(C) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = 2C(2V) = 4CV$.When connected with opposite polarities,the net charge on the system is

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(C) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = 2C(2V) = 4CV$.When connected with opposite polarities,the net charge on the system is $Q_{net} = Q_2 - Q_1 = 4CV - CV = 3CV$.The equivalent capacitance of the parallel combination is $C_{eq} = C + 2C = 3C$.The common potential $V'$ is given by $V' = Q_{net} / C_{eq} = 3CV / 3C = V$.The final energy $U_f$ of the configuration is $U_f = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} (3C) V^2 = \frac{3}{2} CV^2$.

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

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$A$ parallel plate capacitor of capacitance $C$ is charged to a potential difference $V$ by connecting it to a battery. Another capacitor of capacitance $2C$ is charged to a potential difference $2V$ by connecting it to another battery. Now,the charging batteries are removed and the capacitors are connected in parallel such that the positive plate of one is connected to the negative plate of the other and the negative plate of the first is connected to the positive plate of the second. Find the final energy of this configuration.

A
Zero
B
$25 CV^2/6$
C
$3 CV^2/2$
D
$9 CV^2/2$

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected by a thin wire,then the loss of energy will be given by

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$A$ capacitor of $4\,\mu F$ charged to $50\,V$ is connected to another capacitor of $2\,\mu F$ charged to $100\,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $10^{-2}\,J$ is:

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Two conducting spheres of radii $5\, cm$ and $10\, cm$ are given a charge of $15\,\mu C$ each. After the two spheres are joined by a conducting wire, the charge on the smaller sphere is.......$\mu C$.

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Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

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In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

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The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected by a thin wire,then the loss of energy will be given by

$A$ capacitor of $4\,\mu F$ charged to $50\,V$ is connected to another capacitor of $2\,\mu F$ charged to $100\,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $10^{-2}\,J$ is:

Two conducting spheres of radii $5\, cm$ and $10\, cm$ are given a charge of $15\,\mu C$ each. After the two spheres are joined by a conducting wire, the charge on the smaller sphere is.......$\mu C$.

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

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