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Alternating Current Test - 65

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Alternating Current Test - 65
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The inductance of a coil is directly proportional to 
    Solution

  • Question 2
    1 / -0
    An $$8\mu F$$ capacitor is connected to the terminals of an A.C. source whose $$V_{rms}$$ is $$150$$ volt and the frequency is $$60$$ Hz, the capacitive reactance is?
    Solution

  • Question 3
    1 / -0
    An oscillator circuit contains an inductor $$0.05H$$ and a capacitor of  capacity $$80\mu F$$.When the maximum voltage across the capacitor is $$200V$$, the maximum current (in amperes) in the circuit is 
    Solution
    Given
    $$L=0.05H,C=80\mu F$$
    $$V_{max}=200V$$
    $$\because$$ Voltage equation $$V(t)=V_m\sin \omega t$$
    $$\therefore$$ Current (i) $$=\dfrac{cdV}{dt}=c\dfrac{d(V_m\sin \omega t)}{dt}=CV_m\omega \cos \omega t$$

    $$\because \omega =\dfrac{1}{\sqrt{LC}}$$

    $$\therefore i=V_m\sqrt{\dfrac{C}{L}}\cos\omega t$$
    $$\therefore$$ Maximum current $$(i_m)=V_m\sqrt{\dfrac{C}{L}}=200\times \sqrt{\dfrac{80\mu}{0.05}}$$
    $$\Rightarrow i_m=8A$$
  • Question 4
    1 / -0
    Choose the incorrect statement.
    Solution
    Even if it were current-limited, alternating current (AC) would not be appropriate for electrolysis. Because the "cathode" and "anode" will be constantly switching places as the polarities will change with time. Therefore, direct current (DC) is used for electrolysis.
  • Question 5
    1 / -0
    In a series LCR circuit $$R=300\Omega, L=0.9H, C=2\mu F, \omega =1000\ rad/s$$. The impedance of the circuit is?
    Solution
    The correct option is (a)
    $$Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})}$$
         $$=\sqrt{(300)^2+\left(1000\times 0.9-\dfrac{1}{1000\times 2\times 10^{-6}}\right)^2}$$
    $$Z=\sqrt{(300)^2+(400)^2}\Rightarrow 500\Omega $$
  • Question 6
    1 / -0
    The frequency at which $$1\ H$$ inductor will have a reactance of $$2500\ \Omega$$ is:
    Solution

  • Question 7
    1 / -0
    A series AC circuit has a resistance of $$4\Omega $$ and a reactance of $$3\Omega$$. The impedance of the circuit is 
    Solution

  • Question 8
    1 / -0
    The natural frequency of the circuit shown in figure is 

    Solution
    $$\omega=\dfrac{1}{\sqrt{L_{eq}L_{eq}}}$$

    $$=\dfrac{1}{\sqrt{2LC/2}}$$

    $$=\dfrac{1}{\sqrt{LC}}$$
  • Question 9
    1 / -0
    A current $$I=3+8\sin 100t$$ is passing through a resistor of resistance $$10\Omega$$. The effective value of current is?
    Solution

  • Question 10
    1 / -0
    The frequency of oscillation of current in the inductor  is 

    Solution
    Equivalent inductance 

    $$L_{eq}=L+2 L=3L$$

    $$,C_{eq}=C+2C=3C$$

    $$\therefore$$ Frequency of oscillation 

    $$f=\dfrac{1}{2 \pi \sqrt{L_{eq}C_{eq}}}$$

    $$=\dfrac{1}{6 \pi \sqrt{LC}}$$
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