Question

Two capacitors connected in series C1: 100uF, 25V and C2: 300uF, 35V, both are supplied by E= 100V. Verify which capacitor will be damaged

Answer 1

Capacitor
`C_(1)` would be damaged because the voltage across it, 75V, is far too greater than the required value, 25V.

Step-by-step explanation:

A capacitor is an electronic device that can be used to store charges.

From the question,
`C_(1)` = 100uF, 25V and
`C_(2)` = 300uF, 35V, where voltage 'E' across the circuit is 100V.

Since the two capacitors are connected in series, the total capacitance,
`C_(T)`, is;

`C_(T)` =
`(1)/(C_(1) )` +
`(1)/(C_(2) )`

=
`(C_(1) * C_(2) )/(C_(1) + C_(2) )`

=
`(100uF * 300uF)/(100uF + 300uF)`

= 75uF

∴
`C_(T)` = 75uF

Thus, the voltage drop across each capacitor are;

`V_(c1)` =
`(C_(T) )/(C_(1) )` × E

=
`(75uF)/(100uF)` × 100

= 75V

`V_(C2)` =
`(C_(T) )/(C_(2) )` × 100

=
`(75uF)/(300uF)` × 100

= 25V

The capacitor that would be damage is
`C_(1)` because the voltage across it, 75V, is more than the required value, 25V.