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Electric Circuits Test 1
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Electric Circuits Test 1
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  • Question 1/10
    1 / -0

    The product of y – parameters of the two port network shown below is _________

    Solutions

    By applying KVL in the first loop,

    V1 = I1 + 10I2 + 2 (I1 + I2)

    ⇒ V1 = 3I1 + 12I2

    By applying KVL in the second loop,

    V2 = 6I2 + 2(I1 + I2)

    ⇒ V2 = 2I1 + 8I2

    Z=[31228]

    |z| = 24 – 24 = 0

    ⇒ y – Parameter matrix doesn’t exist for the given circuit.

  • Question 2/10
    1 / -0

    For the circuit shown in the figure, V(t)=2202sin377t volts. The value of line power factor is

    Solutions

    V(t)=2202sin377t

    ω = 377 rad/sec.

    inductive reactance XL = jωL = j 0.096 × 377

    = j36.192 Ω

    Capacitive reactance Xc=1jωC=1j×377×73.29×106=j36.192Ω

    |XL|=|Xc|

    So, the circuit is operating at resonant frequency. At resonance, the circuit acts as purely resistive and hence the power factor in unity. 

  • Question 3/10
    1 / -0

    Two networks, N1 and N2, are described graphically in terms of their i-v relations, and connected together through a single resister, as shown below. The current flowing through the circuit after connection is

    Solutions

    Drawing the Thevenin equivalent of network N1 and N2 with proper polarities of voltage v1 and v2 and directions of current i1 and i2 is:

    From the given graph

    |v1|=5v=|v2|

    |i1|=5mA=|i2|

    Substituting the value the given circuit radius is

    Apply KVL

    I=105k=2mA

  • Question 4/10
    1 / -0

    If N is no of nodes and B is the total number of branches in a network, then the number of twigs and number of links of a tree respectively are
    Solutions

    The number of twigs on a tree is always one less than the number of nodes.

    Number of twigs = N – 1

    Number of links, L = B – N + 1
  • Question 5/10
    1 / -0

    The value of I0 (in A), in the circuit shown below is _______

    Solutions

    The given diagram can be simplified as

    By applying nodal analysis at node V1,

    V112+V1V26+V1362=0

    ⇒ V1 + 2V1 – 2V2 + 6V1 – 216 = 0

    ⇒ 9V1 – 2V2 = 216      ----(1)

    By applying nodal analysis at node V2,

    V218+V2369+V2V16=0

    ⇒ V2 + 2V2 – 72 + 3V2 – 3V1 = 0

    ⇒ 6V2 – 3V1 = 72      ----(2)

    By solving the equations (1) and (2), we get

    V1 = 30 V, V2 = 27 V

    i0=36V12=36302=3A.

  • Question 6/10
    1 / -0

    In the circuit shown in the figure, find the turns ratio of the ideal transformer that will match the output of the transistor amplifier to the speaker represented by the 16 Ω load.

    Solutions

    Thevenin equivalent resistance across the terminals A – B is

    Thevenin resistance is the equivalent resistance after removing all the independent sources.

    Vx = 0 V

    Rth = 10 kΩ

    The given network can be represented as

    By considering the expression for the turns ratio in the terms of impedances.

    (1a)2=1610×103

    a = 25

  • Question 7/10
    1 / -0

    The circuit shown in the figure is operating in the sinusoidal steady state. Find the value of ω (in rad/s) if

    i = 40 sin (ωt + 21.87°) mA

    vs = 40 cos (ωt – 15°) V

    Solutions

    Vs = 40 cos (ωt – 15°) V

    = 40 sin (ωt + 75°) V

    i = 40 sin (ωt + 21.87°) mA

    z=Vsi=100053.13=600+j800Ω(1)

    From the given circuit,

    z=600+jω(3.2)+1jω(2.5×106)

    =600+j(3.2ω0.4×106ω)(2)

    From equations (1) and (2)

    3.2ω1.4×106ω=800

    ⇒ ω2 – 250ω - 125000 = 0

    ⇒ ω = 500 and ω = -250.

    Negative value of ω is not possible.

    ⇒ ω = 500 rad/sec.
  • Question 8/10
    1 / -0

    For the network shown in the figure, the current i0(t) at t = 5.33 sec is _______ (in mA)

    Solutions

    At t = 0-, the circuit is as shown below

    i0(0)=10(22+2)=5mA

    i0 (0+) = i0 (0-) = 5 mA

    At t = ∞, the circuit is as shown below

    iA = 0,

    i0()=10(22+2)=5mA

    i0(t)=i0()+[i0(0+)i0()]et/τ

    = 5 mA + [5 - 5] e-t/τ

    = 5 mA

    At any time t > 0, i0 = 5 mA 

  • Question 9/10
    1 / -0

    For the circuit shown in the figure below the rms value of voltage a cross the switch when switch S is opened.

    Solutions

    Line voltage (VL) = 400 V

    Phase voltage (Vph)=4003V

    When the switch is open

    Ian=4003(40+j80)

    =103(1+2j)

    |Ian|=1015A

    rms value of voltage across the switch is

    Vs=1015×80=206.56V

  • Question 10/10
    1 / -0

    In the circuit shown below the average power consumed by the 1 Ω resistor is

    Solutions

    By applying superposition theorem,

    I1=102sin(1000t)1+j(1000)(1×103)=102sin(1000t)1+j=102sin(1000t)245  

    = 10 sin (1000 t – 45°)

    I2=102cos(3000t)1+j(3000)(1×103)=102cos(3000t)1+3j

    =102cos(3000t)1071.56=20cos(3000t71.56)

    I = I1 + I2

    I=10sin(1000t45)+20cos(3000t71.56)

    Irms=(102)2+(202)2

    =50+10=60

    Powerdissipated(P)=Irms2R

    =(60)2(1)=60Watts

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