\[2\mu F\]capacitance has potential difference across its two terminals \[200\ volts\]. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes\[20\ volts\]. Then the capacity of another capacitance will be [CPMT 1991; DPMT 2001]

• A.                    \[2\mu F\]
• B.                                      \[4\mu F\]
• C.                    \[18\mu F\]                  
• D.            \[10\mu F\]