(A) When two capacitors are connected in parallel,they reach a common potential $V_c$.The total charge $Q_{total} = Q + 0 = Q$.The total capacitance $

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(A) When two capacitors are connected in parallel,they reach a common potential $V_c$.The total charge $Q_{total} = Q + 0 = Q$.The total capacitance $C_{total} = C + 2C = 3C$.The common potential is given by $V_c = \frac{Q_{total}}{C_{total}} = \frac{Q}{3C}$.The final charge on the first capacitor is $Q_1 = C \cdot V_c = C \cdot \frac{Q}{3C} = \frac{Q}{3}$.The final charge on the second capacitor is $Q_2 = 2C \cdot V_c = 2C \cdot \frac{Q}{3C} = \frac{2Q}{3}$.Thus,the final charges are $\frac{Q}{3}$ and $\frac{2Q}{3}$.

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

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$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

KCET 2019 All KCET

A
$ \frac{Q}{3}, \frac{2Q}{3} $
B
$ \frac{Q}{2}, \frac{Q}{2} $
C
$ \frac{Q}{5}, \frac{4Q}{5} $
D
$ \frac{Q}{4}, \frac{3Q}{4} $

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

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$A$ $10\,\mu F$ capacitor is charged to a potential difference of $50\;V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20\;V$. The capacitance of the second capacitor is....$\mu F$

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Consider the situation shown in the figure. The switch $S$ is kept open for a long time and then closed. Find the charge that flows through the battery when $S$ is closed.

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$A$ capacitor of capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

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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 $.........\,\mu J$.

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$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

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$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

$A$ $10\,\mu F$ capacitor is charged to a potential difference of $50\;V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20\;V$. The capacitance of the second capacitor is....$\mu F$

Consider the situation shown in the figure. The switch $S$ is kept open for a long time and then closed. Find the charge that flows through the battery when $S$ is closed.

$A$ capacitor of capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

$A$ $600\,pF$ 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 $.........\,\mu J$.

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