(D) Initial energy of the system,$U_i = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{1}{2} C(V_1^2 + V_2^2)$.When the capacitors are connected in p

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(D) Initial energy of the system,$U_i = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{1}{2} C(V_1^2 + V_2^2)$.When the capacitors are connected in parallel,the common potential is $V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}$.Final energy of the system,$U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{1}{4} C(V_1 + V_2)^2$.Decrease in energy,$\Delta U = U_i - U_f = \frac{1}{2} C(V_1^2 + V_2^2) - \frac{1}{4} C(V_1 + V_2)^2$.$\Delta U = \frac{1}{4} C [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)]$.$\Delta U = \frac{1}{4} C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4} C(V_1 - V_2)^2$.

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

Easy

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

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A
$\frac{1}{4} C(V_1^2 + V_2^2)$
B
$\frac{1}{4} C(V_1^2 - V_2^2)$
C
$\frac{1}{4} C(V_1 + V_2)^2$
D
$\frac{1}{4} C(V_1 - V_2)^2$

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Electrostatic Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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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 to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

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$A$ $600\,pF$ capacitor is charged by a $200\,V$ supply. It is then disconnected from the supply and connected to another uncharged $600\,pF$ capacitor. The electrostatic energy lost in the process is $.........\,\mu J$.

DifficultJEE MAIN 2023

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$A$ capacitor of capacitance $6\ \mu F$ is charged to $100\, V$. It is then connected to another uncharged capacitor of capacitance $14\ \mu F$. Find the ratio of charges on the $6\ \mu F$ and $14\ \mu F$ capacitors and the potential across the $6\ \mu F$ capacitor.

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$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

DifficultJEE MAIN 2022

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Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

Similar Questions

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

$A$ $600\,pF$ capacitor is charged by a $200\,V$ supply. It is then disconnected from the supply and connected to another uncharged $600\,pF$ capacitor. The electrostatic energy lost in the process is $.........\,\mu J$.

$A$ capacitor of capacitance $6\ \mu F$ is charged to $100\, V$. It is then connected to another uncharged capacitor of capacitance $14\ \mu F$. Find the ratio of charges on the $6\ \mu F$ and $14\ \mu F$ capacitors and the potential across the $6\ \mu F$ capacitor.

$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

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