Network theory miscellaneous

Network theory miscellaneous

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Network Theory

Network Elements and the Concept of Circuit
Network theory miscellaneous
  1. Consider the circuit as shown below which has a current dependent current source. The value V2 /V1 is—










  1. View Hint View Answer Discuss in Forum

    The given circuit:

    Applying KCL at node A, we get

    VA - V1
    		 +
    VA
    		= αi ........…(i)
    RR

    From figure:
    VA = V2 and
    V1 - VA
    		= i
    R

    Now,
    V2 - V1
    		 +
    V2
    		= α
    V1 - VA
    		 = i
    RRR

    or
    V2
    1
    		 +
    1
    		 +
    α
    		=
    α
    		 V1 +
    V1
    RRRRR

    or
    V2
    2 + α
    		 = V1
    α + 1
    RR

    or
    V2
    		 =
    1 + α
    V12 + α

    Hence alternative (C) is the correct choice.

    Correct Option: C

    The given circuit:

    Applying KCL at node A, we get

    VA - V1
    		 +
    VA
    		= αi ........…(i)
    RR

    From figure:
    VA = V2 and
    V1 - VA
    		= i
    R

    Now,
    V2 - V1
    		 +
    V2
    		= α
    V1 - VA
    		 = i
    RRR

    or
    V2
    1
    		 +
    1
    		 +
    α
    		=
    α
    		 V1 +
    V1
    RRRRR

    or
    V2
    2 + α
    		 = V1
    α + 1
    RR

    or
    V2
    		 =
    1 + α
    V12 + α

    Hence alternative (C) is the correct choice.

  1. The laplace transform of the waveform shown in the figure is
    F(s) =
    1
    		 (1 + Ae-e + Be-4s + Ce-6s + De-8s)
    s2

    What is the value of D?










  1. View Hint View Answer Discuss in Forum

    The given waveform:

    Given, F(s) =
    1
    		[1 + Ae–s + Be–4s + Ce–6s + De–8s]
    s2

    The slope of D is 1/2 = 0·5
    Hence alternative (C) is the correct choice.

    Correct Option: C

    The given waveform:

    Given, F(s) =
    1
    		[1 + Ae–s + Be–4s + Ce–6s + De–8s]
    s2

    The slope of D is 1/2 = 0·5
    Hence alternative (C) is the correct choice.

  1. Which one of the following is the transfer function of an electrical low pass filter using R and C elements?








  1. View Hint View Answer Discuss in Forum

    The low pass filter using R and C components:

    V0(s) =Vi(s)
    1
    Cs
    R +
    1
    Cs

    V0(s)
    		= H(s) =
    1
    Vi(s)(1 + RCs)

    Hence alternative (B) is the correct choice.

    Correct Option: B

    The low pass filter using R and C components:

    V0(s) =Vi(s)
    1
    Cs
    R +
    1
    Cs

    V0(s)
    		= H(s) =
    1
    Vi(s)(1 + RCs)

    Hence alternative (B) is the correct choice.

  1. In the circuit shown in the figure below, for what value of C will the current I be in phase with the sinusoidal source voltage Vs = sin 2t?










  1. View Hint View Answer Discuss in Forum

    The given circuit:

    Given, Vs = sin 2t, means = 2 rad/sec.
    In order to make the current I in phase with the sinusoidal source voltage, the imaginary part of the net impedance offered by the voltage source must be zero.
    So,

    Zeq =
    1
    		|| (1 + j1)
    j C

    or
    Zeq =
    1
    		× (1 + j)
    j2C
    1
    		 + (1 + j)
    j2C

    or
    Zeq =
    1
    		× [1 + j]
    2C
    2C + j (2C - 1)

    2C – 1 = 0 (Since Imaginary part must be zero)
    or
    C = 1/2 F
    Hence alternative (B) is the correct choice.

    Correct Option: B

    The given circuit:

    Given, Vs = sin 2t, means = 2 rad/sec.
    In order to make the current I in phase with the sinusoidal source voltage, the imaginary part of the net impedance offered by the voltage source must be zero.
    So,

    Zeq =
    1
    		|| (1 + j1)
    j C

    or
    Zeq =
    1
    		× (1 + j)
    j2C
    1
    		 + (1 + j)
    j2C

    or
    Zeq =
    1
    		× [1 + j]
    2C
    2C + j (2C - 1)

    2C – 1 = 0 (Since Imaginary part must be zero)
    or
    C = 1/2 F
    Hence alternative (B) is the correct choice.

  1. In the circuit shown below, what is the voltage across 5Ω resistor i.e. (V1 – V2)










  1. View Hint View Answer Discuss in Forum

    The given circuit:

    From figure, KCL at node 1

    V1
    		 +
    V1 – V2
    		 = 176 …. .......(i)
    75

    Again KCL at node 2
    V2
    		 +
    V2 – V1
    		 = 110 …. .......(i)
    105

    From equation (i) and (ii), we get
    V1 = 1190V and V2 = 1160V
    Now, V1 – V2 = 1190 – 1160 = 30V

    Correct Option: B

    The given circuit:

    From figure, KCL at node 1

    V1
    		 +
    V1 – V2
    		 = 176 …. .......(i)
    75

    Again KCL at node 2
    V2
    		 +
    V2 – V1
    		 = 110 …. .......(i)
    105

    From equation (i) and (ii), we get
    V1 = 1190V and V2 = 1160V
    Now, V1 – V2 = 1190 – 1160 = 30V