91Ó°ÊÓ

• Potential difference of the battery, V = 10 V
• Capacitors are first connected in series, then in parallel.
• Energy stored in the capacitors from least to greatest,${\mathrm{U}}_1=75\mathrm{μ}J$${\mathrm{U}}_2=100\mathrm{μ}J$role="math" localid="1661341160485" ${\mathrm{U}}_3=300\mathrm{μ}J$${\mathrm{U}}_4=400\mathrm{μ}J$

When the capacitors ${\mathbf{C}}_{\mathbf{1}}$and ${\mathbf{C}}_2$are connected in series, the equivalent capacitance is given as,

$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}$

When the capacitors ${\mathbf{C}}_{\mathbf{1}}$and ${\mathbf{C}}_2$ are connected in parallel, the equivalent capacitance is given as,

${\mathit{C}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{C}}_{\mathbf{1}}\mathbf{+}{\mathit{C}}_{\mathbf{2}}$

Formulae:

The charge stored between the capacitor plates, q=CV …(i)

The energy stored within the capacitor plates, $U=\frac{1}{2}C{V}^2$ …(¾±¾±)

When the capacitors are connected in series, the equivalent capacitance is smaller than either one individually. In parallel, their equivalent capacitance is more than that of an individual capacitor. The energy stored is directly proportional to the capacitance. Hence, for parallel combinations, the energy stored is larger than the series combination.Thus, the middle two values quoted in the problem must correspond to the individual capacitors.

Using the required stored energy, we can get the smallest capacitance of the capacitor plates by using the smaller value of the middle two values and equation (ii) as follows:

$$\begin{gathered} C_1=\frac{2U_2}{V^2} \\ =\frac{2×100\mathrm{μ}J}{{\left(10V\right)}^2} \\ =\frac{200\mathrm{μ}J}{100V^2} \\ =2\mathrm{μ}F \end{gathered}$$

Hence, the value of the capacitance is $2\mathrm{μ¹ó}$.

Similarly, greater value of the capacitance is found using the greater value of the capacitor of the two middle values in equation (ii) as follows:

$$\begin{gathered} C_2=\frac{2U_3}{V^2} \\ =\frac{2×300\mathrm{μ}J}{{\left(10V\right)}^2} \\ =\frac{600\mathrm{μ}J}{100V^2} \\ =6\mathrm{μ}F \end{gathered}$$

Hence, the value of the capacitance is $6\mathrm{μ}F$.