We know that, in series combination of capacitors, net potential difference is equal to sum of the individual potential difference of each capacitor ‌ i.e., ‌‌‌V‌=V1+V2+V3.......‌‌[V1=V2=V3=200V] ⇒‌‌600‌=n×200 ⇒‌‌n‌=3 So, there should be 3 capacitors in series to obtain the required potential difference. The equivalent capacitance of 3 capacitors in series is given as ‌‌
1
Ceq
=‌
1
6
+‌
1
6
+‌
1
6
=‌
3
6
=‌
1
2
‌⇒‌‌Ceq=2µF Now, we require 18µF capacitance for this, we need 9 number of such capacitors in parallel. ∴ Total number of capacitors (condensers) =9×3=27