somebody please help me!

 

An electrical technician requires a capacitance of 2microfarad in a circuit across a potential difference of 1KV. a large no of 1 microF capacitor are available to him each of which can withstand a potential difference of not more than 400V.

suggest a possible arrangement that requires minimum no of capacitors.

Ans:"18 . 6 parallelrows .3 in each row.

u can solve it as-

 

as the capacitors can't withstand a p.d of not more than 400v,this implies that no. of capacitors in series  should be >=1000/400 (p.d. gets distributed in series) =2.5=3(bcoz 2.5 capacitors r impossible to get). now let there r m branches s.t there r 3 capacitors in each branch. eq.capacitance in one branch=1/3 microfarad.(series connection)

now as there r m branches so equivalent capacitance should be=

 

1/3*m=2  ( bcoz m branches r in parallel)

 

m=6.

 

So ther r 6 rows & 3 capacitors in each row&that's the answer!!!!!

 

rate me plllllllzzzzzzzzzzzzzz...