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AC Fundamentals Test 5
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AC Fundamentals Test 5
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  • Question 1/30
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    The potential difference measured across a coil is 4.5 V, when it carries a direct current of 9 A. The same coil when carries an alternating current of 9 A at 25 Hz, the potential difference is 24 V. Find the inductive reactance.
    Solutions

    Let R be the D.C resistance and L be the inductance of the coil.

    ∴ R = V/I = 4.5/9 = 0.5 Ω

    With A.C current of 25 Hz, Z = V/I = 24/9 = 2.66 Ω.

     

  • Question 2/30
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    The sign convention of susceptance is __________ that of the reactance.
    Solutions

    The sign convention of susceptance is exactly opposite to that of the reactance

    Consider the expression for the impedance in the rectangular form

    Z = R ± jX

    The admittance  

    Rationalize the expression for Y = 

    But Z2 = R2 + X2

     

     

     

     

  • Question 3/30
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    What is the time constant of the circuit?

    Solutions

    Given circuit is RC circuit

    Time constant of RC Circuit = ReqC

    Find Req of the circuit

     

    ∴ Time constant = 2 × 2 = 4 sec

  • Question 4/30
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    A three phase delta load draws _______ power as it would draw when connected as star.
    Solutions

    Let line voltage = x volts

    STAR Connection

     DELTA Connection 

    Vph = VL = x

    PSTAR = 3 (Iph)2R

     

     

    PDELTA = 3(Iph)2 × R

     

    Hence, we can say

    PDelta = 3 PSTAR
  • Question 5/30
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    An A.C. current is given by i = 100 sin100πt. It will achieve a value of 50 A after_______ second.
    Solutions

    V = Vm sin ωt

    100 sin100πt = 50

    100πt = sin-1(1/2) = π/6

    t = 1/600

  • Question 6/30
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    A voltage e(t) = 100 sin 314 t is applied to series circuit consisting of 10-ohm resistance, 0.0318 henry inductance and a capacitor of 63.6 μF. Calculate impedance of the circuit
    Solutions

    ω = 314 rad/s

    XL = ωL = 314 × 0.0318 = 10 Ω

    XC = 1/ω C = 1/314 × 63.6 × 10−6 = 50 Ω

    X = XL − XC = (10 − 50) = − 40 Ω (capacitive)

    Z = 10 − j40

  • Question 7/30
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    A balanced star-connected load of (8 + j6) Ω per phase is connected to a balanced 3-phase 400 V supply. Find the line current.
    Solutions

    Per phase impedance = 

     

     

    In star connection Iph = IL = 23.1 A
  • Question 8/30
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    The reactance offered by a capacitor to alternating current of frequency 50 Hz is 10 Ω. If frequency is increased to 100 Hz reactance becomes ____ohm.
    Solutions

    We know that

     

    XC = Capacitive reactance

     

     

  • Question 9/30
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    A 3ϕ, star connected balanced load is supplied by 400 V three phase 50 Hz supply. The load takes leading current of  and a power of 20 kW. Find the power factor of the load.
    Solutions

    We know that for 3ϕ system

    Total power = √3 VLIL cos ϕ

    PF = cos ϕ = 0.5 leading

    Important point related to 3 phase system:

    Sr. No.

    Description

    Expression

    1.

    Phase shift between adjacent phase voltages in a three-phase system

    120° or 2π/3 rad

    2.

    Mathematical representation of the three phase voltages

    VR = Vm sin ωt

    VY = Vm sin (ωt - 120°)

    VB = Vm sin (ωt - 240°)

    Or Vm sin (ωt - 120°)

    3.

    Phasor representation of three phase voltages

    VR + VY + VB = 0

    4.

    Vector addition of the three phase voltages, at any instant is zero

     

    VR + VY + VB = 0

    5.

    Frequency of phase or line voltages

    50 Hz

     

    Star connected load, voltage and current relations:

    1.

    Relation between line and phase voltage

    Line voltage = √3 × Phase voltage

    2.

    Relation between line and phase voltage

    3.

    Relation between line and phase current

    Iph = IL

    4.

    Total power consumed

    PT = 3 Vph Iph cos ϕ or PT = √3 VL IL cos ϕ

    Delta connected load, voltage and current relations:

    1.

    Relation between line and phase voltage

    Line voltage = Phase voltage

    2.

    Relation between phase and line currents

    IL = √3 Iph or Iph = IL / √3

    3.

    Total power consumed

    PT = 3 Pph = 3 Vph Iph cos ϕ or

    PT = √3 VL IL cos ϕ

    Different types of power and their relations:

    1.

    Total apparent power

    S = √3 VL IL Volt amp.

    2.

    Total active power

    P = √3 VL IL cos ϕ watt.

    3.

    Total reactive power

    Q = √3 VL IL sin ϕ VAR

    4.

    Relation between apparent, active and reactive power

  • Question 10/30
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    Three 100 Ω resistance are connected in star across a 400V 3 ϕ supply. The total power consumed by it is _________.
    Solutions

    = 1600 W
  • Question 11/30
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    Consider an R-L circuit in which a current i = 10 e-2t A is flowing. The voltage across the RL circuit is given by

    Solutions

    I = 10e-2t

    Voltage across resistor VR = 10e-2t × 8 = 80e-2t V

     

    v(t) = VR + VL = 80e-2t – 40e-2t = 40 e-2t V

  • Question 12/30
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    Add the following current as vectors:

    i1 = 7 sin ωt and i2 = 10 sin (ωt + π/3)

    Solutions

    Vector diagram shown in Figure below

    Resolving the vectors into their horizontal and vertical component
    X = Component = 7 + 10 cos 60° = 12

    Y = Component = 0 + 10 sin 60° = 8.66

    Hence the resultant equation = ir = 14.8 sin (ωt + 35.8°)
  • Question 13/30
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    A coil having an inductance of 50 mH and resistance 10 Ω is connected in series with a 25 μF capacitor across a 200 V AC supply. Calculate the value of quality factor.
    Solutions

    We know that resonant frequency

    Quality factor = 

  • Question 14/30
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    The quality factor of an RLC circuit is also known as ______
    Solutions

    The quality factor of an RLC circuit is also known as figure of merit it represents the efficiency of the resonant circuit to store energy

    The Q factor is defined as the ratio of energy stored per cycle to the energy lost per cycle

    Q = 2π × Maximum energy stored per cycle/Energy dissipated per cycle
  • Question 15/30
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    A delta-connected balanced load is connected to 3Φ, 400 V supply. The load PF is 0.8 lagging. The line current is 34.64 ampere. Find the resistance
    Solutions

    Given: VL = 400 V, PF = 0.8 lagging IL = 34.64 A

    We know that

    Here Rph = per phase resistance

    Zph = per phase impedance

    Rph = |Zph| cos ϕ = 20 × 0.8 = 16 Ω
  • Question 16/30
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    Mathematically a _______ voltage is expressed as V = Vm sin (ωt - ϕ)
    Solutions

    If the phase angle ϕ is negative, then the phase difference is said to be a lagging phase difference.

    If the phase angle ϕ is positive, then the phase difference is said to be a leading phase difference.

    Mathematically a sinusoidal lagging voltage is expressed as V = Vm sin (ωt - ϕ)
  • Question 17/30
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    Rms value is also called as _______ of AC current.
    Solutions
    Rms value is also called as heat producing component of ac current. The RMS value of AC (current) is the Direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.
  • Question 18/30
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    A coil having a resistance of 50 ohm and inductances 10 mH is connected in series with a capacitor and is supplied at constant voltage and variable frequency source. The maximum current is 1 A at 750 Hz. Determine the bandwidth.
    Solutions

    Quality factor of coil 

    Bandwidth = 

  • Question 19/30
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    An R-L circuit has Z = (6 + j8) ohm. Its susceptance is _________ siemens.
    Solutions

    We know that

    Admittance = 1/Z = 1/ (6 + j8) = 0.06 – 0.08j

    Conductance = 0.06

    Susceptance = -0.08

  • Question 20/30
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    In an alternating circuit, the impressed voltage is given by V = (100 − j50) volts and the current in the circuit is I = (3 – j4) A. Determine the real power in the circuit.
    Solutions

    We know that power will be found by the conjugate method. Using current conjugate, we have

    Real power = 500 W

    Imaginary power = 250 W

    NOTE: If voltage conjugate is used, then capacitive VARs are positive and inductive VARs negative. If current conjugate is used, then capacitive VARs are negative and inductive VARs are positive.

  • Question 21/30
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    A coil of resistance 20 Ω, inductance 0.8 H is connected to a 400 V DC supply. The initial rate of change of current is
    Solutions

    i(0-) = i(0+) = OA

    Apply kVL

     

     

     

  • Question 22/30
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    In a certain application, it is necessary to have a constant time rate of flow of electron i.e. constant current. There a ______is used instead of a ______.
    Solutions

    In a certain application, it is necessary to have a constant time rate of flow of electron i.e. constant current. There a current source is used instead of a voltage source

    An electric current is defined as rate of flow of electron. For voltage source this rate is not constant
  • Question 23/30
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    For high selectivity, the resistance must be?
    Solutions

    We know that

    Quality factor

    For high selectivity, the Q factor should be large and for Q factor to be large, the resistance would be small because Q is inversely proportional to the resistance.
  • Question 24/30
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    The conductance is defined as G =
    Solutions

    Consider the expression for the impedance in the rectangular form

    Z = R ± jX

    The admittance Y = 1/Z = 1/(R ± jX) =

    Rationalize the expression for Y = 

    But Z2 = R2 + X2

     

     

     

  • Question 25/30
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    A series circuit has R = 10 Ω, L = 0.01 μH and C = 10 μF. Calculate Q - factor of the circuit
    Solutions

    We know that

    For series resonant circuit,

    Quality factor
  • Question 26/30
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    The projection of a phasor on the Y axis represents _______.
    Solutions

    • You can see in the above diagram as the vector rotates in an anticlockwise direction at a speed of rad/sec
    • If we take the projection of this vector in Y-axis, then we get the instantaneous value of this sinusoidal signal
  • Question 27/30
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    Obtaining the voltage, current and power associated with each circuit elements is called as the ____
    Solutions

    Circuit analysis is the process of finding all the currents, voltages and power in a network of connected components

    We look at the basic elements used to build circuits and find out what happens when elements are connected into a circuit
  • Question 28/30
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    Find the total voltage applied in a series RLC circuit when I = 3 mA, VL = 30 V, VC = 18 V and R = 1000 ohms.
    Solutions

    Total voltage = 

    Here VR = voltage across resistor, VL = Voltage across inductor, VC = Voltage across capacitor

    VR = 1000 × 3 × 10-3 = 3 V

    Therefore, total voltage = .

  • Question 29/30
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    A two-element series circuit is connected across an A.C source E = 200√2 sin (ωt + 20°) V. The current in the circuit then is found to be i = 10√2 cos (314 t − 25°) A. Determine the power factor of the circuit.
    Solutions

    E = 200√2 sin (ωt + 20°) V

    i = 10√2 cos (314 t − 25°) A

    It is seen that applied voltage leads by 20° and current lags by 25° with regards to the reference quantity, their mutual phase difference is {20 – (-25)} = 45°

    Hence power factor = cos 45° = 0.707

  • Question 30/30
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    A periodic voltage having the following values for equal time intervals changing suddenly from one value to the next: 0, 5, 10, 20, 50, 60, 50, 20, 10, 5, 0, − 5, −10 V etc. What would be the r.m.s value of sine wave.
    Solutions

    The waveform of the alternating voltage is shown in Fig. It is not sinusoidal, but it is symmetrical.

    Mean value of 

    ∴ RMS value = √965 = 31.04 V

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