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Alternating Current Test - 24

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Alternating Current Test - 24
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  • Question 1
    1 / -0
    An inductance of negligible resistance whose reactance is $$22\Omega$$ at $$200Hz$$ is connected to $$200$$volt. If the line frequency is known to be $$50$$ cycles/second, the equation for the line voltage is :
    Solution
    $${X}_{L}=\omega L=2\pi nL$$
    $$\therefore$$ $$L=\cfrac{{X}_{L}}{2\pi n}=\cfrac{22\times 7}{2\times 22\times 200}H=0.0175H$$
  • Question 2
    1 / -0
    In series combination of $$R,L,C$$ with an A.C source at resonanace, if $$R=20$$ ohm, then impedence $$Z$$ of the combination is
    Solution
    We know at resonance, reactance (resistance due to inductor and capacitor) be zero (0).
    At resonance, Impedance (Z) $$=$$ Resistance (R)
    Therefore, $$Z = 20$$ ohm
  • Question 3
    1 / -0
    For the circuit shown in the fig., the current through the inductor is $$0.9A$$ while the current through the condenser is $$0.4A$$. Then

    Solution
    The current drawn by inductor and capacitor will be in opposite phase.

    Hence net current drawn from generator

    $$={I}_{L}-{I}_{C}=0.9-0.4=0.5amp$$
  • Question 4
    1 / -0
    In a pure capacitive A.C circuit current and voltage differ in phase by
    Solution
    $$i=i_o Sin(\omega t + \pi /2)$$
    current leads voltage by $$\pi /2 $$ i.e., current and voltage differ in phase by $$90^o$$


  • Question 5
    1 / -0
    In a RLC circuit capacitance is changed from $$C$$ to $$2C$$. For the resonant frequency to remain unchanged, the inductance should be changed from $$L$$ to
    Solution
    $$f=\cfrac { 1 }{ 2\pi \sqrt { (LC) }  } $$
    when $$C$$ is doubled, $$L$$ should be halved so that resonant frequency remains unchanged.
  • Question 6
    1 / -0
    If the frequency of an A.C is made 4 times of its initial value, the inductive reactance will
    Solution
    inductive reactance  $$= 2 \pi f L$$
    therefore when f is made 4 times, inductive reactance also becomes 4 times.
  • Question 7
    1 / -0
    If a current $$I$$ given by $$I={ I }_{ 0 }\sin { \left( \omega t-\pi /2 \right)  } $$ flows in inductance in an AC circuit which an A.C potential $$E={ E }_{ 0 }\sin{\omega t}$$ has been applied, then power consumption $$P$$ in the circuit will be
    Solution
    $$P={V}_{r.m.s}\times {I}_{r.m.s}\times \cos { \phi  } =\cfrac{1}{2}{E}_{0}\times {I}_{0}\cos { \pi/2  } =0$$
  • Question 8
    1 / -0
    In series $$L-C-R$$ circuit, the voltages across $$R, L$$ and $$C$$ are $${V}_{R}, {V}_{L}$$ and $${V}_{C}$$ respectively. Then the voltage of applied a.c. source must be
    Solution
    $$\quad V =V_R + \left(V_L-V_C  \right) j$$
    So value of apply voltage $$V=\sqrt { { \quad { V }_{ R } }^{ 2 }+{ \left( { { V }_{ L }-{ V }_{ C } } \right)  }^{ 2 } } $$

  • Question 9
    1 / -0
    Resonance frequency of LCR series a.c. circuit is $${f}_{0}$$. Now the capacitance is made $$4$$ times, then the new resonance frequency will become
    Solution
    In LCR series circuit, resonance frequency $${f}_{0}$$ is given by:
    $$L\omega

    =\cfrac { 1 }{ C\omega  } \quad \Rightarrow { \omega  }^{ 2 }=\cfrac { 1

    }{ LC } \quad $$

    $$\therefore \quad \omega =\sqrt { \cfrac { 1 }{ LC }  }

    =2\pi { f }_{ 0 }$$
    $$\therefore \quad { f }_{ 0 }=\cfrac { 1 }{ 2\pi \sqrt { LC }  } \quad \implies { f }_{ 0 }\propto \cfrac { 1 }{ \sqrt { C }  } $$

    When the capacitance of the circuit is made $$4$$ times, its resonant frequency become $${f}_{0}'$$
    $$\therefore

    \quad \cfrac { { f }_{ 0 }' }{ { f }_{ 0 } } =\cfrac { \sqrt { C }  }{

    \sqrt { 4C }  } \quad \implies { f }_{ 0 }'=\cfrac { { f }_{ 0 } }{ 2 } $$
  • Question 10
    1 / -0
    The AC produced in India changes its direction in every :
    Solution
    Answer is A.

    In India, the frequency of AC voltage is $$50\ Hz$$.
    It means $$50$$ waves will be produced in $$1\ s$$.
    In one wave, the direction is changed $$2$$ times.
    Thus, in 50 waves, the direction will be altered $$50\times 2=100$$ times.
    i.e. the direction is changed $$100$$ times in $$1\ s$$.
    Thus the direction is changed in every $$\dfrac{1}{100}\ s$$.

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