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Alternating Current Test - 33

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Alternating Current Test - 33
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  • Question 1
    1 / -0
    A condenser and a 30 Ω\Omega resistance are connected in series. When they are connected to 120V A.C. source then the current flowing in the circuit is 1A The p.d. across the ends of the condenser will be nearly
    Solution
    i=VZi = \dfrac{V}{Z}

    Z=Vi=1201=120ΩZ = \dfrac{V}{i} = \dfrac{120}{1} = 120\Omega

    Z=R2+XC2Z = \sqrt{R^2+X_C^2}

    (120)2=(30)2+XC2(120)^2 = (30)^2+X_C^2

    XC2=13500X_C^2 = 13500

    XC=116ΩX_C = 116\Omega

    VC=iXCV_C = i X_C

    =1(116)= 1(116)

    =116V= 116V
  • Question 2
    1 / -0
    In the circuit shown in the figure, (neglecting source resistance) the voltmeter and ammeter readings will respectively be

    Solution
    Voltmeter reading is zero since voltage in both Capacitor and inductor is same in magnitude but opposite in sign because current in series is same and both have same reactance.

    So, i=VRi = \dfrac{V}{R}

    =24030= \dfrac{240}{30}

    =8A= 8A
  • Question 3
    1 / -0
    In a series LCR circuit the voltage across resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be
    Solution

  • Question 4
    1 / -0
    In the following circuit, the potential of source is E0=200E_0=200 volts, R=200ΩR=200\Omega, L=0.1L=0.1 henry, C=10.6C=10.6 farad and frequency is variable, then the current at f=0 and f=\infty is :

    Solution
    Z=R2+(ωL1ωC)2Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}

    I=VZ=VωCR2+(ω 2LC1)2 I=\dfrac { V }{ Z } =\dfrac { V\omega C }{ \sqrt { { R }^{ 2 }+{ ({ \omega  }^{ 2 }LC-1) }^{ 2 } }  }

    when w0w\rightarrow0   then 

    I=VωCR2+12 I=\dfrac { V\omega C }{ \sqrt { { R }^{ 2 }+{ 1 }^{ 2 } }  }  I0\Rightarrow I\rightarrow 0

     or  when   ww\rightarrow\infty ,  

    R2+(ω 2LC1)2or{ R }^{ 2 }+{ ({ \omega  }^{ 2 }LC-1) }^{ 2 }\rightarrow \infty \quad \quad or\\

    1R2+(ω 2LC1)2 0\dfrac { 1 }{ \sqrt { { R }^{ 2 }+{ ({ \omega  }^{ 2 }LC-1) }^{ 2 } }  } \rightarrow 0\\

    I0\Rightarrow I\rightarrow 0

    therefore, current is zero in both case.
  • Question 5
    1 / -0
    An LCR series circuit with 100 Ω\Omega resistance is connected to an ac source of 200V and of frequency of 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60060^{0} . When only the inductance is removed, the current leads the voltage by 60060^{0} the current through the circuit is:
    Solution

    Since current lead and lag are same So, circuit is in resonance, so, circuit is purely resistive circuit.

    So, i=VRi = \dfrac{V}{R}

    =200100= \dfrac{200}{100}

    =2A= 2A

  • Question 6
    1 / -0
    In the given figure as shown, the reading of the ammeter in ampere is

    Solution
    I=ER2+(XLXC)2 I = \dfrac{E}{\sqrt{R^2 + (X_L - X_C)^2}}

    =110552+(22)2= \dfrac{ 110}{\sqrt{ 55^2 + ( 2-2)^2}}

    =11055=2A= \dfrac{110}{55} = 2 A
  • Question 7
    1 / -0
    An alternating voltage V=2002sin100tV =200\sqrt{2} \sin{100t}, Where VV is in volt and tt is in seconds, is connected to a series combination of 1μF\mu\text{F} capacitor and 10 kΩ10\ \text{k}\Omega resistor through an AC ammeter. The reading of the ammeter will be_____
    Solution
    i=VrmsR2+(1ωCωL) 2 i=\dfrac { { V }_{ rms } }{ \sqrt { { R }^{ 2 }+{ \left( \dfrac { 1 }{ \omega C } -\omega L \right)  }^{ 2 } }  } \\

    i=200212  100002+(1100×106100×0) 2 =2002×100002  \Rightarrow i=\dfrac { 200\sqrt { 2 } \dfrac { 1 }{ \sqrt { 2 }  }  }{ \sqrt { { 10000 }^{ 2 }+{ \left( \dfrac { 1 }{ 100\times { 10 }^{ -6 } } -100\times 0 \right)  }^{ 2 } }  } =\dfrac { 200 }{ \sqrt { { 2\times 10000 }^{ 2 } }  } \\

    $$ =10\sqrt { 2 }\  mA$$
  • Question 8
    1 / -0
    The capacitor of an oscillatory circuit of negligible resistance is enclosed in a evacuated container. The frequency of the circuit is 150 kHZ and when the container is filled with a gas, the frequency changes by 100 HZ. The dielectric constant of the gas.
    Solution
    We know frequency is given by:

    n1=12πLC1   n_1 = \dfrac { 1 }{ 2\pi \sqrt { LC_1 }  }  

    And 
    n2=12πLKC1n_2=\dfrac{1}{2\pi \sqrt{LKC_1}}

    Thus we get the ratio of frequencies as

    n1n2=150000149900=k\dfrac{n_1}{n_2}=\dfrac{150000}{149900}=\sqrt{k}

    Solving the above equation we get, k1.0012k \approx 1.0012
  • Question 9
    1 / -0
    In the AC circuit shown, XL=7Ω,R=4Ω and Xc=4Ω.X_{L}=7\Omega ,R=4\Omega\ and\ X_{c}=4\Omega . The reading of the ideal voltmeter V2 is 82V_{2}\ is\ 8\sqrt{2}. The reading of the ideal ammeter will be:

    Solution
    From the given figure,

    Given,

    XL=7ΩX_L=7\Omega

    XC=4ΩX_C=4\Omega

    R=4ΩR=4\Omega

    The reading of ideal voltmeter is given by

    V2=82VV_2=8\sqrt{2}V

    From Ohm's law,

    V2=I(R+jXC)V_2=I( {R+jX_C}) (Capacitor and resistor are in series)

    I=V2R+jXCI=\dfrac{V_2}{R+jX_C}

    I=V2(R)2+(XC)2I=\dfrac{V_2}{\sqrt{(R)^2+(X_C)^2}}

    I=82(4)2+(4)2I=\dfrac{8\sqrt{2}}{\sqrt{(4)^2+(4)^2}}

    I=8242I=\dfrac{8\sqrt{2}}{4\sqrt{2}}

    I=2AI=2A

    The correct option is B.
  • Question 10
    1 / -0
    A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be:-
    Solution
    f = 12πLC\dfrac{1} {2 \pi \sqrt{LC}} & f^{'} = 12π2L(4C )\dfrac{1} {2 \pi \sqrt{2L\left (4C  \right )}}
    Therefore f^{'} = (122 )12πLC\left (\dfrac{1} {2 \sqrt{2}}  \right ) \dfrac{1} {2 \pi \sqrt{LC}} = f22\dfrac{f} {2 \sqrt{2}}
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