Two capacitors of capacitance 2 mu F and 4 | vedclass

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(A) When capacitors are connected in series,the charge $Q$ stored on each capacitor is the same.The energy stored in a capacitor is given by the formu

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$2 : 1$

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(A) When capacitors are connected in series,the charge $Q$ stored on each capacitor is the same.The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$.Since $Q$ is constant for both capacitors in a series connection,the energy $U$ is inversely proportional to the capacitance $C$ $(U \propto \frac{1}{C})$.Therefore,the ratio of the energy stored in the two capacitors is:$\frac{U_1}{U_2} = \frac{C_2}{C_1}$Given $C_1 = 2 \ \mu F$ and $C_2 = 4 \ \mu F$:$\frac{U_1}{U_2} = \frac{4 \ \mu F}{2 \ \mu F} = \frac{2}{1}$Thus,the ratio of the energy stored is $2 : 1$.

Two capacitors of capacitance $2 \ \mu F$ and $4 \ \mu F$ are connected in series. If a potential difference of $10 \ V$ is applied across this combination,what will be the ratio of the energy stored in these capacitors?

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