060525

Question : An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 KV. A large number of 1µF capacitors are available to him each of which can withstand a potenrial difference of not more than 400 V. Suggest a possible arrangement that requires a minimum number of capacitors. 

Doubt by Gurdev

Solution : 

Here 
Total Capacitance Required = 2µF

Total Potential applied = 1 KV = 1000 V

Capacitance of each capacitor required = 1µF

Maximum Potential withstand by each 1µF capacitor = 400 V

We have to do this question by hit and trial method.

Let us connect 'n' capacitors in series