Electrostatics Question 109
Question: A body of capacity $ 4\mu F $ is charged to $ 80V $ and another body of capacity $ 6\mu F $ is charged to 30V. When they are connected the energy lost by $ 4\mu F $ capacitor is [EAMCET 2001]
Options:
A) 7.8 mJ
B) 4.6 mJ
C) 3.2 mJ
D) 2.5 mJ
Show Answer
Answer:
Correct Answer: A
Solution:
Initial energy of body of capacitance 4 mF is $ U _{i}=\frac{1}{2}\times (4\times {{10}^{-6}}){{(80)}^{2}}=0.0128J $
Final potential on this body after connection is $ V=\frac{4\times 80+6\times 30}{4+6}=50V. $
So final energy on it $ U _{f}=\frac{1}{2}\times 4\times {{10}^{-6}}{{(50)}^{2}}=0.005J $
Energy lost by this body $ =U _{i}U _{f}=\text{ 7}.\text{8}mJ $
BETA
