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Two capacitors C1 = 2\mu F and C2 = 6\mu F in series, are connected in parallel to a third capacitorC3 = 4\mu F. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is lost by the battery in charging the capacitors ?     

  • Option 1)

    22\times 10^{-6}J

  • Option 2)

    11\times 10^{-6}J

  • Option 3)

    (32/3)\times 10^{-6}J

  • Option 4)

    (16/3)\times 10^{-6}J

 

Answers (1)

best_answer

As we have learned

Energy Stored -

U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}

-

 

 

Equivalent capacitance       C_{eq}= \frac{C_1C_2}{C_1+C_2} + C_3= \frac{2*6}{8}+4= 5.5 \mu F

\                                       U= 1/2 C_{eq}V^2= 1/2*5.5*2^2= 11*10^{-6}J

 


Option 1)

22\times 10^{-6}J

Option 2)

11\times 10^{-6}J

Option 3)

(32/3)\times 10^{-6}J

Option 4)

(16/3)\times 10^{-6}J

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Aadil

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