Question

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:
A. Zero
B. \[\left[ {\dfrac{9}{2}} \right]C{V^2}\]
C. \[\left[ {\dfrac{{25}}{6}} \right]C{V^2}\]
D. \[\left[ {\dfrac{3}{2}} \right]C{V^2}\]

Answer 1

Hint: When the capacitors are rearranged after removing the battery then the final net charge should be equal to the net initial charge on individual capacitors because charge follows law of conservation, the charge can neither be created nor be destroyed but it can be transferred from one body to other, i.e. one capacitor to other.


Formula used:
\[Q = CV\], here Q is the charge stored on the plate of capacitor of capacitance C when a potential difference V is applied across it.
\[E = \dfrac{{C{V^2}}}{2}\], here E is the energy stored in a capacitor of capacitance C when a potential difference V is applied across it.


Complete answer:


Figure: Initial circuit diagram

Figure: Final circuit diagram
The initial charges stored in the capacitors are,
\[{Q_1} = CV\]
\[{Q_2} = \left( {2C} \right)\left( {2V} \right)\]
\[{Q_2} = 4CV\]
Let the final charges on the capacitors are \[Q_{1}^{'}\]and \[Q_{2}^{'}\] and as the capacitors are connected in parallel so the potential difference across both the capacitors will be same. Let the common potential difference is \[V'\]
Then the charges on the capacitors will be,
\[Q_{1}^{'} = CV'\]
\[{Q_{2}^{'}} = 2CV'\]
Using the conservation of charge,
\[\sum {{Q_{final}}} = \sum {{Q_{initial}}} \]
As the positive plate of the capacitor 2 is connected with the negative plate of the capacitor 1, so there will be negative charge on capacitor 1 will get added to the positive charge on the capacitor 2.
\[Q_{1}^{'} +Q_{2}^{'} = - Q_{1} + Q_{2}\]
\[CV' + 2CV' = - CV + 4CV\]
\[3V' = 3V\]
\[V' = V\]
So, the energy stored in the first capacitor is,
\[E_{1}=\dfrac{C\left ( V_{1}^{'} \right )^2}{2}\]
\[{E_1} = \dfrac{{C{{\left( V \right)}^2}}}{2}\]
\[{E_1} = \dfrac{{C{V^2}}}{2}\]
And the energy stored in the second capacitor is,
\[E_{2}=\dfrac{2C\left ( V_{1}^{'} \right )^2}{2}\]
\[{E_2} = \dfrac{{2C{{\left( V \right)}^2}}}{2}\]
\[{E_2} = \dfrac{{2C{V^2}}}{2}\]
Hence, total energy stored in this configuration is,
\[E = {E_1} + {E_2}\]
\[E = \dfrac{{C{V^2}}}{2} + \dfrac{{2C{V^2}}}{2}\]
\[E = \dfrac{{\left( {1 + 2} \right)C{V^2}}}{2}\]
\[E = \left( {\dfrac{3}{2}} \right)C{V^2}\]
Hence, the energy stored in the final configuration is \[\left( {\dfrac{3}{2}} \right)C{V^2}\]
Therefore, the correct option is D.


Note: We should be careful about the order of the plates connected in the final configuration. If the positive plate of one capacitor is connected to the positive plate of another capacitor then the polarity of the charges would not be changed, i.e. \[Q_{1}^{'} + Q_{2}^{'} = {Q_{1}} + {Q_{2}}\]and the rest of the steps will be same.