Electrostatics Question 540
Question: Three capacitors of capacity $ C _{1},\ C _{2}\ C _{3} $ are connected in series. Their total capacity will be [MP Board 1977; MP PET/PMT 1988; CPMT 1996]
Options:
A) $ C _{1}+C _{2}+C _{3} $
B) $ 1/(C _{1}+C _{2}+C _{3}) $
C) $ {{(C _{1}^{-1}+C _{2}^{-1}+C _{3}^{-1})}^{-1}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1}{C _{R}}=\frac{1}{C _{1}}+\frac{1}{C _{2}}+\frac{1}{C _{3}} $
therefore $ C _{R}={{(C _{1}^{-1}+C _{2}^{-1}+C _{3}^{-1})}^{-1}} $
BETA
