Electrostatics Question 599
Three capacitors each of capacitance 6 μF are connected in series. The total capacitance of the combination will be [MH CET 2000; CPMT 2001]
Options:
A) $ 9\times {{10}^{-12}}F $
B) $ 6\times {{10}^{-12}}F $
C) $ 3\times {{10}^{-12}}F $
D) $ 2\times {{10}^{-12}}F $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {C}’=C$
$n=\frac{6\times {{10}^{-12}}}{3}=2\times {{10}^{-12}}F $
BETA
