In the circuit shown in the figure, every capacitor has a...

From the given circuit, capacitors $b$ and $c$ are connected in parallel, and their combination is in series with capacitors $a$ and $d$.

The equivalent capacitance of $b$ and $c$ is:
$C_{bc} = C_b + C_c = 10\mu\text{F} + 10\mu\text{F} = 20\mu\text{F}$

The total equivalent capacitance of the circuit $C_{eq}$ is:
$\dfrac{1}{C_{eq}} = \dfrac{1}{C_a} + \dfrac{1}{C_{bc}} + \dfrac{1}{C_d}$

$\dfrac{1}{C_{eq}} = \dfrac{1}{10} + \dfrac{1}{20} + \dfrac{1}{10} = \dfrac{5}{20}\mu\textF^{-1}$

$C_{eq} = 4\mu\text{F}$

The total charge supplied by the battery is:
$Q = C_{eq}V = 4\mu\text{F} \times 100\text{ V} = 400\mu\text{C}$

Since capacitors $a$ and $d$ are in series with the main circuit, the charge on each is:
$Q_a = Q_d = 400\mu\text{C}$

Since capacitors $b$ and $c$ are in parallel and have equal capacitance, the total charge splits equally between them:
$Q_b = Q_c = \dfrac{400\mu\text{C}}{2} = 200\mu\text{C}$

The energy stored in a capacitor is given by $U = \dfrac{Q^2}{2C}$.

Energy stored in capacitors $a$ and $d$:
$U_a = U_d = \dfrac{(400 \times 10^{-6})^2}{2 \times 10 \times 10^{-6}} = \dfrac{160000 \times 10^{-12}}{20 \times 10^{-6}} = 8 \times 10^{-3}\text{ J} = 8\text{ mJ}$

Energy stored in capacitors $b$ and $c$:
$U_b = U_c = \dfrac{(200 \times 10^{-6})^2}{2 \times 10 \times 10^{-6}} = \dfrac{40000 \times 10^{-12}}{20 \times 10^{-6}} = 2 \times 10^{-3}\text{ J} = 2\text{ mJ}$

Answer: $8 \text{ mJ}$ in (a) and (d), $2 \text{ mJ}$ in (b) and (c)