91Ó°ÊÓ

a) Capacitance,$C_1=2.0\mathrm{μ¹ó}$

b) Capacitance,$C_2=4.0\mathrm{μ¹ó}$

c) The capacitors are connected in parallel.

d) Potential difference, V =300 V

We use the formula of energy stored in the capacitor to find the energy of each capacitor. The total energy stored will be the sum of individual energy stored as the capacitances are in parallel connection.

Formulae:

The energy stored in the capacitor,$U=\frac{1}{2}C{V}^2$ ...(i)

The equivalent capacitance of a parallel connection of capacitors,

$C_{eqivalant}=\underset{}{∑}{C}_i$ ...(ii)

Since the two capacitors are connected in parallel, the potential difference across both the capacitors is the same. So, the total energy stored is given using equation (ii) in equation (i) as follows:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{total}}=\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){\mathrm{V}}^2 \\ =\frac{1}{2}\left(2×{10}^{-6}\mathrm{F}+4×{10}^{-6}\mathrm{F}\right){\left(300\mathrm{V}\right)}^2 \\ =\frac{1}{2}\left(2+4\right)×{10}^{-6}\mathrm{F}×{\left(300\mathrm{V}\right)}^2 \\ =270000×{10}^{-6}\mathrm{J} \\ =0.27\mathrm{J} \end{gathered}$$

Hence, the value of the energy stored is 0.27 J.