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QJEEM13Ph94ab: JEE Main 2013 Physics - Class 12 JEEM13Ph94ab Solution: Two capacitors and are charged to and respectively. It is found …

Two capacitors

{C_1}

and

{C_2}

are charged to

120

V

and

200

V

respectively. It is found that connecting them together the potential on each one can be made zero. Then

Step-by-Step Verified Solution

For potential to be made zero, after connection

120{C_1} = 200{C_2}

\left[ {\,\,} \right.

\left. {C = {q \over v}\,\,} \right]

\Rightarrow 3{C_1} = 5{C_2}