Question

How would you find the total energy stored in the

Answer 1

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = $2.0\micro F = 2.0\times 10^{- 6} F$

C' = $4.0\micro F = 4.0\times 10^{- 6} F$

Potential difference, V = 300 V

Solution:

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

$C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F$

The energy of the capacitor, E is given by;

$E = \frac{1}{2}C_{eq}V^{2}$

$E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J$