Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]
C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]
The energy of the capacitor, E is given by;
[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]
[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]
