Question

Two capacitors of $2\mu F$ and $4\mu F$ are connected in series and 9 volt applied across their across their free ends. What is the ratio of the energies stored in them.

• A. 4 : 1
• B. 1 : 4
• C. 1 : 2
• D. 2 : 1

Answer 1

The correct option is D 2 : 1

$\begin{array}{c}{C}_{eq}=\frac{4\times 2}{2\times 4}=\frac{8}{6}=\frac{4}{3}\\ Q=C_{eq}V\\ =\frac{4}{3}\times 9=12\mu C\\ \therefore E_1=\frac{Q^2}{2C_1}=\frac{12\times 12}{2\times 2}=36\mu J\\ E_2=\frac{Q^2}{2C_2}=\frac{12\times 12}{2\times 4}=18\mu J\\ \therefore \frac{E_1}{E_2}=2\\ Here,\\ optionDiscorrectanswer.\end{array}$