$\therefore C_{234}=C_2+C_3+C_4$
$\therefore C_{234}=6\mu F$
Capacitors $C_1$,$C_{234}$ $\text{and}$ $C_5$ are in series$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_{234}}+\frac{1}{C_5}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}$
$=\frac{7}{6}\mu F$
$C_{equivalent}=\frac{6}{7}\mu F$
Charge drawn from the source$Q=C_{eq}V,$
$=\frac{6}{7}\times7\mu C=6\mu C$
Energy stored $U=\frac{Q^2}{2C}$$=\frac{6\times6\times10^{-12}\times7}{2\times6\times10^{-6}}\text{J}$
$=21\mu \text{J}$.
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$^{2}_{1}\text{H} +^{2}_{1}\text{H}\rightarrow^{3}_{2}\text{He} + \text{n} + 3.27\text{MeV}$
although number of nucleons is conserved, yet energy is released. How? Explain.
Given
$\text{m}(^{226}_{88}\text{Ra})=226.02540\text{ u}.$ $\text{m}(^{222}_{86}\text{Rn})=222.01750\text{ u}.$
$\text{m}(^{222}_{86}\text{Rn})=220.01137\text{ u}.$ $\text{m}(^{216}_{84}\text{Po})=216.00189\text{ u}.$