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Alternating Current Test - 74

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Alternating Current Test - 74
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  • Question 1
    1 / -0
    When 100 V100\space V dc is applied across a coil, a current of 1 A1\space A flows through it and when 100 V100\space V ac of 50 Hz50\space Hz is applied to the same coil, only 0.5 A0.5\space A flows. The inductance of coil is
    Solution
    R=VDCIDC=1001=100Ω R = \frac{V_{DC}}{I_{DC} } = \frac{100}{1} = 100 \Omega
    VAC2IAC2=R2+XL210020.52=1002+XL2 XL=30000=1003 Ω\frac{V_{AC}^2}{I_{AC}^2} = R^2 + X_L^2 \Rightarrow \frac{100^2}{0.5^2}= 100^2+ X_L^2  \Rightarrow X_L = \sqrt{30000}= 100\sqrt{3} \Omega
    XL=L2πf1003=L2π50L=3πHX_L= L 2 \pi f \Rightarrow 100\sqrt{3} = L 2 \pi 50 \Rightarrow L =\frac{\sqrt{3}}{\pi} H
  • Question 2
    1 / -0
    A charged capacitor discharges through a resistance R with time constant τ \tau . The two are now placed in series across an AC source of angular frequency ω=1τ\displaystyle \omega = \frac {1} {\tau}  . The impedance of the circuit will be:
    Solution
    In RC circuit, impedance Z=R2+1ω2C2Z=\sqrt{R^2+\dfrac{1}{\omega^2C^2}} and time constant τ=CR\tau=CR
    Since ω=1τ\omega=\dfrac{1}{\tau} so Z=R2+τ2C2=R2+C2R2C2=2RZ=\sqrt{R^2+\dfrac{\tau^2}{C^2}}=\sqrt{R^2+\dfrac{C^2R^2}{C^2}}=\sqrt{2} R
  • Question 3
    1 / -0
    The resonant frequency of an L-C circuit is
    Solution

  • Question 4
    1 / -0
    A series LCR circuit is connected across a source of alternating emf of changing frequency and resonates at frequency fo\text{f}_o. Keeping capacitance constant, if the inductance (L) is increased by 3\sqrt{3} times and resistance(R) is increased  by 1.4 times, the resonant frequency now is:
    Solution
    Resonant frequency of a series LCR circuit is given by the formula 

    f=12πLC  f=\dfrac { 1 }{ 2\pi \sqrt { LC }  } 

    Hence, f0=12πLC  { f }_{ 0 }=\dfrac { 1 }{ 2\pi \sqrt { LC }  } 

    In second case, f=12πLC =12π3LC =1314 f0{ f }'=\dfrac { 1 }{ 2\pi \sqrt { L'C }  } =\dfrac { 1 }{ 2\pi \sqrt { \sqrt { 3 } LC }  } =\dfrac { 1 }{ { 3 }^{ \frac { 1 }{ 4 }  } } { f }_{ 0 }
  • Question 5
    1 / -0
    In a LR circuit of 3  mH3\;mH inductane and 4  Ω4\;\Omega resistance, emf E=4  cos  (1000  t)E=4\;cos\;(1000\;t) volt is applied. The amplitude of current is:
    Solution
    Given :        L=3mH =3×103 HL = 3mH  = 3 \times 10^{-3}  H                     R=4ΩR =4 \Omega
    Emf       E= 4cos1000tE =  4 cos1000t
    \therefore   Eo=4E_o = 4 volt        and      ω=1000\omega = 1000
    Inductive reactance      XL=ωL =1000×3×103 =3 ΩX_L = \omega L  = 1000 \times 3 \times 10^{-3}  = 3  \Omega
    Impedance        Z=R2+XL2 =42+32 =5 ΩZ = \sqrt{R^2 + X_L^2}  = \sqrt{4^2 + 3^2}  = 5  \Omega
    Current amplitude         I= EoZ =45 =0.8 AI=  \dfrac{E_o}{Z}  = \dfrac{4}{5}  = 0.8  A
  • Question 6
    1 / -0
    In a purely inductive circuit, the current is
  • Question 7
    1 / -0
    If the power factor in a circuit is unit, then the impedance of the circuit is
    Solution

  • Question 8
    1 / -0
    Inductive reactance of a coil is expressed in
    Solution

  • Question 9
    1 / -0
    An AC generator producing 10V10V (rms) at 200rad/s200rad/s is connected in series with a 50Ω50\Omega resistor, a 400mH400mH inductor and a 200μF200\mu F capacitor. The rms voltage across the inductor is 
    Solution
    Given :  $$E=10V, \  \ \ w=200, \ \ \ \ R=50\Omega , \ \ \ \ L=400mH, \ \ \ C=200\mu F\quad $$
    So, capacitive reactance  XC=1wC=1(200)(200×106)=25ΩX_C = \dfrac{1}{wC} = \dfrac{1}{(200)(200\times 10^{-6})} = 25\Omega
    Inductive reactance  XL=wL =(200)(400×103)=80ΩX_L = wL  =(200)(400\times 10^{-3}) =80\Omega
    Impedance  Z=R2+(XLXC) 2=502+(8025)2\quad Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right)  }^{ 2 } } =\sqrt { { 50 }^{ 2 }+{ (80-25) }^{ 2 } }
    Z=74.3Ω\quad Z=74.3\Omega
    I=EZ=1074.3=0.13459AI=\cfrac { E }{ Z } =\cfrac { 10 }{ 74.3 } =0.13459A
    EL=IXL=0.1345×80=10.76V{ E }_{ L }=I{ X }_{ L }=0.1345\times 80=10.76V
  • Question 10
    1 / -0
    An inductor and a resistor are connected to an ac supply of 50V50 V and 50Hz50 Hz. If the voltage across the resistor is 40V40 V the voltage across the inductor will be:
    Solution
    Given: Supply voltage, VS=50 VV_S=50\ V
                Voltage across resistor, VR=40 VV_R=40\ V

    For an LR circuit:
    Applied voltage VS=VR2+VL2V_S=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ L }^{ 2 } }   
     VL=502402=30 V\Rightarrow  { V }_{ L }=\sqrt { { 50 }^{ 2 }-{ 40 }^{ 2 } } =30\ V
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