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Alternating Current Test - 57

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Alternating Current Test - 57
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  • Question 1
    1 / -0
    For the series LCR circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency.

    Solution
    Hint: At resonance current and voltage are in same phase\textbf{Hint: At resonance current and voltage are in same phase}
    Step1: Find resonating angular frequency\textbf{Step1: Find resonating angular frequency}
    ω=1LC=18×103×20×106=2500rad/sec\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{8\times 10^{-3}\times 20 \times 10^{-6}}} = 2500 rad/sec
    Step2: Find amplitude of current at resonance\textbf{Step2: Find amplitude of current at resonance}
    Resonance current =VR=22044=5A= \dfrac{V}{R} = \dfrac{220}{44} = 5A
    Hence option B correct\textbf{Hence option B correct}
  • Question 2
    1 / -0
    In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected. When switch S'S' is closed then there is a phase difference of π4\dfrac {\pi}{4} between current and applied voltage and voltage across resister is 1002V\dfrac {100}{\sqrt {2}}V. When switch is open current and applied voltage are in same phase. Neglecting resistance of connecting wire answer the following questions:
    Resonance frequency of circuit is

    Solution

  • Question 3
    1 / -0
    In the L -C circuit shown in figure,the current is in direction shown in the figure and charges on the capacitor plates have sign shown in the figure A this time:-

    Solution

  • Question 4
    1 / -0
    An alternating current is that current which changes regularly its
    Solution

  • Question 5
    1 / -0
    The voltage across a pure inductor is represented in figure. Which one of the following curves in the figure will represent the current?

    Solution

  • Question 6
    1 / -0
    In a series L-C circuit , if L=103L= 10^{-3}H and C=3×107C=3 \times 10^{-7} F  is connected to a 100V50Hz100V-50Hz a.c. source, the impedance of the circuit is 
    Solution
    Impedance Z=R2+(XLXC)2 Z= \sqrt {R^2 +(X_L - X_C)^2}
    Where XL=ωL X_L = \omega L
    XC=1ωC X_C = \cfrac{1}{\omega C}
    and ω=2πf \omega = 2 \pi f
    =2π×50=100π =2 \pi \times 50 = 100\pi 
    XL=100π ×103=π10 \therefore X_L = 100 \pi  \times 10^{-3} = \cfrac{\pi}{10}
    XC=1100π×3×107=1053π X_C = \cfrac{1}{100 \pi \times 3 \times 10^{-7} } = \cfrac{10^5}{3\pi}
    R=0R=0
    Z=(π101053π)2\therefore Z=\sqrt{(\cfrac{\pi}{10} - \cfrac{10^5}{3\pi})^2}
    =1053ππ10 = \cfrac {10^5}{3\pi} - \cfrac{\pi}{10}      As, (1053π>π10 \cfrac {10^5}{3\pi} > \cfrac{\pi}{10})

  • Question 7
    1 / -0
    A 22 FF capacitor is initially charged to 2020 VV and then shorted across a 55 mHmH inductor. The angular frequency of oscillation is.
    Solution
    ωo=1LC\omega_o=\cfrac{1}{\sqrt{LC}}
    ωo=12×5×103\omega_o=\cfrac{1}{\sqrt{2\times 5\times 10^{-3}}}
    ωo=10\omega_o=10 rad/sec{rad/sec}
  • Question 8
    1 / -0
    A inductance capacitance circuit is in the state of resonance. if the C=0.1μFC = 0.1 \mu F and L=0.25L = 0.25 Henry. Neglecting ohmic resistance of circuit what is the frequency of oscillations
    Solution
    From the formula, the frequency of oscillation is

    f=12πLCf=\dfrac{1}{2\pi\sqrt{LC}}

    =12π0.25×0.1×106=\dfrac{1}{2\pi\sqrt{0.25\times 0.1\times 10^{-6}}}

    =1007Hz=1007\,Hz
  • Question 9
    1 / -0
    In a series LCR circuit 
    Vl=3VRV_l = 3V_R and VC=2VRV_C = 2V_R
    Volatge across capacitance

  • Question 10
    1 / -0
    In the given circuit the maximum current can be

    Solution

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