MCQ
$2\,\mu F$ 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.......$\mu F$
- A$2$
- B$4$
- ✓$18$
- D$10$
