Electrostatics Question 612
Question: Three capacitors of $ 2\mu F,3\mu F $ and $ 6\mu F $ are joined in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the $ 6\mu F $ capacitor is [MP PMT 2002]
Options:
A) 4 volt
B) 6 volt
C) 8 volt
D) 10 volt
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{C _{eq}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\Rightarrow C _{eq}=1\mu F $
Total charge Q = Ceq.V = 1 × 24 = 24 mC
So p.d. across 6 mF capacitor = $ \frac{24}{6}=4volt $
BETA
