In figure, ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{}\mathit{\mu }\mathit{F}$, ${\mathit{C}}_2\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{F}$and ${\mathit{C}}_3\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{F}$If no capacitor can withstand a potential difference of more than 100 Vwithout failure, (a)What is the magnitude of the maximum potential difference that can exist between points A and B?(b)What is the maximum energy that can be stored in the three capacitor arrangement?

1. The capacitance of the three capacitors connection in series connection are: $C_1=10\mu F$, $C_2=20\mu F$, and $C_3=25.0\mu F$.
2. Maximum potential difference that any two capacitors can withstand is V = 100 V.

We know that the capacitance of the capacitor is defined as the total charge per potential difference and from this formula we can find the charge stored in each capacitor by the relation. Also, to find the energy stored in the capacitor, we have the relation, we need to find the voltage drop across each capacitor, and then adding those we will get the maximum voltage drop.

Formulae:

The capacitance between two capacitor plates, $C=\frac{Q}{V}$ …(i)

The amount of energy stored between the plates, $U=\frac{1}{2}C·V^2$ …(ii)

The maximum voltage across each capacitor is 100 V.

The maximum voltage occurs across smallest value of capacitor.

Thus, the charge on first capacitor can be found using equation (i) as follows:

$\begin{array}{ll}{Q}_1=C_1.V& \\ =(10\times {10}^8& F)\times (100\hspace{0.17em}V)\\ =1\times {10}^{-3}C& \end{array}$.

The charges remain same in series.

Charge on second and third capacitor will be same as of charge on capacitor 1.

Now, the values of the potential difference across the three capacitors are given using equation (i) as follows:

$$\begin{gathered} \begin{array}{l}{V}_1=\frac{Q_1}{C_1}\end{array} \\ =\frac{1\times {10}^{-3}C}{10\times {10}^{-6}F} \\ =100V \\ \begin{array}{l}{V}_2=\frac{Q_2}{C_2}\end{array} \\ =\frac{1\times {10}^{-3}C}{10\times {10}^{-6}F} \\ =50V \\ \begin{array}{l}{V}_3=\frac{Q_3}{C_3}\end{array} \\ =\frac{1\times {10}^{-3}C}{10\times {10}^{-6}F} \\ =40V \end{gathered}$$

Thus, the maximum potential difference can be given as:

$$\begin{gathered} V_{maximum}=\left(100+50+40\right)V \\ =190V \end{gathered}$$

Hence, the value of the maximum potential difference is 190 V.

Now, the energy stored by each capacitor in a three-arrangement capacitor system is given using equation (ii) as follows:

$$\begin{gathered} \begin{array}{lll}{\mathrm{U}}_1=\frac{1}{2}\times (10\times {10}^{-6}& \mathrm{F})\times (100& {\mathrm{V}}^2)\\ =0.05\mathrm{J}& & \end{array} \\ \begin{array}{lll}{\mathrm{U}}_2=\frac{1}{2}\times (20\times {10}^{-6}& \mathrm{F})\times (50& {\mathrm{V}}^2)\\ =0.025\mathrm{J}& & \end{array} \end{gathered}$$

$\begin{array}{lll}{\mathrm{U}}_3=\frac{1}{2}\times (25.0\times {10}^{-6}& \mathrm{F})\times (40& {\mathrm{V}}^2)\\ =0.02\mathrm{J}& & \end{array}$

Thus, the value of the maximum stored energy in the three-arrangement system is given by,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{maximum}}=\left(0.05+0.025+0.02\right)\mathrm{J} \\ =0.095\mathrm{J} \end{gathered}$$

Hence, the value of the maximum energy is 0.095 J.