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Alternating Current Test - 8
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Alternating Current Test - 8
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  • Question 1/10
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    A pure resistance R, pure capacitance C and pure inductance L are connected in series and the impedance of the circuit at resonance is Z0. If they are connected in parallel to each other, the maximum impedance at resonance will be:​

    Solutions

    If series LCR circuit is present then z={R2+(xl-xc)2}1/2
    if resonance is present then XC=CL or VC=CL then Z= Zo
    if LCR circuit is in parallel form then in which circuit resonance is also present firstly 1/Z ={(1/R)2+ ((1/xc) –(1/xl))2}1/2 and in resonance conditions XC=CL according to this Z= Zo then option C is the correct answer

  • Question 2/10
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    At resonance the impedance of RLC circuit is

    Solutions

    Resonance is a condition in LCR circuit where the inductive reactance ( XL) and capacitive reactance (Xc) is equal.
    Impedance at resonance is equal to the Resistance of the resistor because the other two cancel themselves off according to the below formula -
    Z2=R2+(XL2-Xc2)
    As XL = Xc ,
    Z = R
     

  • Question 3/10
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    Ratio  is equal to

    Solutions

    frequency-to-bandwidth ratio of the resonator:
    Q=fr/△f= ωr/△ω
    where fr is the resonant frequency, Δf is the resonance width or full width at half maximum (FWHM) i.e. the bandwidth over which the power of vibration is greater than half the power at the resonant frequency, ωr = 2πfr is the angular resonant frequency, and Δω is the angular half-power bandwidth.

  • Question 4/10
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    When the oscillations produced are of constant amplitude. The are called

    Solutions

    If the losses incurred in the electrical system could be compensated, the amplitude of oscillation would remain constant and as such the oscillation would continue indefinitely against both external disturbances and changes in the initial conditions. This type of oscillation is called undamped oscillation.

  • Question 5/10
    1 / -0

    Find the series resonance frequency when L = 0.09 H, C = 25 F​

    Solutions

    ω=1/√LC
    ω=1√0.09x25
    ω=0.666
    ω=2πf
    f=0.666/(2 x 3.14)
    f=0.106 s-1

  • Question 6/10
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    Measure of sharpness in an ac circuit is given by

  • Question 7/10
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    In a series RCL circuit at resonance the applied ac voltage is 220 V. The potential drop across the inductance is 110V, what is the potential drop across the capacitor?​

    Solutions

    At resonance, the potential drop across the capacitance is equal and opposite to that of inductance i.e. 110V.

  • Question 8/10
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    Acceptor circuit is

    Solutions

    The series resonance circuit is known as an acceptor circuit. The impedance of the acceptor circuit is minimum but the voltage can be magnified.
    The Acceptor circuit provides the maximum response to currents at its resonant frequency.
    Series resonance circuit is known as acceptor circuit because the impedance at the resonance is at its minimum so as to accept the current easily such that the frequency of the accepted current is equal to the resonant frequency.
     

  • Question 9/10
    1 / -0

    An ac circuit contains a resistance R, capacitance C and inductance L in series with a source of emf ε = ε0sin (ωt – Ø). The current through the circuit is maximum when:​

  • Question 10/10
    1 / -0

    Resonant circuits are used in tuning of

    Solutions

    The most common application of tank circuits is tuning radio transmitters and receivers. For example, when we tune a radio to a particular station, the LC circuits are set at resonance for that particular carrier frequency. A parallel resonant circuit can be used as load impedance in output circuits of RF amplifiers.
     

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