A) \[\frac{1}{4}C(V_{1}^{2}-V_{2}^{2})\]
B) \[\frac{1}{4}C(V_{1}^{2}+V_{2}^{2})\]
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 :
Initial energy of the system \[{{U}_{i}}=\frac{1}{2}C{{V}_{1}}^{2}+\frac{1}{2}C{{V}_{2}}^{2}\] When the capacitors are joined, common potential \[V=\frac{C{{V}_{1}}+C{{V}_{2}}}{2C}=\frac{{{V}_{1}}+{{V}_{2}}}{2}\] Final energy of the system \[{{U}_{f}}=\frac{1}{2}(2C){{V}^{2}}=\frac{1}{2}2C\,{{\left( \frac{{{V}_{1}}+{{V}_{2}}}{2} \right)}^{2}}=\frac{1}{4}C{{({{V}_{1}}+{{V}_{2}})}^{2}}\] Decrease in energy = \[{{U}_{i}}-{{U}_{f}}=\frac{1}{4}C{{({{V}_{1}}-{{V}_{2}})}^{2}}\]You need to login to perform this action.
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