Network theory miscellaneous

Network theory miscellaneous

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Network Theory

Network Elements and the Concept of Circuit
Network theory miscellaneous
  1. For the circuit shown below, the voltage across 30Ω resistor is 45 volts. The reading of ammeter A will be—










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    The given circuit:

    From figure,

    I30Ω =
    45
    		 = 1·5A
    30

    Voltage across 10Ω resistance is
    = 1·5 × 10 = 15V
    Now,
    VAB = 15V + 45V = 60 V
    Current in branch
    CD =
    60V
    		 = 2A
    30Ω

    Now, current in ammeter
    = 1·5 + 2 = 3·5 A

    Correct Option: B

    The given circuit:

    From figure,

    I30Ω =
    45
    		 = 1·5A
    30

    Voltage across 10Ω resistance is
    = 1·5 × 10 = 15V
    Now,
    VAB = 15V + 45V = 60 V
    Current in branch
    CD =
    60V
    		 = 2A
    30Ω

    Now, current in ammeter
    = 1·5 + 2 = 3·5 A

  1. For a series RLC resonant circuit, which one of the following gives the impedance at the lower and upper half power frequencies, respectively?








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    For a series RLC circuit, at lower half frequency current is 1/2 times and leading in nature whereas at upper half frequency current is 1/2 times and lagging in nature.

    Correct Option: B

    For a series RLC circuit, at lower half frequency current is 1/2 times and leading in nature whereas at upper half frequency current is 1/2 times and lagging in nature.

  1. For the a.c. circuit given below, what is the value of I?










  1. View Hint View Answer Discuss in Forum

    The given circuit:

    I =
    V
    		=
    120
    Zeq(60 + j 60) || – 120j

    or
    I =
    120
    60 (1 + j) × (– 120j)
    60 (1 + j) – 120j

    or
    I =
    120
    60 (1 + j) (– 120j)
    60 (1 – j)

    or
    I =
    (1 – j) j
    		=
    j + 1
    		= 0
    (1 + j)1 + j

    or
    I = 0 + j 0
    Hence alternative (D) is the correct choice.

    Correct Option: D

    The given circuit:

    I =
    V
    		=
    120
    Zeq(60 + j 60) || – 120j

    or
    I =
    120
    60 (1 + j) × (– 120j)
    60 (1 + j) – 120j

    or
    I =
    120
    60 (1 + j) (– 120j)
    60 (1 – j)

    or
    I =
    (1 – j) j
    		=
    j + 1
    		= 0
    (1 + j)1 + j

    or
    I = 0 + j 0
    Hence alternative (D) is the correct choice.

  1. In figure, the admittance values of the elements in siemens are YR = 0·5 + j0, YL = 0 – J1·5, YC = 0 + j0·3 respectively. The value of I as a phasor when the voltage E across the elements is 10 ∠ 0° V is—










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    The given figure:

    I = V Yeq
    where, Yeq = YR + YL + YC
    = 0·5 + j0 + 0 – j1·5 + 0·j 0·3
    = 0·5 – j1·2
    Now, I = 10   0° (0·5 + j1·2)
    or
    I = 5 – j12
    Hence alternative (D) is the correct choice.

    Correct Option: D

    The given figure:

    I = V Yeq
    where, Yeq = YR + YL + YC
    = 0·5 + j0 + 0 – j1·5 + 0·j 0·3
    = 0·5 – j1·2
    Now, I = 10   0° (0·5 + j1·2)
    or
    I = 5 – j12
    Hence alternative (D) is the correct choice.

  1. In the circuit shown in the given figure, power dissipated in the 5Ω resistor is—










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    The given circuit:

    The circuit can be modified as shown

    – 5I + 20 – 10 I – 20 – 4 I = 0
    or
    19 I = 0
    or I = 0
    Therefore, the power dissipated in 5Ω resistor is zero.

    Correct Option: A

    The given circuit:

    The circuit can be modified as shown

    – 5I + 20 – 10 I – 20 – 4 I = 0
    or
    19 I = 0
    or I = 0
    Therefore, the power dissipated in 5Ω resistor is zero.