Two capacitors of capacitance 2 mu F and 4 mu | vedclass

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(A) In a series connection,the charge $Q$ stored on each capacitor is the same.The energy $U$ stored in a capacitor is given by the formula $U = \frac

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$2: 1$

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(A) In a series connection,the charge $Q$ stored on each capacitor is the same.The energy $U$ stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$.For the two capacitors with capacitances $C_1 = 2 \mu F$ and $C_2 = 4 \mu F$,the energies stored are $U_1 = \frac{Q^2}{2C_1}$ and $U_2 = \frac{Q^2}{2C_2}$ respectively.The ratio of the energies is $\frac{U_1}{U_2} = \frac{Q^2 / 2C_1}{Q^2 / 2C_2} = \frac{C_2}{C_1}$.Substituting the values,we get $\frac{U_1}{U_2} = \frac{4 \mu F}{2 \mu F} = \frac{2}{1}$.Thus,the ratio of energies stored is $2: 1$.

Two capacitors of capacitance $2 \mu F$ and $4 \mu F$ respectively are connected in series. The combination is connected across a potential difference of $10 \ V$. The ratio of energies stored by the capacitors will be . . . . . . .

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