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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
    $$\textbf{Hint: At resonance current and voltage are in same phase}$$
    $$\textbf{Step1: Find resonating angular frequency}$$
    $$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{8\times 10^{-3}\times 20 \times 10^{-6}}} = 2500 rad/sec$$
    $$\textbf{Step2: Find amplitude of current at resonance}$$
    Resonance current $$= \dfrac{V}{R} = \dfrac{220}{44} = 5A$$
    $$\textbf{Hence option B correct}$$
  • Question 2
    1 / -0
    In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected. When switch $$'S'$$ is closed then there is a phase difference of $$\dfrac {\pi}{4}$$ between current and applied voltage and voltage across resister is $$\dfrac {100}{\sqrt {2}}V$$. 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 $$L= 10^{-3}$$H and $$C=3 \times 10^{-7}$$ F  is connected to a $$100V-50Hz$$ a.c. source, the impedance of the circuit is 
    Solution
    Impedance $$ Z= \sqrt {R^2 +(X_L - X_C)^2}$$
    Where $$ X_L = \omega L$$
    $$ X_C = \cfrac{1}{\omega C}$$
    and $$ \omega = 2 \pi f$$
    $$ =2 \pi \times 50 = 100\pi$$ 
    $$ \therefore X_L = 100 \pi  \times 10^{-3} = \cfrac{\pi}{10}$$
    $$ X_C = \cfrac{1}{100 \pi \times 3 \times 10^{-7} } = \cfrac{10^5}{3\pi}$$
    $$R=0$$
    $$\therefore Z=\sqrt{(\cfrac{\pi}{10} - \cfrac{10^5}{3\pi})^2}$$
    $$ = \cfrac {10^5}{3\pi} - \cfrac{\pi}{10}$$      As, ($$ \cfrac {10^5}{3\pi} > \cfrac{\pi}{10}$$)

  • Question 7
    1 / -0
    A $$2$$ $$F$$ capacitor is initially charged to $$20$$ $$V$$ and then shorted across a $$5$$ $$mH$$ inductor. The angular frequency of oscillation is.
    Solution
    $$\omega_o=\cfrac{1}{\sqrt{LC}}$$
    $$\omega_o=\cfrac{1}{\sqrt{2\times 5\times 10^{-3}}}$$
    $$\omega_o=10$$ $${rad/sec}$$
  • Question 8
    1 / -0
    A inductance capacitance circuit is in the state of resonance. if the $$C = 0.1 \mu F$$ and $$L = 0.25$$ Henry. Neglecting ohmic resistance of circuit what is the frequency of oscillations
    Solution
    From the formula, the frequency of oscillation is

    $$f=\dfrac{1}{2\pi\sqrt{LC}}$$

    $$=\dfrac{1}{2\pi\sqrt{0.25\times 0.1\times 10^{-6}}}$$

    $$=1007\,Hz$$
  • Question 9
    1 / -0
    In a series LCR circuit 
    $$V_l = 3V_R$$ and $$V_C = 2V_R$$
    Volatge across capacitance

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

    Solution

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