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Alternating Current Test - 56

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Alternating Current Test - 56
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A resistance of $$ 10\Omega $$, a
    capacitance of $$ 0.1\mu F$$ and an inductance of $$2mH$$ are connected in
    series  across a source of alternating
    emf of  variab frequency. At what frequency  does maximum current flow?

    Solution

  • Question 2
    1 / -0
    The frequancy of oscillation of current in the inductor is-

    Solution
    Frequency of oscilation$$=\cfrac { 1 }{ \sqrt { 3C(3L) }  } =\cfrac { 1 }{ 3\sqrt { LC }  } $$

  • Question 3
    1 / -0
    The equation of an alternating voltage is $$V=100\sqrt {2}\sin {100\pi t}$$ volt. The RMS value of voltage and frequency will be respectively
    Solution
    Angular frequency is nothing but $$coefficient$$ of $$t$$ in the equation,
    so we have $$\omega =100\pi$$ or $$2\pi f=100\pi$$, we get $$f=50Hz$$ 
    rms voltage is $$\dfrac{\text{peak voltage}}{\sqrt {2}}=100 \sqrt{2}/ \sqrt{2}=100 V$$.

    Option A is correct.
  • Question 4
    1 / -0
    An alternating power supply of $$220 V$$ is applied across a series circuit of resistance $$10\sqrt{3}\Omega$$, capacitive reactance $$40 \Omega$$ and inductive reactance $$30 \Omega$$. The respective current in the circuit for zero and infinite frequencies are
    Solution
    Case I: When the frequency is zero. 
    $$f=0$$ implies $$\omega = 0$$
    Therefore, the capacitive reactance becomes infinite
    $$X_C= \dfrac{1}{\omega C}  \to \infty$$ 
    As a result the impedance(Z) of the circuit becomes infinite and the current becomes zero. 

    Case I: When the frequency is infinite. 
    $$f \to \infty$$ implies $$\omega \to \infty$$
    Therefore, the inductive reactance becomes infinite
    $$X_L= \omega L \to \infty$$ 
    As a result the impedance(Z) of the circuit becomes infinite and the current becomes zero. 
  • Question 5
    1 / -0
    A coil has resistance $$30 \Omega $$ and inductive reactance $$20 \Omega $$ at $$50 Hz$$ frequency.If an ac source of $$200 V,100Hz$$ is connected across the coil ,the current in the coil in the coil will be 
    Solution
    Given,
    $$X_L=20\Omega$$
     frequency   $$ f=50Hz$$
    So,  $$2\pi fL=20$$
    $$\Rightarrow 2\pi\times 50L=20\Rightarrow L=\dfrac{1}{5\pi}$$
    When the frequency is $$100Hz$$
    $$X_L=2\pi\times\dfrac{1}{5\pi}\times100=40\Omega$$
    Impedance $$Z= \sqrt{R^2+X_L^2}=\sqrt{900+1600}=50\Omega$$
    $$V=200volt$$
    Thus the current $$I=\dfrac{200}{50}=4A$$
  • Question 6
    1 / -0
    Figure shows a long straight conductor carrying current i. A square loop of side a is kept at a distance r from it. Which of the following is correct?

    Solution

  • Question 7
    1 / -0
    A L - C resonant circuit contains a 200 pF capacitor and a $$ 100 \mu $$ H inductor it is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is
    Solution

    Given that,

       $$ C=200pF $$

     $$ C=200\times {{10}^{-12}}F $$

      $$ L=100\mu H $$

     $$ L=100\times {{10}^{-6}}H $$

     We know that,

    Frequency of wave =   $$ \nu =\dfrac{1}{2\pi \sqrt{LC}} $$

     $$ \nu =\dfrac{1}{2\times 3.14\sqrt{200\times {{10}^{-12}}\times 100\times {{10}^{-6}}}} $$

     $$ \nu =\dfrac{1}{2\times 3.14\times 1.4142\times {{10}^{-7}}} $$

    Now, the wavelength is

      $$ \lambda =\dfrac{3\times {{10}^{8}}}{\nu } $$

     $$ \lambda =\dfrac{3\times {{10}^{8}}}{\dfrac{1}{2\times 3.14\times 1.4142\times {{10}^{-7}}}} $$

     $$ \lambda =3\times {{10}^{8}}\times 2\times 3.14\times 1.4142\times {{10}^{-7}} $$

     $$ \lambda =26.643\times 10 $$

     $$ \lambda =266.43\,m $$

    Hence, the wavelength of radiated electromagnetic wave is $$266$$ m.

  • Question 8
    1 / -0
    In L-C-R series circuit consists of a resistance of 10 $$\Omega$$ a capacitor of reactance 6.0 $$\Omega$$ and an inductor coil. The circuit is found to resonate when put across a 300 V, 100 Hz supply. The inductance of coil is (Take $$\pi$$ = 3)
    Solution

  • Question 9
    1 / -0
    The sinusoidal potential difference $$V_1$$ shown in figure applied across a resistor $$R$$ produces heat at a rate $$W$$. what is the rate of heat dissipation when the square waves potential different $$V_2$$ as shown in figure is applied across the resistor?

    Solution

  • Question 10
    1 / -0
    In a series LCR circuit,the inductive reactance is twice the resistance and the capacitance reactance is $${\frac{1}{3}^{rd}}$$ the inductive reactance. The power factor of the circuit is:
    Solution
    $$\begin{array}{l}\omega L = 2R\\\frac{1}{{\omega C}} = \frac{1}{3}\left( {\omega L} \right)\\\omega L - \frac{1}{{\omega C}} = 2R - \frac{{2R}}{3} = \frac{{4R}}{3}\\\tan \phi  = \frac{{4R}}{3} \times \frac{1}{R} = \frac{4}{3}\\\cos \phi  = \frac{1}{{\sqrt {1 + {{\tan }^2}\phi } }} = \frac{1}{{\sqrt {1 + \frac{{{4^2}}}{{{3^2}}}} }} = \frac{3}{5} = 0.6\end{array}$$
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