Electrostatics Question 631
Question: Three capacitors 2, 3 and 6 mF are joined in series with each other. What is the minimum effective capacitance [Orissa PMT 2004]
Options:
A) $ \frac{1}{2}\mu F $
B) 1 mF
C) 2 mF
D) 3 mF
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1}{C _{eq}}=\frac{1}{C _{1}}+\frac{1}{C _{2}}+\frac{1}{C _{3}} $
$ =\frac{1}{2}+\frac{1}{3}+\frac{1}{6} $
$ =\frac{3+2+1}{6}=\frac{6}{6}=1\mu F $
