Answer:
[tex]0.11[/tex] J
Explanation:
[tex]C_{1}[/tex] = capacitance of first capacitor = 2.8 µF
[tex]C_{2}[/tex] = capacitance of second capacitor = 5.7 µF
The capacitors are connected in parallel and their parallel combination is given as
[tex]C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}[/tex]
[tex]C = \frac{(2.8) (5.7)}{2.8 + 5.7}[/tex]
[tex]C = \frac{(2.8) (5.7)}{2.8 + 5.7}[/tex]
[tex]C = 1.88[/tex] μF
[tex]C = 1.88\times 10^{-6}[/tex] F
[tex]V[/tex] = Potential difference across the parallel combination = 340 Volts
Total energy stored by capacitors is given as
[tex]U = (0.5) C V^{2}[/tex]
inserting the values
[tex]U = (0.5) (1.88\times 10^{-6}) (340)^{2}[/tex]
[tex]U = 0.11[/tex] J
