How to find out the breakdown voltage?

https://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"#