s / 3

    Yc =
    1
    		+
    1
    		=
    1
    		 + 6s
    51 / 6 s5

    Y11 = Ya + Yc =
    1
    		+
    3
    		+
    1
    		 + 6s
    25s5

  1. Find VC (0+) for the circuit—










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    Applying potential divider rule

    VA = 12.
    R1
    		= 6V
    R1 + R1

    VB = 12.
    2R1
    		= 8V
    2R1 + R1

    VC (0+) = VB – VA = 8 – 6 = 2 V

    Correct Option: A

    Applying potential divider rule

    VA = 12.
    R1
    		= 6V
    R1 + R1

    VB = 12.
    2R1
    		= 8V
    2R1 + R1

    VC (0+) = VB – VA = 8 – 6 = 2 V