Electrostatics Question 605
Question: Two capacitors $ C _{1}=2\mu F $ and $ C _{2}=6\mu F $ in series, are connected in parallel to a third capacitor $ C _{3}=4\mu F $ . This arrangement is then connected to a battery of e.m.f. = 2V, . How much energy is lost by the battery in charging the capacitors [MP PET 2001]
Options:
A) $ 22\times {{10}^{-6}}J $
B) $ 11\times {{10}^{-6}}J $
C) $ ( \frac{32}{3} )\times {{10}^{-6}}J $
D) $ ( \frac{16}{3} )\times {{10}^{-6}}J $
Show Answer
Answer:
Correct Answer: B
Solution:
$ C _{eq}=\frac{C _{1}C _{2}}{C _{1}+C _{2}}+C _{3}=\frac{2\times 6}{2+6}+4=5.5\mu F $
Energy supplied $ (E)=QV=CV^{2}=22\times {{10}^{-6}}J $
P.E. stored$ (U)=\frac{1}{2}C _{eq}V^{2}=\frac{1}{2}\times 5.5\times {{(2)}^{2}}=11\times {{10}^{-6}}J $
therefore Energy lost$ =E-U=11\times {{10}^{-6}}J $
BETA
