Question

A capacitor 1 with breakdown voltage 120 volt is connected with two other capacitors 2 and 3 (2 and 3 are in parallel connected with 1 as series) .the breakdown voltage of 2 is 120 V but that of 3 is 150 V. The maximum potential across the plates of 2 so that all capacitors remain safe (all capacitors have same capacity)​

Answer 1

Given:

Three capacitors each of capacity C have been connected in a circuit as shown. Breakdown voltage of capacitor 1 is 120 V , breakdown voltage of capacitor 2 is 120 V and that of capacitor 3 is 150 V.

To find:

Max voltage that can be applied across the plates of capacitor 2 in order to keep all the three capacitors safe.

Calculation:

Let us assume that breakdown voltage of approx. 120 V has been obtained on the plates of capacitor 1.

So charge on capacitor 1 will be

$\therefore \: q = CV = (C \times 120)$

Now capacitors 2 and 3 (in parallel) are in series with capacitor 1.

Equivalent capacitance (considering capacitor 2 and 3) is (C + C) = 2C.

So potential difference across 2C will be:

$V = \dfrac{q}{C} = \dfrac{120C}{2C} = 60 \: volt$

So maximum potential difference that can be obtained across the plates of capacitor 2 is 60 Volts , in order to keep all the three capacitors safe.