Question

How to find out the breakdown voltage?

answer is 4 KV

Answer 1

`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"`