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A capacitor C_{1}=1.0 µF is charged up to a voltage V =60 V by connecting it to battery B through switch (1). Now C_{1} is disconnected from battery and connected to a circuit consisting of two uncharged capacitors C_{2}=3.0 µF and C_{3}=6.0 µ F through switch (2), as shown in the figure. The sum of final charges on C_{2} and C_{3} is :

  • Option 1)

    40 µ C

  • Option 2)

    36 µ C

  • Option 3)

    20 µ C

  • Option 4)

    54 µ C

 

Answers (2)

best_answer

As we learnt

Common Potential -

V=\frac{Total\: charge}{Total\: capacity}=\frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}

V=\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}

-

 

 Initial Charge =q=60\mu C

Let final potential across c2 &c3 is v;

\left ( \frac{c_{2}c_{3}}{c_{2}+c_{3}} \right ).v+c_{1}v=60\mu C

\therefore v =\frac{ 60\mu C}{c_{1}+\frac{c_{2}c_{3}}{c_{2}+c_{3}}}=\frac{60}{1+2}=20V

Total charge on c_{2} \: and\: c_{3} =\left ( \frac{c_{2}c_{3}}{c_{2}+c_{3}} \right ).v

                                           =40\mu C

 

 


Option 1)

40 µ C

Option 2)

36 µ C

Option 3)

20 µ C

Option 4)

54 µ C

Posted by

Avinash

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