Question

A parallel plate capacitor of capacitance $C$ is charged to a potential $V$ and then disconnected from the battery. The capacitor is now connected to an identical capacitor, charged to a potential $2V$ such that the positive polarity plates are connected together. At steady state, the common potential of the capacitors will be equal to

$[$0.77 Mark$]$

• A. $1.5V$
• B. $2V$
• C. $3V$
• D. $\frac{V}{2}$

Answer 1

The correct option is A $1.5V$


$V_{common}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

$=\frac{CV+C\left(2V\right)}{C+C}$

$=\frac{3CV}{2C}=\frac{3V}{2}$

$=1.5V$