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Electrostatic Potential and Capacitance Test - 54

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Electrostatic Potential and Capacitance Test - 54
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  • Question 1
    1 / -0
    Six capacitors each of capacitance of 2μF2\mu F are connected as shown in the figure. The effective capacitance between AA and BB is:

    Solution
    All six capacitors are connected in parallel across the terminal AA and BB.
    Therefore, the equivalent capacitance is given by,
    Ceq=(2+2+2+2+2+2)μFC_{eq}=(2+2+2+2+2+2)\, \mu F
    Ceq=12μF.C_{eq}=12\, \mu F.
  • Question 2
    1 / -0
    The capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then-

    Solution

    Given,

    Both capacitors are in series

    Capacitance of A & B, CA=2μF  &  CB=3μF{{C}_{A}}=2\mu F\,\,\And \,\,{{C}_{B}}=3\mu F

    Total Potential is VT=(VA+VB)=10V{{V}_{T}}=({{V}_{A}}+{{V}_{B}})=10V

     

    Let,

    Potential difference on capacitor A & B, VA,VB{{V}_{A}},\,{{V}_{B}}  

    In series, Capacitor have equal charge

    QA=QB{{Q}_{A}}={{Q}_{B}}

    CAVA=CBVB{{C}_{A}}{{V}_{A}}={{C}_{B}}{{V}_{B}}

    VAVB=CBCA\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}

    VA+VBVB=CB+CACA\dfrac{{{V}_{A}}+{{V}_{B}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}+{{C}_{A}}}{{{C}_{A}}}

    VB=CA(VA+VB)CB+CA=2×105=4V{{V}_{B}}=\dfrac{{{C}_{A}}({{V}_{A}}+{{V}_{B}})}{{{C}_{B}}+{{C}_{A}}}=\dfrac{2\times 10}{5}=4V

    VB=4V{{V}_{B}}=4V

    Similarly, VAVB=CBCA   VA=VB×CBCA  \dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}\Rightarrow \,\,\,{{V}_{A}}=\dfrac{{{V}_{B}}\times {{C}_{B}}}{{{C}_{A}}}\,\,

    VA=4×32=6V\Rightarrow \,{{V}_{A}}=\dfrac{4\times 3}{2}=6V

    Potential difference on capacitor A & B, VA=6V,    VB=4V{{V}_{A}}=6V,\,\,\,\,{{V}_{B}}=4V   

  • Question 3
    1 / -0
    If potential of AA is 10V10 V, then potential of BB is:

    Solution
    Step 1: Redraw the circuit in simpler way\textbf{Step 1: Redraw the circuit in simpler way} [Ref. Fig. 1]\textbf{[Ref. Fig. 1]}

    Step 2: Equivalent Capacitance:\textbf{Step 2: Equivalent Capacitance:}[Ref. Fig.1 and 2]\textbf{[Ref. Fig.1 and 2]}
     C1C_1 and C2C_2 are in parallel       

      C=C1+C2C = C_1 + C_2=1μF+1μF=2μF=1\mu F + 1 \mu F = 2\mu F

      CC and C3C_3 are in series
    Equivalent Capacitance Ceq=C×C3C+C3=2×12+1=23μFC_{eq} = \dfrac{C \times C_3}{ C + C_3 }= \dfrac{2\times 1}{2 + 1} = \dfrac{2}{3}\mu F

    Step 3: Charge on capacitor C3\textbf{Step 3: Charge on capacitor }C_3
     As CC and C3C_3 are in series, the charge will be the same.

                    Q=Ceq×V=23×10=203μCQ = C_{eq} \times V = \dfrac{2}{3} \times 10 = \dfrac{20}{3} \mu C

    Step 4: Voltage across capacitor C3\textbf{Step 4: Voltage across capacitor} \ C_3
               
                    V3=QC3=20μC3×1μF=203VV_3 = \dfrac{Q}{C_3} = \dfrac{20\mu C}{3\times 1\mu F} = \dfrac{20}{3}V

    Step 5: Potential at B\textbf{Step 5: Potential at B}   
     V3=VBVAV_3 = V_B - V_A
    203=  VB10\Rightarrow \dfrac{20}{3} =   V_B - 10                 (VA=10V)(V_A=10V)
    VB=503V\Rightarrow V_B = \dfrac{50}{3}V

    Hence, Option (B) is correct.

  • Question 4
    1 / -0
    Two capacitors of 1μF1\mu F and 2μF2\mu F are connected in series and this combination is changed upto a potential difference of 120120 volt. What will be the potential difference across 1μF1 \mu F capacitor:
    Solution
    Given, c1=1μf,c2=2μf,PD=120vc_1=1\mu f,c_2=2\mu f,PD=120v

    Ceq=c1c2c1+c2=2×12+1=23μfC_{eq}=\dfrac{c_1c_2}{c_1+c_2}=\dfrac{2\times1}{2+1}=\dfrac{2}{3}\mu f

    We know,  Q=cvQ=cv Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.

    Now, QnetQ_{net} in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD

    Qnet=12×23=80Q_{net}=12\times\dfrac{2}{3}=80

    Q=cv=1μfv=80v=80vQ=cv=1\mu fv=80\Rightarrow v=80v
  • Question 5
    1 / -0
    Three capacitors 3μF,9F3\mu F,9 F and 18μF18\mu F are connected first in series and then in parallel. The ratio of the equivalent capacitance in two cases is :
    Solution
    Given, Three capacitors 3μf,9μf,18μf3\mu f,9\mu f,18\mu f

    So, the capacitor connected in the series,

    1Ceq=1c1+1C21C3=13+19+118\dfrac{1}{C_{eq}}=\dfrac{1}{c_1}+\dfrac{1}{C_2}\dfrac{1}{C_3}\Rightarrow =\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{18}

    =6+2+118=918=12Ceq=2=\dfrac{6+2+1}{18}=\dfrac{9}{18}=\dfrac{1}{2}\Rightarrow C_{eq}=2

    And the capacitor is connected in parallel Ceq=3+9+18=30C_{eq}=3+9+18=30

    So, the ratio of the series and the parallel is:

    CeqCeq=302=15:1\dfrac{C_{eq}}{C_{eq}}=\dfrac{30}{2}=15:1
  • Question 6
    1 / -0
    The effective capacitance between the point PP and QQ of the arrangement shown in the figure is:

    Solution

    The given circuit can be made like above figure.

    This forms a Wheatstone bridge so the capacitor of $5\,\mu F$gets removed. (fig.1)

    Capacitors on the upper branch is in series so their equivalent is (fig.2)

    1Ceq=12+12 \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{2}+\dfrac{1}{2}

    Ceq=1μF {{C}_{eq}}=1\,\mu F

    Similarly, for the lower branch equivalent is Ceq=1μF{{C}_{eq}}=1\,\mu F   (fig.2)

    Capacitors  1μF1\,\mu F and 1μF1\,\mu F are in parallel. So equivalent is 2μF2\,\mu F (fig.3)

    Now, 2μF2\,\mu F and 2μF2\,\mu F are in series.

    So,

    1Ceq=12+12 \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{2}+\dfrac{1}{2}

    Ceq=1  μF {{C}_{eq}}=1\,\,\mu F

  • Question 7
    1 / -0
    Calculate effective capacitance:

    Solution
    In parallel(v is same) 
    then, in series Q is same 
    so: 3V1±6V23{V_1} \pm 6{V_2}
    and V1+V2{V_1} + {V_2}= 2424
    = 7V17{V_1}= 1616
    = V2=8{V_2}=8

  • Question 8
    1 / -0
    In the arrangement of the capacitors shown in the figure, each C1C_1 capacitor has capacitance of 3μF3\mu F and each C2C_2 capacitor has capacitance of 2μF2\mu F then,
    (i) Equivalent capacitance of the network between the points a and b is 

    Solution

  • Question 9
    1 / -0
    The effective capacitance between the points AA and BB will be

    Solution

  • Question 10
    1 / -0
    An uncharged metal object MM is insultated from its surroundings.A positively charges metal sphere SS is then brought near to MM. Which diagram illustrate the resultant distribution of charge on SS and MM
    Solution
    Since all negative charges attracted towards SS so the move towards left and leaving positive charges on right.

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