How to find out the breakdown voltage?
https://useruploads.socratic.org/bktEI7aGSUqLpJXp7gLy_New%20Imagex.jpg answer is 4 KV
https://useruploads.socratic.org/bktEI7aGSUqLpJXp7gLy_New%20Imagex.jpg answer is 4 KV
1 Answer
Apr 23, 2018
Explanation:
In a series combination of capacitors, the voltage is divided in inverse ratio of the capacitance.
If
- the voltage applied across
6\ mu"F" is2/(2+6)V=1/4V - that across
2\ mu"F" is6/(2+6)V=3/4V - that across
3\ mu"F" is1/(3+1)V=1/4V - that across
1\ mu"F" is3/(3+1)V=3/4V
Now,
- For the
6\ mu"F" capacitor to break down, the applied voltage must be4times 1\ "kV"=4\ "kV" -
For the
2\ mu"F" capacitor to break down, the applied voltage must be4/3times 4\ "kV"=16/3\ "kV" -
For the
3\ mu"F" capacitor to break down, the applied voltage must be4times 4\ "kV"=16\ "kV" -
For the
1\ mu"F" capacitor to break down, the applied voltage must be4/3times 5\ "kV"=20/3\ "kV"
The breakdown voltage for the circuit is the least of these values :