Electrostatics Question 546
Question: The capacities of two conductors are $ C _{1} $ and $ C _{2} $ and their respective potentials are $ V _{1} $ and$ V _{2} $ . If they are connected by a thin wire, then the loss of energy will be given by [MP PMT 1986]
Options:
A) $ \frac{C _{1}C _{2}(V _{1}+V _{2})}{2(C _{1}+C _{2})} $
B) $ \frac{C _{1}C _{2}(V _{1}-V _{2})}{2(C _{1}+C _{2})} $
C) $ \frac{C _{1}C _{2}{{(V _{1}-V _{2})}^{2}}}{2(C _{1}+C _{2})} $
D) $ \frac{(C _{1}+C _{2})(V _{1}-V _{2})}{C _{1}C _{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Initial energy
$ U _{i}=\frac{1}{2}C _{1}V _{1}^{2}+\frac{1}{2}C _{2}V _{2}^{2} $ ,
Final energy $ U _{f}=\frac{1}{2}(C _{1}+C _{2})V^{2} $ (where $ V=\frac{C _{1}V _{1}+C _{2}V _{2}}{C _{1}C _{2}}) $
Hence energy loss $ \Delta U=U _{i}-U _{f}=\frac{C _{1}C _{2}}{2(C _{1}+C _{2})}{{(V _{1}-V _{2})}^{2}} $
BETA
