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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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Verified Answer
The correct answer is:
\(1.33 \times 10^{-2} \mathrm{~J}\)
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, \(E=\frac{1}{2} \frac{q^2}{C}\)
\(\begin{aligned}
& =\frac{1}{2} \times \frac{\left(\frac{8}{3} \times 10^{-4}\right)^2}{\frac{8}{3} \times 10^{-6}}=\frac{4}{3} \times 10^{-2} \mathrm{~J} \\
\Rightarrow E & =1.33 \times 10^{-2} \mathrm{~J}
\end{aligned}\)
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, \(E=\frac{1}{2} \frac{q^2}{C}\)
\(\begin{aligned}
& =\frac{1}{2} \times \frac{\left(\frac{8}{3} \times 10^{-4}\right)^2}{\frac{8}{3} \times 10^{-6}}=\frac{4}{3} \times 10^{-2} \mathrm{~J} \\
\Rightarrow E & =1.33 \times 10^{-2} \mathrm{~J}
\end{aligned}\)
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