(D) The common potential $V$ of two capacitors connected in parallel is given by the formula:$V = \frac{Q_1 + Q_2}{C_1 + C_2}$Since $Q = CV$,we have:$

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(D) The common potential $V$ of two capacitors connected in parallel is given by the formula:$V = \frac{Q_1 + Q_2}{C_1 + C_2}$Since $Q = CV$,we have:$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$Given:$C_1 = 2 \mu F, V_1 = 50 \ V$$C_2 = 3 \mu F, V_2 = 100 \ V$Substituting the values:$V = \frac{(2 \times 10^{-6} \times 50) + (3 \times 10^{-6} \times 100)}{2 \times 10^{-6} + 3 \times 10^{-6}}$$V = \frac{(100 + 300) \times 10^{-6}}{5 \times 10^{-6}}$$V = \frac{400}{5} = 80 \ V$

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

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$A$ $2 \mu F$ capacitor is connected to a $50 \ V$ supply,and a $3 \mu F$ capacitor is connected to a $100 \ V$ supply. After removing the batteries,if the two plates of the same type of charge are connected to each other,the new potential difference is . . . . . . $V$.

GUJCET 2020 All GUJCET

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$75$
B
$333$
C
$200$
D
$80$

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Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

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$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

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$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

EasyTS EAMCET 2001

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When two isolated conductors $A$ and $B$ are connected by a conducting wire,positive charge will flow from:

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$A$ sphere of radius $4 \ cm$ has a charge of $80 \ \mu C$ and a sphere of radius $6 \ cm$ has a charge of $40 \ \mu C$. If they are connected by a conducting wire,how much charge will flow from the $4 \ cm$ sphere to the $6 \ cm$ sphere?

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$A$ $2 \mu F$ capacitor is connected to a $50 \ V$ supply,and a $3 \mu F$ capacitor is connected to a $100 \ V$ supply. After removing the batteries,if the two plates of the same type of charge are connected to each other,the new potential difference is . . . . . . $V$.

Similar Questions

There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

When two isolated conductors $A$ and $B$ are connected by a conducting wire,positive charge will flow from:

$A$ sphere of radius $4 \ cm$ has a charge of $80 \ \mu C$ and a sphere of radius $6 \ cm$ has a charge of $40 \ \mu C$. If they are connected by a conducting wire,how much charge will flow from the $4 \ cm$ sphere to the $6 \ cm$ sphere?

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