A $100 \mathrm{~V}$ battery is connected across the series combination of the two capacitors of $4 \mu \mathrm{F}$ and $8 \mu \mathrm{F}$. The energy stored in the series combination is

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Aicharya · Accepted Answer

Potential difference across terminals of battery, $V=100 \mathrm{~V}$

Equivalent capacitance of series combination is given as $\begin{aligned} & \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{4}+\frac{1}{8}=\frac{2+1}{8} \\ & \Rightarrow \quad \frac{1}{C}=\frac{3}{8} \Rightarrow C=\frac{8}{3} \mu \mathrm{F} \end{aligned}$ Same charge will be flow in series combination which is given as $\begin{aligned} q & =C V=\frac{8}{3} \times 10^{-6} \times 100 \\ & =\frac{8}{3} \times 10^{-4} \mathrm{C} \end{aligned}$ $\therefore$ Energy stored, $E=\frac{1}{2} \frac{q^2}{C}$ $\begin{aligned} & =\frac{1}{2} \times \frac{\left(\frac{8}{3} \times 10^{-4}\right)^2}{\frac{8}{3} \times 10^{-6}}=\frac{4}{3} \times 10^{-2} \mathrm{~J} \\ \Rightarrow E & =1.33 \times 10^{-2} \mathrm{~J} \end{aligned}$

A \(100 \mathrm{~V}\) battery is connected across the series combination of the two capacitors of \(4 \mu…

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