Question

Two air-filled, parallel-plate capacitors are to be connected to a 10 V battery, first individually, then in series, and then in parallel. Inthose arrangements, the energy stored in the capacitors turns out tobe, listed least to greatest: $\mathbf{75}\mathbf{}\mathbf{\mu J}\mathbf{}$,$\mathbf{100}\mathbf{}\mathbf{\mu J}$ ,$\mathbf{300}\mathbf{}\mathbf{\mu J}$ , and$\mathbf{400}\mathbf{}\mathbf{\mu J}$ . Of the two capacitors, what is the (a) smaller and (b) greater capacitance?

Short answer

1. The smaller capacitance is$2\mathrm{\mu F}$
2. The greater capacitance is $6\mathrm{\mu F}$

Step-by-step solution

The given data

• 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\mu J$${\mathrm{U}}_2=100\mu J$role="math" localid="1661341160485" ${\mathrm{U}}_3=300\mu J$${\mathrm{U}}_4=400\mu J$

Understanding the concept of the capacitance in series and parallel

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, $U=\frac{1}{2}C{V}^2$ …(ii)

(a) Calculation of the smallest capacitance

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:

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

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

(b) Calculation of the greatest capacitance

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:

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

Hence, the value of the capacitance is $6\mu F$.