Question

How would you find the total energy stored in the

capacitorsif.....

A 2.0 microF capacitor and a 4.0 microF capacitor are connected
inPARALLEL across a 300V potential difference.

Answer 1

The energy of the capacitors connected in parallel is 0.27 J

Given:

C =
`2.0\micro F = 2.0* 10^(- 6) F`

C' =
`4.0\micro F = 4.0* 10^(- 6) F`

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

`C_(eq) = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0* 10^(- 6) F`

The energy of the capacitor, E is given by;

`E = (1)/(2)C_(eq)V^(2)`

`E = (1)/(2)* 6.0* 10^(- 6)* 300^(2) = 0.27 J`