question_answer

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

A) \[\frac{1}{4}C\left( V_{1}^{2}-V_{2}^{2} \right)\]

B)        \[\frac{1}{4}C\left( V_{1}^{2}+V_{2}^{2} \right)\]

C) \[\frac{1}{4}C{{\left( V_{1}^{{}}-{{V}_{2}} \right)}^{2}}\]    

D) \[\frac{1}{4}C{{\left( V_{1}^{{}}+{{V}_{2}} \right)}^{2}}\]

Correct Answer: C

Solution :

Initially \[Initial\,\,energy=\frac{1}{2}C(V_{1}^{2}+V_{2}^{2})\]                     ... (i) \[{{q}_{1}}'+{{q}_{2}}'=C{{V}_{1}}+C{{V}_{2}}\]                          ... (ii) (charge conservation) Potential difiference across both the capacitors will be the same. \[\frac{{{q}_{1}}'}{C}=\frac{{{q}_{2}}'}{C}\,\,\Rightarrow \,\,{{q}_{1}}'\,+\,\,{{q}_{2}}'\] From equation (ii) \[{{q}_{1}}'=\frac{C({{V}_{1}}+{{V}_{2}})}{2}\] \[\therefore \,\,Final\,\,energy=\left[ \frac{{{C}^{2}}{{({{V}_{1}}+{{V}_{2}})}^{2}}}{4\times 4C} \right]\times 2\] \[\left( U\sin g\,\,U=\frac{{{Q}^{2}}}{2C} \right)\] \[Final\,\,energy=\frac{C{{({{V}_{1}}+{{V}_{2}})}^{2}}}{4}\] \[\therefore \, Change in energy=Initial energy\,- Final energy\]\[=\,\,\,\frac{1}{2}C(V_{1}^{2}+V_{2}^{2})-\frac{C}{4}(V_{1}^{2}+V_{2}^{2}+2{{V}_{1}}{{V}_{2}})\] \[=\,\,\,\frac{C}{4}\,[2V_{1}^{2}+V_{2}^{2})-V_{1}^{2}-V_{2}^{2}-2{{V}_{1}}{{V}_{2}}]\] \[=\,\,\frac{C}{4}{{({{V}_{1}}-{{V}_{2}})}^{2}}\]