(B) Initial charges on the capacitors are:$q_1 = C \times V = CV$$q_2 = 2C \times 2V = 4CV$When connected with opposite polarities,the net charge on t

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(B) Initial charges on the capacitors are:$q_1 = C \times V = CV$$q_2 = 2C \times 2V = 4CV$When connected with opposite polarities,the net charge on the system is:$Q_{net} = |q_2 - q_1| = |4CV - CV| = 3CV$The equivalent capacitance of the capacitors in parallel is:$C_{eq} = C + 2C = 3C$The common potential $V'$ across the combination is:$V' = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$The final energy stored in the configuration is:$U = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} (3C) (V)^2 = \frac{3}{2} CV^2$

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

Medium

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A
zero
B
$\frac{3}{2} CV^2$
C
$\frac{25}{6} CV^2$
D
$\frac{9}{2} CV^2$

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Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

MediumAIPMT 2002

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$A$ $0.2 \, F$ capacitor is charged to $600 \, V$ by a battery. After removing the battery,it is connected to another uncharged parallel plate capacitor of $1 \, F$. The potential decreases to $V$ equal to:

Medium

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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:

Difficult

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Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

Medium

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Two identical capacitors are connected in parallel across a potential difference $V$. After they are fully charged,the positive plate of the first capacitor is connected to the negative plate of the second,and the negative plate of the first is connected to the positive plate of the other. The loss of energy will be

Easy

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$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

$A$ $0.2 \, F$ capacitor is charged to $600 \, V$ by a battery. After removing the battery,it is connected to another uncharged parallel plate capacitor of $1 \, F$. The potential decreases to $V$ equal to:

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:

Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

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