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Alternating Current Test - 44

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Alternating Current Test - 44
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  • Question 1
    1 / -0
    220 V, 50 Hz, AC source is connected to an inductance of 0.2 H and a resistance of 20 Ω\Omega in series. What is the current in the circuit?
    Solution
    The frequency of AC source =f=50Hz=f=50Hz
    Thus the impedance of the inductor, XL=ωL=(2πf)L=20πΩX_L=\omega L=(2\pi f)L=20\pi\Omega
    Resistance of the resistor =20Ω=20\Omega
    Thus, the net impedance of the circuit =XL2+R2=65.94Ω=\sqrt{X_L^2+R^2}=65.94\Omega
    Thus, the current in the circuit =VX=220V65.94Ω=3.33A=\dfrac{V}{X}=\dfrac{220V}{65.94\Omega}=3.33A
  • Question 2
    1 / -0
    In L-C-R circuit power of 3 mH inductance and 4Ω4\Omega resistance, EMF E=4cos1000tE=4\,cos\,1000t volt is applied. The amplitude of current is
    Solution
    Impedance of inductor=XL=ωH=1000×3mH=3ΩX_L=\omega H=1000\times 3mH=3\Omega
    Thus the impedance of the circuit=X=XL2+R2=5ΩX=\sqrt{X_{L}^2+R^2}=5\Omega
    Thus amplitude of current Imax=EmaxX=4V5Ω=0.8AI_{max}=\dfrac{E_{max}}{X}=\dfrac{4V}{5\Omega}=0.8A
  • Question 3
    1 / -0
    The impedance of a circuit, when a resistance RR and an inductor of inductance LL are connected in series in an AC circuit of frequency ff, is
    Solution
    The impedance due to the inductor is 2πfL2\pi fL and voltage drop across inductor leads the voltage drop across resistor by 90°90°
    So, the net impedance will be: Z=R2+(2πfL)2Z=\sqrt { { R }^{ 2 }+{ (2\pi fL) }^{ 2 } }
  • Question 4
    1 / -0
    The current through a coil of self-inductance L=2mHL=2mH is given by i=t2eti=t^2e^{-t} at time t. How long it will take to make the emf zero?
    Solution
    The current is given by the relation i=t2eti=t^2e^{-t}.
    Hence the emf would be equal to
    e=Ldidt=2×103×(2tett2et)e=-L\dfrac{di}{dt}=-2\times10^{-3}\times(2te^{-t}-t^2e^{-t})
    If the emf is zero, e=0e=0, hence
    2tet=t2ett=2s2te^{-t}=t^2e^{-t}\Rightarrow\,t=2\,s
  • Question 5
    1 / -0
    In an LCR circuit inductance is changed from LL to L2\dfrac{L}{2}. To keep the same resonance frequency, CC should be changed to
    Solution
    Resonance frequency, (f)=12πLC(f)=\dfrac{1}{2\pi \sqrt{LC}}
    For ff to be constant the product LCLC must be constant.
    So, if we half the value of inductance then the value of capacitance must be doubleddoubled.
    CC should be changed to 2C2C.
  • Question 6
    1 / -0
    The resonant frequency of an L-C circuit is
    Solution
    Resonance frequency ff of an L-C circuit can be written as 

    Resonance frequency, f=12πLCf =\dfrac{1}{2 \pi \sqrt{LC}} where LL= inductance and CC is capacitance 
  • Question 7
    1 / -0
    Resonance frequency of a circuit is ff. If the capacitance is made 44 times the initial value, then the resonance frequency will become
    Solution
    As  we know,
    Resonance frequency of a circuit
    f=12πLC f=\dfrac { 1 }{ 2\pi \sqrt { LC }  }
    ie, f1C f\propto \dfrac { 1 }{ \sqrt { C }  }
    f f=C 4C =12\Rightarrow \dfrac { { f }^{ \prime  } }{ f } =\dfrac { \sqrt { C }  }{ \sqrt { 4C }  } =\dfrac { 1 }{ 2 }
    f =f2\Rightarrow { f }^{ \prime  }=\dfrac { f }{ 2 }
  • Question 8
    1 / -0

    An oscillating circuit contains an inductor of inductance 106 { 10 }^{ -6 } H and two capacitor each of capacitance 5×1065\times { 10 }^{ -6 } farad connected in parallel. Then the resonance frequency of the circuit is

    Solution
    ω=1LC =1106×2×5×106 \omega =\dfrac { 1 }{ \sqrt { LC }  } =\dfrac { 1 }{ \sqrt { { 10 }^{ -6 }\times 2\times 5\times { 10 }^{ -6 } }  }

    =10113×105=\sqrt { { 10 }^{ 11 } } \approx 3\times { 10 }^{ 5 }\\

    f=ω 2π =3×1052π  f=\dfrac { \omega  }{ 2\pi  } =\dfrac { 3\times { 10 }^{ 5 } }{ 2\pi  }

    =1.5×105π =\dfrac { { 1.5\times 10 }^{ 5 } }{ \pi  }
  • Question 9
    1 / -0
    An alternating emf given by equation E=300sin[(100π)t]E = 300 sin [(100\pi )t] voltvolt is applied to a resistance 100 ohms100\ ohmsThe rms current through the circuit is (in amperesamperes):
    Solution
    Alternating emf equation            E=300E = 300 sin(100πt) sin(100\pi t)
    \therefore   Peak emf       Eo=300E_o = 300 voltsvolts
    The rms emf       Erms =Eo2 =3002E_{rms}  = \dfrac{E_o}{\sqrt{2}}  = \dfrac{300}{\sqrt{2}} voltsvolts
    Value of resistance        R=100 ΩR = 100\ \Omega

    \therefore Value of rms current          Irms =ErmsR =3002×100 =32I_{rms}  = \dfrac{E_{rms}}{R}  = \dfrac{300}{\sqrt{2} \times 100}   =\dfrac{3}{\sqrt{2}} AA
  • Question 10
    1 / -0

    When 100 volt DC100\ volt\ DC is applied across a solenoid, a current of 1.0 amp1.0\ amp flows in it. When 100 volt AC100\ volt\ AC is applied across the same coil, the current drops to 0.5 amp.0.5\ amp. If the frequency of the AC source is 50 Hz50\ Hz the impedance and inductance of the solenoid are:

    Solution
    We know that
    i=VR,forDCR=1001=100Ωi=VLω ,whenACisappliedLω=Vi=1000.5=200ΩL=2002π×50=0.055Hi=\frac { V }{ R } ,\quad for\quad DC\Rightarrow R=\frac { 100 }{ 1 } =100\Omega \\ i=\frac { V }{ L\omega  } ,\quad when\quad AC\quad is\quad applied\Rightarrow L\omega =\frac { V }{ i } =\frac { 100 }{ 0.5 } =200\Omega \\ L=\frac { 200 }{ 2\pi \times 50 } =0.055H
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