(D) Initial energy of the first capacitor,$U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05 J$. When connected

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(D) Initial energy of the first capacitor,$U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05 J$. When connected to an uncharged capacitor,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{10 \times 100 + 30 \times 0}{10 + 30} = 25 V$. The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (40 \times 10^{-6}) \times (25)^2 = 0.0125 J$. The energy lost is $\Delta U = U_i - U_f = 0.05 - 0.0125 = 0.0375 J = 3.75 \times 10^{-2} J$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

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$A$ $10 \mu F$ capacitor is charged by a $100 V$ battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu F$. During this process,the electrostatic energy lost by the first capacitor is

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A
$5 \times 10^{-2} J$
B
$1.25 \times 10^{-2} J$
C
$2.75 \times 10^{-2} J$
D
$3.75 \times 10^{-2} J$

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be

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If $n$ drops, each of capacitance $C$, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

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$A$ capacitor of $20\,\mu F$ is charged to $500\,V$ and connected in parallel with another capacitor of $10\,\mu F$ which is charged to $200\,V$. The common potential is.......$V$.

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An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

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$A$ capacitor of $2\,\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ......$\%$

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$A$ $10 \mu F$ capacitor is charged by a $100 V$ battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu F$. During this process,the electrostatic energy lost by the first capacitor is

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Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be

If $n$ drops, each of capacitance $C$, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

$A$ capacitor of $20\,\mu F$ is charged to $500\,V$ and connected in parallel with another capacitor of $10\,\mu F$ which is charged to $200\,V$. The common potential is.......$V$.

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