Question

A$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$capacitor and a $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$ capacitor are connected in parallel across a 300 Vpotential difference. Calculate the total energy stored in the capacitors.

Short answer

The total energy stored in the capacitors is 0.27 J.

Step-by-step solution

The given data

a) Capacitance,$C_1=2.0\mathrm{\mu F}$

b) Capacitance,$C_2=4.0\mathrm{\mu F}$

c) The capacitors are connected in parallel.

d) Potential difference, V =300 V

Understanding the concept of the stored energy

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,

$C_{eqivalant}=\sum _{}{C}_i$ ...(ii)

Step 3: Calculation of the total stored energy

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:

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

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