How to find out the breakdown voltage?

https://useruploads.socratic.org/bktEI7aGSUqLpJXp7gLy_New%20Imagex.jpghttps://useruploads.socratic.org/bktEI7aGSUqLpJXp7gLy_New%20Imagex.jpg answer is 4 KV

1 Answer
Apr 23, 2018

4\ "kV"

Explanation:

In a series combination of capacitors, the voltage is divided in inverse ratio of the capacitance.

If V is the applied voltage,then

  • the voltage applied across 6\ mu"F" is 2/(2+6)V=1/4V
  • that across 2\ mu"F" is 6/(2+6)V=3/4V
  • that across 3\ mu"F" is 1/(3+1)V=1/4V
  • that across 1\ mu"F" is 3/(3+1)V=3/4V

Now,

  • For the 6\ mu"F" capacitor to break down, the applied voltage must be 4times 1\ "kV"=4\ "kV"
  • For the 2\ mu"F" capacitor to break down, the applied voltage must be 4/3times 4\ "kV"=16/3\ "kV"

  • For the 3\ mu"F" capacitor to break down, the applied voltage must be 4times 4\ "kV"=16\ "kV"

  • For the 1\ mu"F" capacitor to break down, the applied voltage must be 4/3times 5\ "kV"=20/3\ "kV"

The breakdown voltage for the circuit is the least of these values : 4\ "kV"