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QJEEM20Ph8b5f: JEE Main 2020 Physics - Class 12 JEEM20Ph8b5f Solution: A capacitor C is fully charged with voltage V 0 . After…

A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C2{C \over 2} . The energy loss in the process after the charge is distributed between the two capacitors is :

Multiple Choice Options

  • A 12CV02{1 \over 2}CV_0^2
  • B 14CV02{1 \over 4}CV_0^2
  • C 13CV02{1 \over 3}CV_0^2
  • D 16CV02{1 \over 6}CV_0^2

Step-by-Step Verified Solution

Heat loss = 12(C1C2C1+C2)V02{1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2

= 12(C×C2C+C2)V02{1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2

= 16CV02{1 \over 6}CV_0^2

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